Subproblem 0: Each of the two Magellan telescopes has a diameter of $6.5 \mathrm{~m}$. In one configuration the effective focal length is $72 \mathrm{~m}$. Find the diameter of the image of a planet (in $\mathrm{cm}$ ) at this focus if the angular diameter of the planet at the time of the observation is $45^{\prime \prime}$.

Solution:
Start with:
\[
s=\alpha f \text {, }
\]
where $s$ is the diameter of the image, $f$ the focal length, and $\alpha$ the angular diameter of the planet. For the values given in the problem:
\[
s=\frac{45}{3600} \frac{\pi}{180} 7200=\boxed{1.6} \mathrm{~cm}
\]

Answer: 1.6


Subproblem 0: A white dwarf star has an effective temperature, $T_{e}=50,000$ degrees Kelvin, but its radius, $R_{\mathrm{WD}}$, is comparable to that of the Earth. Take $R_{\mathrm{WD}}=10^{4} \mathrm{~km}\left(10^{7} \mathrm{~m}\right.$ or $\left.10^{9} \mathrm{~cm}\right)$. Compute the luminosity (power output) of the white dwarf. Treat the white dwarf as a blackbody radiator. Give  your answer in units of ergs per second, to two significant figures.

Solution:
\[
\begin{aligned}
L=4 \pi R^{2} \sigma T_{e}^{4} &=4 \pi\left(10^{9}\right)^{2}\left(5.7 \times 10^{-5}\right)(50,000)^{4} \operatorname{ergs~s}^{-1} \\
L & \simeq \boxed{4.5e33} \mathrm{ergs} \mathrm{s}^{-1} \simeq 1 L_{\odot}
\end{aligned}
\]

Answer: 4.5e33


Preamble: A prism is constructed from glass and has sides that form a right triangle with the other two angles equal to $45^{\circ}$. The sides are $L, L$, and $H$, where $L$ is a leg and $H$ is the hypotenuse. A parallel light beam enters side $L$ normal to the surface, passes into the glass, and then strikes $H$ internally. The index of refraction of the glass is $n=1.5$.

Subproblem 0: Compute the critical angle for the light to be internally reflected at $H$.  Give your answer in degrees to 3 significant figures.

Solution:
From Snell's law we have:
\[
\begin{gathered}
n_{g} \sin \left(\theta_{g}\right)=n_{\text {air }} \sin \left(\theta_{\text {air }}\right) \\
\sin \left(\theta_{\text {crit }}\right)=\frac{1}{1.5} \sin \left(90^{\circ}\right) \Rightarrow \theta_{\text {crit }}=\boxed{41.8}^{\circ}
\end{gathered}
\]

Answer: 41.8


Subproblem 0: A particular star has an absolute magnitude $M=-7$. If this star is observed in a galaxy that is at a distance of $3 \mathrm{Mpc}$, what will its apparent magnitude be?

Solution:
\[
\text { Given: } M=-7 \text { and } d=3 \mathrm{Mpc}
\]
\[
\begin{aligned}
  & \text { Apparent Magnitude: } m=M+5 \log \left[\frac{d}{10 \mathrm{pc}}\right]=-7+5 \log \left[\frac{3 \times 10^{6}}{10}\right]=\boxed{20.39} \\
\end{aligned}
\]

Answer: 20.39


Subproblem 0: Find the gravitational acceleration due to the Sun at the location of the Earth's orbit (i.e., at a distance of $1 \mathrm{AU}$ ).  Give your answer in meters per second squared, and express it to one significant figure.

Solution:
\begin{equation}
F = ma = \frac{GM_{\odot}m}{r^2},
\end{equation}
so
\begin{equation}
a = \frac{GM_{\odot}{r^2}}
\end{equation}

Plugging in values for $G$, $M_{\odot}$, and $r$ gives $a = \boxed{0.006}$ meters per second squared.

Answer: 0.006


Preamble: A collimated light beam propagating in water is incident on the surface (air/water interface) at an angle $\theta_w$ with respect to the surface normal.

Subproblem 0: If the index of refraction of water is $n=1.3$, find an expression for the angle of the light once it emerges from the water into the air, $\theta_a$, in terms of $\theta_w$.

Solution: Using Snell's law, $1.3 \sin{\theta_w} = \sin{\theta_a}$. So $\theta_a = \boxed{\arcsin{1.3 \sin{\theta_w}}}$.

Final answer: The final answer is \arcsin{1.3 \sin{\theta_w}}. I hope it is correct.

Subproblem 1: What is the critical angle, i.e., the critical value of $\theta_w$ such that the light will not emerge from the water?  Leave your answer in terms of inverse trigonometric functions; i.e., do not evaluate the function.

Solution: The relation derived in the previous problem is $\theta_a = \arcsin{1.3 \sin{\theta_w}}$.  The critical angle thus occurs when $1.3 \sin{\theta_w}$ exceeds unity, because then there is no corresponding solution for $\theta_a$.  So the answer is $\boxed{np.arcsin(10/13)}$.

Answer: np.arcsin(10/13)


Subproblem 0: Find the theoretical limiting angular resolution (in arcsec) of a commercial 8-inch (diameter) optical telescope being used in the visible spectrum (at $\lambda=5000 \AA=500 \mathrm{~nm}=5 \times 10^{-5} \mathrm{~cm}=5 \times 10^{-7} \mathrm{~m}$).  Answer in arcseconds to two significant figures.

Solution:
\[
\theta=1.22 \frac{\lambda}{D}=1.22 \frac{5 \times 10^{-5} \mathrm{~cm}}{8 \times 2.54 \mathrm{~cm}}=2.46 \times 10^{-6} \text { radians }=\boxed{0.49} \operatorname{arcsecs}
\]

Answer: 0.49


Subproblem 0: A star has a measured parallax of $0.01^{\prime \prime}$, that is, $0.01$ arcseconds. How far away is it, in parsecs?

Solution: Almost by definition, it is $\boxed{100}$ parsecs away.

Answer: 100


Subproblem 0: An extrasolar planet has been observed which passes in front of (i.e., transits) its parent star. If the planet is dark (i.e., contributes essentially no light of its own) and has a surface area that is $2 \%$ of that of its parent star, find the decrease in magnitude of the system during transits.

Solution:
The flux goes from a maximum of $F_{0}$, when the planet is not blocking any light, to $0.98 F_{0}$ when the planet is in front of the stellar disk. So, the uneclipsed magnitude is:
\[
m_{0}=-2.5 \log \left(F_{0} / F_{\text {ref }}\right) \quad .
\]
When the planet blocks $2 \%$ of the stellar disk, the magnitude increases to:
\[
m=-2.5 \log \left(F / F_{\text {ref }}\right)=-2.5 \log \left(0.98 F_{0} / F_{\text {ref }}\right) \quad .
\]
Thus, the change in magnitude is:
\[
\Delta m=m-m_{0}=-2.5 \log (0.98) \simeq \boxed{0.022} \quad \text { magnitudes }
\]

Answer: 0.022


Subproblem 0: If the Bohr energy levels scale as $Z^{2}$, where $Z$ is the atomic number of the atom (i.e., the charge on the nucleus), estimate the wavelength of a photon that results from a transition from $n=3$ to $n=2$ in Fe, which has $Z=26$. Assume that the Fe atom is completely stripped of all its electrons except for one.  Give your answer in Angstroms, to two significant figures.

Solution:
\[
\begin{gathered}
h \nu=13.6 Z^{2}\left[\frac{1}{n_{f}^{2}}-\frac{1}{n_{i}^{2}}\right] \mathrm{eV} \\
h \nu=13.6 \times 26^{2}\left[\frac{1}{2^{2}}-\frac{1}{3^{2}}\right] \mathrm{eV} \\
h \nu=1280 \mathrm{eV}=1.28 \mathrm{keV} \Rightarrow \boxed{9.6} \AA
\end{gathered}
\]

Answer: 9.6


Subproblem 0: If the Sun's absolute magnitude is $+5$, find the luminosity of a star of magnitude $0$ in ergs/s. A useful constant: the luminosity of the sun is $3.83 \times 10^{33}$ ergs/s.

Solution: The relation between luminosity and absolute magnitude is: $m - n = 2.5 \log (f_n/f_m)$; note the numerator and denominator: brighter objects have numericallly smaller magnitudes.  If a star has magnitude $0$, then since the difference in magnitudes from the sun is $5$, it must have $100$ times the sun's luminosity.  Therefore, the answer is $\boxed{3.83e35}$ ergs/s.

Answer: 3.83e35


Preamble: A spectrum is taken of a single star (i.e., one not in a binary). Among the observed spectral lines is one from oxygen whose rest wavelength is $5007 \AA$. The Doppler shifted oxygen line from this star is observed to be at a wavelength of $5012 \AA$. The star is also observed to have a proper motion, $\mu$, of 1 arc second per year (which corresponds to $\sim 1.5 \times 10^{-13}$ radians per second of time). It is located at a distance of $60 \mathrm{pc}$ from the Earth. Take the speed of light to be $3 \times 10^8$ meters per second.

Subproblem 0: What is the component of the star's velocity parallel to its vector to the Earth (in kilometers per second)?

Solution: To find this longitudinal velocity component, we use the Doppler shift, finding $V_{r}=\frac{\Delta \lambda}{\lambda} c=\frac{5}{5000} c=\boxed{300} \mathrm{~km} / \mathrm{s}$.

Answer: 300


Subproblem 0: The differential luminosity from a star, $\Delta L$, with an approximate blackbody spectrum, is given by:
\[
\Delta L=\frac{8 \pi^{2} c^{2} R^{2}}{\lambda^{5}\left[e^{h c /(\lambda k T)}-1\right]} \Delta \lambda
\]
where $R$ is the radius of the star, $T$ is its effective surface temperature, and $\lambda$ is the wavelength. $\Delta L$ is the power emitted by the star between wavelengths $\lambda$ and $\lambda+\Delta \lambda$ (assume $\Delta \lambda \ll \lambda)$. The star is at distance $d$. Find the star's spectral intensity $I(\lambda)$ at the Earth, where $I(\lambda)$ is defined as the power per unit area per unit wavelength interval.

Solution:
\[
I(\lambda)=\frac{1}{4 \pi d^{2}} \frac{\Delta L}{\Delta \lambda}=\boxed{\frac{2 \pi c^{2} R^{2}}{\lambda^{5}\left[e^{h c /(\lambda k T)}-1\right] d^{2}}}
\]

Answer: \frac{2 \pi c^{2} R^{2}}{\lambda^{5}\left[e^{h c /(\lambda k T)}-1\right] d^{2}}


Preamble: A very hot star is detected in the galaxy M31 located at a distance of $800 \mathrm{kpc}$. The star has a temperature $T = 6 \times 10^{5} K$ and produces a flux of $10^{-12} \mathrm{erg} \cdot \mathrm{s}^{-1} \mathrm{cm}^{-2}$ at the Earth. Treat the star's surface as a blackbody radiator.

Subproblem 0: Find the luminosity of the star (in units of $\mathrm{erg} \cdot \mathrm{s}^{-1}$).

Solution: \[
  L=4 \pi D^{2} \text { Flux }_{\text {Earth }}=10^{-12} 4 \pi\left(800 \times 3 \times 10^{21}\right)^{2}=\boxed{7e37} \mathrm{erg} \cdot \mathrm{s}^{-1}
\]

Final answer: The final answer is 7e37. I hope it is correct.

Subproblem 1: Compute the star's radius in centimeters.

Solution:
\[
  R=\left(L / 4 \pi \sigma T^{4}\right)^{1 / 2}=\boxed{8.7e8} \mathrm{~cm}=0.012 R_{\odot}
\]

Answer: 8.7e8


Subproblem 0: A star is at a distance from the Earth of $300 \mathrm{pc}$. Find its parallax angle, $\pi$, in arcseconds to one significant figure.

Solution:
\[
\begin{aligned}
D &=1 \mathrm{pc} / \pi^{\prime \prime} \\
\pi^{\prime \prime} &=1 \mathrm{pc} / 300 \mathrm{pc} \\
\pi^{\prime \prime} &=\boxed{0.003}^{\prime \prime}
\end{aligned}
\]

Answer: 0.003


Subproblem 0: The Sun's effective temperature, $T_{e}$, is 5800 Kelvin, and its radius is $7 \times 10^{10} \mathrm{~cm}\left(7 \times 10^{8}\right.$ m). Compute the luminosity (power output) of the Sun in erg/s. Treat the Sun as a blackbody radiator, and give your answer to one significant figure.

Solution: Using the standard formula for power output of a blackbody radiator gives $P = \sigma A T^4$, where the area in this case is $4\piR_{sun}^2$.  Plugging in the numbers given in the problem yields that the sun's power output is (to one significant figure) $\boxed{4e33}$ ergs.

Answer: 4e33


Subproblem 0: Use the Bohr model of the atom to compute the wavelength of the transition from the $n=100$ to $n=99$ levels, in centimeters. [Uscful relation: the wavelength of $L \alpha$ ( $\mathrm{n}=2$ to $\mathrm{n}=1$ transition) is $1216 \AA$.]

Solution:
The inverse wavelength of radiation is proportional to the energy difference between the initial and final energy levels.  So for our transition of interest, we have
\begin{equation}
  \lambda^{-1} = R(\frac{1}{99^2} - \frac{1}{100^2}).
\end{equation}
Using the information given in the problem for the $L \alpha$ transition, we get
\begin{equation}
  (1216 \AA)^{-1} = R(\frac{1}{1^2} - \frac{1}{2^2}).
\end{equation}
Combining the above two relations yields $\lambda = \boxed{4.49}$ cm.

Answer: 4.49


Preamble: A radio interferometer, operating at a wavelength of $1 \mathrm{~cm}$, consists of 100 small dishes, each $1 \mathrm{~m}$ in diameter, distributed randomly within a $1 \mathrm{~km}$ diameter circle.

Subproblem 0: What is the angular resolution of a single dish, in radians?

Solution: The angular resolution of a single dish is roughly given by the wavelength over its radius, in this case $\boxed{0.01}$ radians.

Answer: 0.01


Preamble: Orbital Dynamics: A binary system consists of two stars in circular orbit about a common center of mass, with an orbital period, $P_{\text {orb }}=10$ days. Star 1 is observed in the visible band, and Doppler measurements show that its orbital speed is $v_{1}=20 \mathrm{~km} \mathrm{~s}^{-1}$. Star 2 is an X-ray pulsar and its orbital radius about the center of mass is $r_{2}=3 \times 10^{12} \mathrm{~cm}=3 \times 10^{10} \mathrm{~m}$.

Subproblem 0: Find the orbital radius, $r_{1}$, of the optical star (Star 1) about the center of mass, in centimeters.

Solution: \[
\begin{gathered}
v_{1}=\frac{2 \pi r_{1}}{P_{\text {orb }}} \\
r_{1}=\frac{P_{\text {orb }} v_{1}}{2 \pi}=\boxed{2.75e11} \mathrm{~cm}
\end{gathered}
\]

Final answer: The final answer is 2.75e11. I hope it is correct.

Subproblem 1: What is the total orbital separation between the two stars, $r=r_{1}+r_{2}$ (in centimeters)?

Solution:
\[
  r=r_{1}+r_{2}=2.75 \times 10^{11}+3 \times 10^{12}=\boxed{3.3e12} \quad \mathrm{~cm}
\]

Answer: 3.3e12


Subproblem 0: If a star cluster is made up of $10^{4}$ stars, each of whose absolute magnitude is $-5$, compute the combined apparent magnitude of the cluster if it is located at a distance of $1 \mathrm{Mpc}$.

Solution:
The absolute magnitude of one of the stars is given by:
\[
M=-2.5 \log \left(L / L_{\mathrm{ref}}\right)=-5
\]
where $L$ is the stellar luminosity, and $L_{\text {ref }}$ is the luminosity of a zero magnitude star. This equation implies that $L=100 L_{\text {ref }}$. Armed with this fact, we can now compute the combined magnitude of the collection of $10^{4}$ stars:
\[
M_{\text {TOT }}=-2.5 \log \left[\left(10^{4} \times 100 L_{\text {ref }}\right) / L_{\text {ref }}\right]=-2.5 \log \left(10^{6}\right)=-15
\]
Finally, the distance modulus corresponding to $1 \mathrm{Mpc}$ is $5 \log \left(10^{6} / 10\right)=25$. Therefore, the apparent magnitude of the star cluster at this distance is:
\[
m=M+\text { distance modulus } \Rightarrow m=-15+25=+\boxed{10} .
\]

Answer: 10


Subproblem 0: A galaxy moves directly away from us with a speed of $3000 \mathrm{~km} \mathrm{~s}^{-1}$. Find the wavelength of the $\mathrm{H} \alpha$ line observed at the Earth, in Angstroms. The rest wavelength of $\mathrm{H} \alpha$ is $6565 \AA$.  Take the speed of light to be $3\times 10^8$ meters per second.

Solution:
We have that the velocity of the galaxy is $0.01$ times $c$, the speed of light.  So, using Doppler effect formulas,
\begin{equation}
\lambda_{obs} = (6565 \AA)(1 + v/c) = (6565 \AA)(1.01)
\end{equation}
So the answer is $\boxed{6630}$ Angstroms.

Answer: 6630


Subproblem 0: The Spitzer Space Telescope has an effective diameter of $85 \mathrm{cm}$, and a typical wavelength used for observation of $5 \mu \mathrm{m}$, or 5 microns. Based on this information, compute an estimate for the angular resolution of the Spitzer Space telescope in arcseconds.

Solution: Using the formula for angular resolution $\theta$ in terms of the effective size $d$ and the wavelength $\lambda$, namely $\theta = \lambda/d$, gives \boxed{1.2} arcseconds.

Answer: 1.2


Subproblem 0: It has long been suspected that there is a massive black hole near the center of our Galaxy. Recently, a group of astronmers determined the parameters of a star that is orbiting the suspected black hole. The orbital period is 15 years, and the orbital radius is $0.12$ seconds of arc (as seen from the Earth). Take the distance to the Galactic center to be $8 \mathrm{kpc}$. Compute the mass of the black hole, starting from $F=m a$. Express your answer in units of the Sun's mass; i.e., answer the question `what is the ratio of masses between this black hole and our Sun'? Give your answer to 1 significant figure. (Assume that Newton's law of gravity is applicable for orbits sufficiently far from a black hole, and that the orbiting star satisfies this condition.)

Solution:
The force of gravitational attraction between the black hole (of mass $M_{BH}$) and the star (of mass $M_s$) is given by
\begin{equation}
F = \frac{G M_{BH} M_s}{R^2},
\end{equation}
where $R$ is the distance between the star and black hole (assuming a circular orbit).  Equating this to the centripetal force gives
\begin{equation}
F = \frac{G M_{BH} M_s}{R^2} = \frac{M_s v^2}{R},
\end{equation}
where $v$, the (linear) orbital velocity, is related to the orbital period $P$ by
\begin{equation}
v = \frac{2\pi R}{P}.
\end{equation}
Combining the above equations, we get
\begin{equation}
\frac{G M_{BH} M_s}{R^2} = \frac{M_s 4 \pi^2 R^2}{RP^2},
\end{equation}
or
\begin{equation}
G M_{BH} = 4 \pi^2 R^3 / P^2
\end{equation}
Since this equation should also be valid for Earth's orbit around the Sun, if we replace $M_{BH}$ by the Sun's mass, $R$ by the Earth-sun distance, and $P$ by the orbital period of 1 year, we find that the ratio of masses between the black hole and our Sun is given by $(R / 1 \mathrm{year})^3 / (P / 1 \mathrm{a.u.})^2$.
To evaluate the above expression, we need to find $R$ from the information given in the problem; since we know the angle its orbital radius subtends ($0.12$ arcseconds) at a distance of $8 \mathrm{kpc}$, we simply multiply these two quantities to find that $R = 900~\mathrm{a.u.}$.  So $M_{BH}/M_{sun} = (900)^3/(15)^2$, or $\boxed{3e6}$.

Answer: 3e6


Preamble: A very hot star is detected in the galaxy M31 located at a distance of $800 \mathrm{kpc}$. The star has a temperature $T = 6 \times 10^{5} K$ and produces a flux of $10^{-12} \mathrm{erg} \cdot \mathrm{s}^{-1} \mathrm{cm}^{-2}$ at the Earth. Treat the star's surface as a blackbody radiator.

Subproblem 0: Find the luminosity of the star (in units of $\mathrm{erg} \cdot \mathrm{s}^{-1}$).

Solution:
\[
  L=4 \pi D^{2} \text { Flux }_{\text {Earth }}=10^{-12} 4 \pi\left(800 \times 3 \times 10^{21}\right)^{2}=\boxed{7e37} \mathrm{erg} \cdot \mathrm{s}^{-1}
\]

Answer: 7e37


Subproblem 0: A large ground-based telescope has an effective focal length of 10 meters. Two astronomical objects are separated by 1 arc second in the sky. How far apart will the two corresponding images be in the focal plane, in microns?

Solution:
\[
s=f \theta=1000 \mathrm{~cm} \times \frac{1}{2 \times 10^{5}} \text { radians }=0.005 \mathrm{~cm}=\boxed{50} \mu \mathrm{m}
\]

Answer: 50


Subproblem 0: The equation of state for cold (non-relativistic) matter may be approximated as:
\[
P=a \rho^{5 / 3}-b \rho^{4 / 3}
\]
where $P$ is the pressure, $\rho$ the density, and $a$ and $b$ are fixed constants. Use a dimensional analysis of the equation of hydrostatic equilibrium to estimate the ``radius-mass'' relation for planets and low-mass white dwarfs whose material follows this equation of state. Specifically, find $R(M)$ in terms of $G$ and the constants $a$ and $b$. You should set all constants of order unity (e.g., $4, \pi, 3$, etc.) to $1.0$. [Hint: solve for $R(M)$ rather than $M(R)$ ]. You can check your answer by showing that for higher masses, $R \propto M^{-1 / 3}$, while for the lower-masses $R \propto M^{+1 / 3}$.

Solution:
\[
\begin{gathered}
\frac{d P}{d r}=-g \rho \\
\frac{a \rho^{5 / 3}-b \rho^{4 / 3}}{R} \sim\left(\frac{G M}{R^{2}}\right)\left(\frac{M}{R^{3}}\right) \\
\frac{a M^{5 / 3}}{R^{6}}-\frac{b M^{4 / 3}}{R^{5}} \sim\left(\frac{G M^{2}}{R^{5}}\right) \\
G M^{2} \sim \frac{a M^{5 / 3}}{R}-b M^{4 / 3} \\
R \frac{a M^{5 / 3}}{G M^{2}+b M^{4 / 3}} \simeq \boxed{\frac{a M^{1 / 3}}{G M^{2 / 3}+b}}
\end{gathered}
\]
For small masses, $R \propto M^{1 / 3}$ as for rocky planets, while for larger masses, $R \propto M^{-1 / 3}$ as for white dwarfs where the degenerate electrons are not yet relativistic.

Answer: \frac{a M^{1 / 3}}{G M^{2 / 3}+b}


Subproblem 0: Take the total energy (potential plus thermal) of the Sun to be given by the simple expression:
\[
E \simeq-\frac{G M^{2}}{R}
\]
where $M$ and $R$ are the mass and radius, respectively. Suppose that the energy generation in the Sun were suddenly turned off and the Sun began to slowly contract. During this contraction its mass, $M$, would remain constant and, to a fair approximation, its surface temperature would also remain constant at $\sim 5800 \mathrm{~K}$. Assume that the total energy of the Sun is always given by the above expression, even as $R$ gets smaller. By writing down a simple (differential) equation relating the power radiated at Sun's surface with the change in its total energy (using the above expression), integrate this equation to find the time (in years) for the Sun to shrink to $1 / 2$ its present radius.  Answer in units of years.

Solution:
\[
\begin{gathered}
L=4 \pi \sigma R^{2} T^{4}=d E / d t=\left(\frac{G M^{2}}{R^{2}}\right) \frac{d R}{d t} \\
\int_{R}^{0.5 R} \frac{d R}{R^{4}}=-\int_{0}^{t} \frac{4 \pi \sigma T^{4}}{G M^{2}} d t \\
-\frac{1}{3(R / 2)^{3}}+\frac{1}{3 R^{3}}=-\left(\frac{4 \pi \sigma T^{4}}{G M^{2}}\right) t \\
t=\frac{G M^{2}}{12 \pi \sigma T^{4}}\left(\frac{8}{R^{3}}-\frac{1}{R^{3}}\right) \\
t=\frac{7 G M^{2}}{12 \pi \sigma T^{4} R^{3}}=2.2 \times 10^{15} \mathrm{sec}=75 \text { million years }
\end{gathered}
\]
So the answer is $\boxed{7.5e7}$ years.

Answer: 7.5e7


Preamble: Once a star like the Sun starts to ascend the giant branch its luminosity, to a good approximation, is given by:
\[
L=\frac{10^{5} L_{\odot}}{M_{\odot}^{6}} M_{\text {core }}^{6}
\]
where the symbol $\odot$ stands for the solar value, and $M_{\text {core }}$ is the mass of the He core of the star. Further, assume that as more hydrogen is burned to helium - and becomes added to the core - the conversion efficiency between rest mass and energy is:
\[
\Delta E=0.007 \Delta M_{\text {core }} c^{2} .
\]

Subproblem 0: Use these two expressions to write down a differential equation, in time, for $M_{\text {core }}$.  For ease of writing, simply use the variable $M$ to stand for $M_{\text {core }}$.  Leave your answer in terms of $c$, $M_{\odot}$, and $L_{\odot}$.

Solution:
\[
L \equiv \frac{\Delta E}{\Delta t}=\frac{0.007 \Delta M c^{2}}{\Delta t}=\frac{10^{5} L_{\odot}}{M_{\odot}^{6}} M^{6}.
\]
Converting these to differentials, we get
\begin{equation}
\frac{0.007 dM c^{2}}{dt}=\frac{10^{5} L_{\odot}}{M_{\odot}^{6}} M^{6}, or
\end{equation}
\begin{equation}
\boxed{\frac{dM}{dt}=\frac{10^{5} L_{\odot}}{0.007 c^{2} M_{\odot}^{6}} M^{6}}
\end{equation}

Answer: \frac{dM}{dt}=\frac{10^{5} L_{\odot}}{0.007 c^{2} M_{\odot}^{6}} M^{6}


Subproblem 0: A star of radius, $R$, and mass, $M$, has an atmosphere that obeys a polytropic equation of state:
\[
P=K \rho^{5 / 3} \text {, }
\]
where $P$ is the gas pressure, $\rho$ is the gas density (mass per unit volume), and $K$ is a constant throughout the atmosphere. Assume that the atmosphere is sufficiently thin (compared to $R$ ) that the gravitational acceleration can be taken to be a constant.
Use the equation of hydrostatic equilibrium to derive the pressure as a function of height $z$ above the surface of the planet. Take the pressure at the surface to be $P_{0}$.

Solution:
Start with the equation of hydrostatic equilibrium:
\[
\frac{d P}{d z}=-g \rho
\]
where $g$ is approximately constant through the atmosphere, and is given by $G M / R^{2}$. We can use the polytropic equation of state to eliminate $\rho$ from the equation of hydrostatic equilibrium:
\[
\frac{d P}{d z}=-g\left(\frac{P}{K}\right)^{3 / 5}
\]
Separating variables, we find:
\[
P^{-3 / 5} d P=-g\left(\frac{1}{K}\right)^{3 / 5} d z
\]
We then integrate the left-hand side from $P_{0}$ to $P$ and the right hand side from 0 to $z$ to find:
\[
\frac{5}{2}\left(P^{2 / 5}-P_{0}^{2 / 5}\right)=-g K^{-3 / 5} z
\]
Solving for $P(z)$ we have:
\[
  P(z)=\boxed{\left[P_{0}^{2 / 5}-\frac{2}{5} g K^{-3 / 5} z\right]^{5 / 2}}=P_{0}\left[1-\frac{2}{5} \frac{g}{P_{0}^{2 / 5} K^{3 / 5}} z\right]^{5 / 2}
\]
The pressure therefore, goes to zero at a finite height $z_{\max }$, where:
\[
z_{\max }=\frac{5 P_{0}^{2 / 5} K^{3 / 5}}{2 g}=\frac{5 K \rho_{0}^{2 / 3}}{2 g}=\frac{5 P_{0}}{2 g \rho_{0}}
\]

Answer: \left[P_{0}^{2 / 5}-\frac{2}{5} g K^{-3 / 5} z\right]^{5 / 2}


Subproblem 0: An eclipsing binary consists of two stars of different radii and effective temperatures. Star 1 has radius $R_{1}$ and $T_{1}$, and Star 2 has $R_{2}=0.5 R_{1}$ and $T_{2}=2 T_{1}$. Find the change in bolometric magnitude of the binary, $\Delta m_{\text {bol }}$, when the smaller star is behind the larger star. (Consider only bolometric magnitudes so you don't have to worry about color differences.)

Solution:
\[
\begin{gathered}
\mathcal{F}_{1 \& 2}=4 \pi \sigma\left(T_{1}^{4} R_{1}^{2}+T_{2}^{4} R_{2}^{2}\right) \\
\mathcal{F}_{\text {eclipse }}=4 \pi \sigma T_{1}^{4} R_{1}^{2} \\
\Delta m=-2.5 \log \left(\frac{\mathcal{F}_{1 \& 2}}{\mathcal{F}_{\text {eclipse }}}\right) \\
\Delta m=-2.5 \log \left(1+\frac{T_{2}^{4} R_{2}^{2}}{T_{1}^{4} R_{1}^{2}}\right) \\
\Delta m=-2.5 \log \left(1+\frac{16}{4}\right)=-1.75
\end{gathered}
\]
So, the binary is $\boxed{1.75}$ magnitudes brighter out of eclipse than when star 2 is behind star 1 .

Answer: 1.75


Preamble: It has been suggested that our Galaxy has a spherically symmetric dark-matter halo with a density distribution, $\rho_{\text {dark }}(r)$, given by:
\[
\rho_{\text {dark }}(r)=\rho_{0}\left(\frac{r_{0}}{r}\right)^{2},
\]
where $\rho_{0}$ and $r_{0}$ are constants, and $r$ is the radial distance from the center of the galaxy. For star orbits far out in the halo you can ignore the gravitational contribution of the ordinary matter in the Galaxy.

Subproblem 0: Compute the rotation curve of the Galaxy (at large distances), i.e., find $v(r)$ for circular orbits.

Solution:
\[
\begin{gathered}
-\frac{G M(<r)}{r^{2}}=-\frac{v^{2}}{r} \quad(\text { from } F=m a) \\
M(<r)=\int_{0}^{r} \rho_{0}\left(\frac{r_{0}}{r}\right)^{2} 4 \pi r^{2} d r=4 \pi \rho_{0} r_{0}^{2} r
\end{gathered}
\]
Note that, in general, $M \neq \rho \times$ volume! You must integrate over $\rho(r)$. From these expressions we find:
\[
v(r)=\boxed{\sqrt{4 \pi G \rho_{0} r_{0}^{2}}}=\text { constant }
\]

Answer: \sqrt{4 \pi G \rho_{0} r_{0}^{2}}


Subproblem 0: The Very Large Array (VLA) telescope has an effective diameter of $36 \mathrm{~km}$, and a typical wavelength used for observation at this facility might be $6 \mathrm{~cm}$.  Based on this information, compute an estimate for the angular resolution of the VLA in arcseconds

Solution: Using the formula for angular resolution $\theta$ in terms of the effective size $d$ and the wavelength $\lambda$, namely $\theta = \lambda/d$, gives \boxed{0.33} arcseconds.

Answer: 0.33


Subproblem 0: A particular star has an absolute magnitude $M=-7$. If this star is observed in a galaxy that is at a distance of $3 \mathrm{Mpc}$, what will its apparent magnitude be?

Solution: \[
\text { Given: } M=-7 \text { and } d=3 \mathrm{Mpc}
\]
\[
\begin{aligned}
  & \text { Apparent Magnitude: } m=M+5 \log \left[\frac{d}{10 \mathrm{pc}}\right]=-7+5 \log \left[\frac{3 \times 10^{6}}{10}\right]=\boxed{20.39} \\
\end{aligned}
\]

Final answer: The final answer is 20.39. I hope it is correct.

Subproblem 1: What is the distance modulus to this galaxy?

Solution:
Distance Modulus: $DM=m-M=20.39+7=\boxed{27.39}$
\end{aligned}

Answer: 27.39


Subproblem 0: Find the distance modulus to the Andromeda galaxy (M31). Take the distance to Andromeda to be $750 \mathrm{kpc}$, and answer to three significant figures.

Solution:
\[
\mathrm{DM}=5 \log \left(\frac{d}{10 \mathrm{pc}}\right)=5 \log (75,000)=\boxed{24.4}
\]

Answer: 24.4


Subproblem 0: The Hubble Space telescope has an effective diameter of $2.5 \mathrm{~m}$, and a typical wavelength used for observation by the Hubble might be $0.6 \mu \mathrm{m}$, or 600 nanometers (typical optical wavelength). Based on this information, compute an estimate for the angular resolution of the Hubble Space telescope in arcseconds.

Solution: Using the formula for angular resolution $\theta$ in terms of the effective size $d$ and the wavelength $\lambda$, namely $\theta = \lambda/d$, gives \boxed{0.05} arcseconds.

Answer: 0.05


Preamble: A collimated light beam propagating in water is incident on the surface (air/water interface) at an angle $\theta_w$ with respect to the surface normal.

Subproblem 0: If the index of refraction of water is $n=1.3$, find an expression for the angle of the light once it emerges from the water into the air, $\theta_a$, in terms of $\theta_w$.

Solution: Using Snell's law, $1.3 \sin{\theta_w} = \sin{\theta_a}$. So $\theta_a = \boxed{\arcsin{1.3 \sin{\theta_w}}}$.

Answer: \arcsin{1.3 \sin{\theta_w}}


Subproblem 0: What fraction of the rest mass energy is released (in the form of radiation) when a mass $\Delta M$ is dropped from infinity onto the surface of a neutron star with $M=1 M_{\odot}$ and $R=10$ $\mathrm{km}$ ?

Solution:
\[
\Delta E=\frac{G M \Delta m}{R}
\]
The fractional rest energy lost is $\Delta E / \Delta m c^{2}$, or
\[
\frac{\Delta E}{\Delta m c^{2}}=\frac{G M}{R c^{2}} \simeq \boxed{0.15}
\]

Answer: 0.15


Preamble: The density of stars in a particular globular star cluster is $10^{6} \mathrm{pc}^{-3}$. Take the stars to have the same radius as the Sun, and to have an average speed of $10 \mathrm{~km} \mathrm{sec}^{-1}$.

Subproblem 0: Find the mean free path for collisions among stars.  Express your answer in centimeters, to a single significant figure.

Solution:
\[
\begin{gathered}
\ell \simeq \frac{1}{n \sigma}=\frac{1}{10^{6} \mathrm{pc}^{-3} \pi R^{2}} \\
\ell \simeq \frac{1}{3 \times 10^{-50} \mathrm{~cm}^{-3} \times 1.5 \times 10^{22} \mathrm{~cm}^{2}} \simeq \boxed{2e27} \mathrm{~cm}
\end{gathered}
\]

Answer: 2e27


Subproblem 0: For a gas supported by degenerate electron pressure, the pressure is given by:
\[
P=K \rho^{5 / 3}
\]
where $K$ is a constant and $\rho$ is the mass density. If a star is totally supported by degenerate electron pressure, use a dimensional analysis of the equation of hydrostatic equilibrium:
\[
\frac{d P}{d r}=-g \rho
\]
to determine how the radius of such a star depends on its mass, $M$.  Specifically, you will find that $R$ is proportional to some power of $M$; what is that power?

Solution:
\[
\begin{gathered}
\frac{K \rho^{5 / 3}}{R} \simeq\left(\frac{G M}{R^{2}}\right)\left(\frac{M}{R^{3}}\right) \\
\rho \sim \frac{M}{R^{3}} \\
\frac{K M^{5 / 3}}{R R^{5}} \simeq \frac{G M^{2}}{R^{5}} \\
R \simeq \frac{K}{G M^{1 / 3}}
\end{gathered}
\]
So the answer is $\boxed{-1./3}$.

Answer: -1./3


Subproblem 0: A galaxy moves directly away from us with speed $v$, and the wavelength of its $\mathrm{H} \alpha$ line is observed to be $6784 \AA$. The rest wavelength of $\mathrm{H} \alpha$ is $6565 \AA$. Find $v/c$.

Solution:
\[
\lambda \simeq \lambda_{0}(1+v / c)
\]
where $\lambda=6784 \AA$ and $\lambda_{0}=6565 \AA$. Rearranging,
\[
\frac{v}{c} \simeq \frac{\lambda-\lambda_{0}}{\lambda_{0}} \simeq \frac{6784-6565}{6565} \Rightarrow v \simeq 0.033 c
\]
So $v/c \simeq \boxed{0.033}$.

Answer: 0.033


Subproblem 0: A candle has a power in the visual band of roughly $3$ Watts. When this candle is placed at a distance of $3 \mathrm{~km}$ it has the same apparent brightness as a certain star. Assume that this star has the same luminosity as the Sun in the visual band $\left(\sim 10^{26}\right.$ Watts $)$. How far away is the star (in pc)?

Solution: The fact that the two sources have the same apparent brightness implies that the flux at the respective distances is the same; since flux varies with distance as $1/d^2$, we find that (with distances in km) $\frac{3}{3^2} = \frac{10^{26}}{d^2}$, so $d = 10^{13}\times\frac{3}{\sqrt{3}}$, or roughly $1.7\times 10^{13}$ kilometers.  In parsecs, this is $\boxed{0.5613}$ parsecs.

Answer: 0.5613


Preamble: A galaxy is found to have a rotation curve, $v(r)$, given by
\[
v(r)=\frac{\left(\frac{r}{r_{0}}\right)}{\left(1+\frac{r}{r_{0}}\right)^{3 / 2}} v_{0}
\]
where $r$ is the radial distance from the center of the galaxy, $r_{0}$ is a constant with the dimension of length, and $v_{0}$ is another constant with the dimension of speed. The rotation curve is defined as the orbital speed of test stars in circular orbit at radius $r$.

Subproblem 0: Find an expression for $\omega(r)$, where $\omega$ is the angular velocity.  The constants $v_{0}$ and $r_{0}$ will appear in your answer.

Solution: $\omega=v / r & \Rightarrow \omega(r)=\boxed{\frac{v_{0}}{r_{0}} \frac{1}{\left(1+r / r_{0}\right)^{3 / 2}}}$

Answer: \frac{v_{0}}{r_{0}} \frac{1}{\left(1+r / r_{0}\right)^{3 / 2}}


Preamble: Orbital Dynamics: A binary system consists of two stars in circular orbit about a common center of mass, with an orbital period, $P_{\text {orb }}=10$ days. Star 1 is observed in the visible band, and Doppler measurements show that its orbital speed is $v_{1}=20 \mathrm{~km} \mathrm{~s}^{-1}$. Star 2 is an X-ray pulsar and its orbital radius about the center of mass is $r_{2}=3 \times 10^{12} \mathrm{~cm}=3 \times 10^{10} \mathrm{~m}$.

Subproblem 0: Find the orbital radius, $r_{1}$, of the optical star (Star 1) about the center of mass, in centimeters.

Solution:
\[
\begin{gathered}
v_{1}=\frac{2 \pi r_{1}}{P_{\text {orb }}} \\
r_{1}=\frac{P_{\text {orb }} v_{1}}{2 \pi}=\boxed{2.75e11} \mathrm{~cm}
\end{gathered}
\]

Answer: 2.75e11


Preamble: The density of stars in a particular globular star cluster is $10^{6} \mathrm{pc}^{-3}$. Take the stars to have the same radius as the Sun, and to have an average speed of $10 \mathrm{~km} \mathrm{sec}^{-1}$.

Subproblem 0: Find the mean free path for collisions among stars.  Express your answer in centimeters, to a single significant figure.

Solution: \[
\begin{gathered}
\ell \simeq \frac{1}{n \sigma}=\frac{1}{10^{6} \mathrm{pc}^{-3} \pi R^{2}} \\
\ell \simeq \frac{1}{3 \times 10^{-50} \mathrm{~cm}^{-3} \times 1.5 \times 10^{22} \mathrm{~cm}^{2}} \simeq \boxed{2e27} \mathrm{~cm}
\end{gathered}
\]

Final answer: The final answer is 2e27. I hope it is correct.

Subproblem 1: Find the corresponding mean time between collisions. (Assume that the stars move in straight-line paths, i.e., are not deflected by gravitational interactions.)  Answer in units of years, to a single significant figure.

Solution: $\tau_{\text {coll }} \simeq \frac{2 \times 10^{27} \mathrm{~cm}}{10^{6} \mathrm{~cm} / \mathrm{sec}} \simeq 2 \times 10^{21} \mathrm{sec} \simeq \boxed{6e13} \text { years }$

Answer: 6e13


Preamble: A radio interferometer, operating at a wavelength of $1 \mathrm{~cm}$, consists of 100 small dishes, each $1 \mathrm{~m}$ in diameter, distributed randomly within a $1 \mathrm{~km}$ diameter circle.

Subproblem 0: What is the angular resolution of a single dish, in radians?

Solution: The angular resolution of a single dish is roughly given by the wavelength over its radius, in this case $\boxed{0.01}$ radians.

Final answer: The final answer is 0.01. I hope it is correct.

Subproblem 1: What is the angular resolution of the interferometer array for a source directly overhead, in radians?

Solution: The angular resolution of the full array is given by the wavelength over the dimension of the array, in this case $\boxed{1e-5}$ radians.

Answer: 1e-5


Subproblem 0: If a star cluster is made up of $10^{6}$ stars whose absolute magnitude is the same as that of the Sun (+5), compute the combined magnitude of the cluster if it is located at a distance of $10 \mathrm{pc}$.

Solution:
At $10 \mathrm{pc}$, the magnitude is (by definition) just the absolute magnitude of the cluster.  Since the total luminosity of the cluster is $10^{6}$ times the luminosity of the Sun, we have that
\begin{equation}
\delta m = 2.5 \log \left( \frac{L_{TOT}}{L_{sun}} \right) = 2.5 \log 10^6 = 15.
\end{equation}
Since the Sun has absolute magnitude +5, the magnitude of the cluser is $\boxed{-10}$.

Answer: -10


Subproblem 0: A certain red giant has a radius that is 500 times that of the Sun, and a temperature that is $1 / 2$ that of the Sun's temperature. Find its bolometric (total) luminosity in units of the bolometric luminosity of the Sun.

Solution: Power output goes as $T^4r^2$, so the power output of this star is $\boxed{15625}$ times that of the Sun.

Answer: 15625


Subproblem 0: Suppose air molecules have a collision cross section of $10^{-16} \mathrm{~cm}^{2}$. If the (number) density of air molecules is $10^{19} \mathrm{~cm}^{-3}$, what is the collision mean free path in cm? Answer to one significant figure.

Solution:
\[
\ell=\frac{1}{n \sigma}=\frac{1}{10^{19} 10^{-16}}=\boxed{1e-3} \mathrm{~cm}
\]

Answer: 1e-3


Subproblem 0: Two stars have the same surface temperature. Star 1 has a radius that is $2.5$ times larger than the radius of star 2. Star 1 is ten times farther away than star 2. What is the absolute value of the difference in apparent magnitude between the two stars, rounded to the nearest integer?

Solution: Total power output goes as $r^2 T^4$, where $r$ is the star's radius, and $T$ is its temperature.  Flux, at a distance $R$ away thus goes as $r^2 T^4 / R^2$.  In our case, the ratio of flux from star 1 to star 2 is $1/16$ (i.e., star 2 is greater in apparent magnitude).  Using the relation between apparent magnitude and flux, we find that that the absolute value of the difference in apparent magnitudes is $2.5 \log{16}$, which rounded to the nearest integer is $\boxed{3}$.

Answer: 3


Subproblem 0: What is the slope of a $\log N(>F)$ vs. $\log F$ curve for a homogeneous distribution of objects, each of luminosity, $L$, where $F$ is the flux at the observer, and $N$ is the number of objects observed per square degree on the sky?

Solution: The number of objects detected goes as the cube of the distance for objects with flux greater than a certain minimum flux. At the same time the flux falls off with the inverse square of the distance. Thus, the slope of the $\log N(>F)$ vs. $\log F$ curve is $\boxed{-3./2}$.

Answer: -3./2


Preamble: Comparison of Radio and Optical Telescopes.

Subproblem 0: The Very Large Array (VLA) is used to make an interferometric map of the Orion Nebula at a wavelength of $10 \mathrm{~cm}$. What is the best angular resolution of the radio image that can be produced, in radians? Note that the maximum separation of two antennae in the VLA is $36 \mathrm{~km}$.

Solution: The best angular resolution will occur at the maximum separation, and is simply the ratio of wavelength to this separation $p$: $\theta = \frac{\lambda}{p}$, or $\frac{0.1}{36\times 10^3}$, which is $\boxed{2.7778e-6}$ radians.

Answer: 2.7778e-6


Subproblem 0: A globular cluster has $10^{6}$ stars each of apparent magnitude $+8$. What is the combined apparent magnitude of the entire cluster?

Solution:
\[
\begin{gathered}
+8=-2.5 \log \left(F / F_{0}\right) \\
F=6.3 \times 10^{-4} F_{0} \\
F_{\text {cluster }}=10^{6} \times 6.3 \times 10^{-4} F_{0}=630 F_{0} \\
m_{\text {cluster }}=-2.5 \log (630)=\boxed{-7}
\end{gathered}
\]

Answer: -7


Preamble: A very hot star is detected in the galaxy M31 located at a distance of $800 \mathrm{kpc}$. The star has a temperature $T = 6 \times 10^{5} K$ and produces a flux of $10^{-12} \mathrm{erg} \cdot \mathrm{s}^{-1} \mathrm{cm}^{-2}$ at the Earth. Treat the star's surface as a blackbody radiator.

Subproblem 0: Find the luminosity of the star (in units of $\mathrm{erg} \cdot \mathrm{s}^{-1}$).

Solution: \[
  L=4 \pi D^{2} \text { Flux }_{\text {Earth }}=10^{-12} 4 \pi\left(800 \times 3 \times 10^{21}\right)^{2}=\boxed{7e37} \mathrm{erg} \cdot \mathrm{s}^{-1}
\]

Final answer: The final answer is 7e37. I hope it is correct.

Subproblem 1: Compute the star's radius in centimeters.

Solution: \[
  R=\left(L / 4 \pi \sigma T^{4}\right)^{1 / 2}=\boxed{8.7e8} \mathrm{~cm}=0.012 R_{\odot}
\]

Final answer: The final answer is 8.7e8. I hope it is correct.

Subproblem 2: At what wavelength is the peak of the emitted radiation? Answer in $\AA$.

Solution:
Using the Wien displacement law:
\[
  \lambda_{\max }=0.29 / T \mathrm{~cm}=\boxed{48} \AA
\]

\section{Information and Entropy (6.050J Spring 2008)}

Answer: 48


Subproblem 0: A Boolean function $F(A, B)$ is said to be universal if any arbitrary boolean function can be constructed by using nested $F(A, B)$ functions. A universal function is useful, since using it we can build any function we wish out of a single part. For example, when implementing boolean logic on a computer chip a universal function (called a 'gate' in logic-speak) can simplify design enormously. We would like to find a universal boolean function. In this problem we will denote the two boolean inputs $A$ and $B$ and the one boolean output as $C$.
First, to help us organize our thoughts, let's enumerate all of the functions we'd like to be able to construct. How many different possible one-output boolean functions of two variables are there? I.e., how many functions are there of the form $F(A, B)=C ?$

Solution:
This particular definition of universality only treats arbitrary functions of two Boolean variables, but with any number of outputs. It appears to be an onerous task to prove universality for an arbitrary number of outputs. However, since each individual output of a multi-output function can be considered a separate one-ouput function, it is sufficient to prove the case of only one-output functions. This is why we begin by listing all one-output functions of one variable.
Each variable $A$ and $B$ has two possible values, making four different combinations of inputs $(A, B)$. Each combination of inputs (four possible) can cause one of two output values. Therefore the number of possible one-output binary functions of two binary variables is $2^{4}$, or \boxed{16}. They are enumerated in the table below.
\begin{tabular}{cc|ccccccccccccccccccc}
$A$ & $B$ & $b_{0}$ & $b_{1}$ & $b_{2}$ & $b_{3}$ & $b_{4}$ & $b_{5}$ & $b_{6}$ & $b_{7}$ & $b_{8}$ & $b_{9}$ & $b_{10}$ & $b_{11}$ & $b_{12}$ & $b_{13}$ & $b_{14}$ & $b_{15}$ &  \\
\hline
0 & 0 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 &  \\
0 & 1 & 0 & 0 & 1 & 1 & 0 & 0 & 1 & 1 & 0 & 0 & 1 & 1 & 0 & 0 & 1 & 1 &  \\
1 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 &  \\
1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 &  \\
\end{tabular}

Answer: 16


Subproblem 0: Unfortunately, a mutant gene can turn box people into triangles late in life. A laboratory test has been developed which can spot the gene early so that the dreaded triangle transformation can be prevented by medications. This test is 95 percent accurate at spotting the gene when it is there. However, the test gives a "false positive" $0.4$ percent of the time, falsely indicating that a healthy box person has the mutant gene. If $0.1$ percent (be careful - that's one-tenth of one percent) of the box people have the mutant gene, what's the probability that a box person actually has the mutant gene if the test indicates that he or she does?

Solution:
We see that the probability that a person has the disease given that the test is positive, is:
\[
\frac{0.001 \times 0.95}{0.001 \times 0.95+0.999 \times 0.004}=19.2 \%
\]
$\begin{array}{ccccc}\text { Have Disease? } & \text { Percent } & \text { Test Results } & \text { Percent } & \text { Total } \\ \text { Yes } & 0.001 & \text { Positive } & 0.95 & 0.00095 \\ & & \text { Negative } & 0.05 & 0.00005 \\ \text { No } & 0.999 & \text { Positive } & 0.004 & 0.003996 \\ & & \text { Negative } & 0.996 & 0.95504\end{array}$
Answer: \boxed{0.192}.

Answer: 0.192


Subproblem 0: Buzz, the hot new dining spot on campus, emphasizes simplicity. It only has two items on the menu, burgers and zucchini. Customers make a choice as they enter (they are not allowed to order both), and inform the cooks in the back room by shouting out either "B" or "Z". Unfortunately the two letters sound similar so $8 \%$ of the time the cooks misinterpret what was said. The marketing experts who designed the restaurant guess that $90 \%$ of the orders will be for burgers and $10 \%$ for zucchini.
The cooks can hear one order per second. The customers arrive at the rate of one per second. One of the chefs says that this system will never work because customers can only send one bit per second, the rate at which orders can be accepted, so you could barely keep up even if there were no noise in the channel. You are hired as an outside consultant to deal with the problem.
What is the channel capacity $\mathrm{C}$ of this communication channel in bits per second?

Solution:
This is a noisy channel with the same probabilities for mixing up $Z$ and $B$. Channel capacity is defined as the maximum mutual information (for any possible input probability) times the rate $W$. The rate of error is $\epsilon=0.08$. So the channel capacity for this channel is given by:
\[
\begin{aligned}
C &=M_{\max } W \\
&=1-\epsilon \log _{2}\left(\frac{1}{\epsilon}\right)-(1-\epsilon) \log _{2}\left(\frac{1}{(1-\epsilon)}\right) \\
&=1-0.08 \log _{2}\left(\frac{1}{0.08}\right)-(0.92) \log _{2}\left(\frac{1}{0.92}\right) \\
&=0.5978 \mathrm{bits} / \mathrm{second}
\end{aligned}
\]
So the final answer is \boxed{0.5978} bits/s.

\section{Ecology I (1.018J Fall 2009)}

Answer: 0.5978


Preamble: Given the following data from an Experimental Forest, answer the following questions. Show your work and units.
$\begin{array}{ll}\text { Total vegetative biomass } & 80,000 \mathrm{kcal} \mathrm{m}^{-2} \\ \text { Detritus and organic matter in soil } & 120,000 \mathrm{kcal } \mathrm{m}^{-2} \\ \text { Total Gross Primary Productivity } & 20,000 \mathrm{kcal } \mathrm{m}^{-2} \mathrm{yr}^{-1} \\ \text { Total Plant Respiration } & 5,000 \mathrm{kcal} \mathrm{m}^{-2} \mathrm{yr}^{-1} \\ \text { Total Community Respiration } & 9,000 \mathrm{kcal} \mathrm{m}^{-2} \mathrm{yr}^{-1}\end{array}$

Subproblem 0: What is the net primary productivity of the forest?

Solution: NPP $=$ GPP $-R_{A}=20,000-5,000=\boxed{15000} \mathrm{kcal} \mathrm{m}^{-2} \mathrm{yr}^{-1}$

Final answer: The final answer is 15000. I hope it is correct.

Subproblem 1: What is the net community production?

Solution: $\mathrm{NCP}=\mathrm{GPP}-\mathrm{R}_{\mathrm{A}}-\mathrm{R}_{\mathrm{H}}=20,000-9000=\boxed{11000} \mathrm{kcal} \mathrm{m}^{-2} \mathrm{yr}^{-1}$

Answer: 11000


Preamble: A population of 100 ferrets is introduced to a large island in the beginning of 1990 . Ferrets have an intrinsic growth rate, $r_{\max }$ of $1.3 \mathrm{yr}^{-1}$.

Subproblem 0: Assuming unlimited resources-i.e., there are enough resources on this island to last the ferrets for hundreds of years-how many ferrets will there be on the island in the year 2000? (Show your work!)

Solution: $N_o = 100$ (in 1990)
\\
$N = ?$ (in 2000)
\\
$t = 10$ yr
\\
$r = 1.3 \text{yr}^{-1}$
\\
$N = N_{o}e^{rt} = 100*e^{(1.3/\text{yr})(10 \text{yr})} = 4.4 x 10^7$ ferrets
\\
There will be \boxed{4.4e7} ferrets on the island in the year 2000.

Final answer: The final answer is 4.4e7. I hope it is correct.

Subproblem 1: What is the doubling time of the ferret population? (Show your work!)

Solution:
$N_o = 100$ (in 1990)
\\
$t = 10$ yr
\\
$r = 1.3 \text{yr}^{-1}$
\\
$t_d = (ln(2))/r = 0.693/(1.3 \text{yr}^{-1}) = 0.53$ years
\\
The doubling time of the ferret population is \boxed{0.53} years.

Answer: 0.53


Preamble: Given the following data from an Experimental Forest, answer the following questions. Show your work and units.
$\begin{array}{ll}\text { Total vegetative biomass } & 80,000 \mathrm{kcal} \mathrm{m}^{-2} \\ \text { Detritus and organic matter in soil } & 120,000 \mathrm{kcal } \mathrm{m}^{-2} \\ \text { Total Gross Primary Productivity } & 20,000 \mathrm{kcal } \mathrm{m}^{-2} \mathrm{yr}^{-1} \\ \text { Total Plant Respiration } & 5,000 \mathrm{kcal} \mathrm{m}^{-2} \mathrm{yr}^{-1} \\ \text { Total Community Respiration } & 9,000 \mathrm{kcal} \mathrm{m}^{-2} \mathrm{yr}^{-1}\end{array}$

Subproblem 0: What is the net primary productivity of the forest?

Solution: NPP $=$ GPP $-R_{A}=20,000-5,000=\boxed{15000} \mathrm{kcal} \mathrm{m}^{-2} \mathrm{yr}^{-1}$

Answer: 15000


Preamble: The Peak District Moorlands in the United Kingdom store 20 million tonnes of carbon, almost half of the carbon stored in the soils of the entire United Kingdom (the Moorlands are only $8 \%$ of the land area). In pristine condition, these peatlands can store an additional 13,000 tonnes of carbon per year.

Subproblem 0: Given this rate of productivity, how long did it take for the Peatlands to sequester this much carbon?

Solution: $20,000,000$ tonnes $C / 13,000$ tonnes $C y^{-1}=\boxed{1538}$ years

Answer: 1538


Preamble: A population of 100 ferrets is introduced to a large island in the beginning of 1990 . Ferrets have an intrinsic growth rate, $r_{\max }$ of $1.3 \mathrm{yr}^{-1}$.

Subproblem 0: Assuming unlimited resources-i.e., there are enough resources on this island to last the ferrets for hundreds of years-how many ferrets will there be on the island in the year 2000? (Show your work!)

Solution:
$N_o = 100$ (in 1990)
\\
$N = ?$ (in 2000)
\\
$t = 10$ yr
\\
$r = 1.3 \text{yr}^{-1}$
\\
$N = N_{o}e^{rt} = 100*e^{(1.3/\text{yr})(10 \text{yr})} = 4.4 x 10^7$ ferrets
\\
There will be \boxed{4.4e7} ferrets on the island in the year 2000.

\section{Differential Equations (18.03 Spring 2010)}

Answer: 4.4e7


Preamble: The following subproblems refer to a circuit with the following parameters. Denote by $I(t)$ the current (where the positive direction is, say, clockwise) in the circuit and by $V(t)$ the voltage increase across the voltage source, at time $t$. Denote by $R$ the resistance of the resistor and $C$ the capacitance of the capacitor (in units which we will not specify)-both positive numbers. Then
\[
R \dot{I}+\frac{1}{C} I=\dot{V}
\]

Subproblem 0: Suppose that $V$ is constant, $V(t)=V_{0}$. Solve for $I(t)$, with initial condition $I(0)$.

Solution: When $V$ is constant, the equation becomes $R \dot{I}+\frac{1}{C} I=0$, which is separable. Solving gives us
\[
I(t)=\boxed{I(0) e^{-\frac{t}{R C}}
}\].

Final answer: The final answer is I(0) e^{-\frac{t}{R C}}
. I hope it is correct.

Subproblem 1: It is common to write the solution to the previous subproblem in the form $c e^{-t / \tau}$. What is $c$ in this case?

Answer: I(0)


Subproblem 0: Consider the following "mixing problem." A tank holds $V$ liters of salt water. Suppose that a saline solution with concentration of $c \mathrm{gm} /$ liter is added at the rate of $r$ liters/minute. A mixer keeps the salt essentially uniformly distributed in the tank. A pipe lets solution out of the tank at the same rate of $r$ liters/minute. The differential equation for the amount of salt in the tank is given by
\[
x^{\prime}+\frac{r}{V} x-r c=0 .
\]
Suppose that the out-flow from this tank leads into another tank, also of volume 1 , and that at time $t=1$ the water in it has no salt in it. Again there is a mixer and an outflow. Write down a differential equation for the amount of salt in this second tank, as a function of time, assuming the amount of salt in the second tank at moment $t$ is given by $y(t)$, and the amount of salt in the first tank at moment $t$ is given by $x(t)$.

Solution: The differential equation for $y(t)$ is $\boxed{y^{\prime}+r y-r x(t)=0}$.

Answer: y^{\prime}+r y-r x(t)=0


Subproblem 0: Find the general solution of $x^{2} y^{\prime}+2 x y=\sin (2 x)$, solving for $y$. Note that a general solution to a differential equation has the form $x=x_{p}+c x_{h}$ where $x_{h}$ is a nonzero solution of the homogeneous equation $\dot{x}+p x=0$. Additionally, note that the left hand side is the derivative of a product.

Solution: We see that $\left(x^{2} y\right)^{\prime}=x^{2} y^{\prime}+2 x y$. Thus, $x^{2} y=-\frac{1}{2} \cos (2 x)+c$, and $y=\boxed{c x^{-2}-\frac{\cos (2 x)}{2 x^{2}}}$.

Answer: c x^{-2}-\frac{\cos (2 x)}{2 x^{2}}


Subproblem 0: An African government is trying to come up with good policy regarding the hunting of oryx. They are using the following model: the oryx population has a natural growth rate of $k$, and we suppose a constant harvesting rate of $a$ oryxes per year.
Write down an ordinary differential equation describing the evolution of the oryx population given the dynamics above, using $x(t)$ to denote the oryx population (the number of individual oryx(es)) at time $t$, measured in years.

Solution:
The natural growth rate is $k$, meaning that after some short time $\Delta t$ year(s) passes, we expect $k x(t) \Delta t$ new oryxes to appear. However, meanwhile the population is reduced by $a \Delta t$ oryxes due to the harvesting. Therefore, we are led to
\[
x(t+\Delta t) \simeq x(t)+k x(t) \Delta t-a \Delta t,
\]
and the unit on both sides is oryx $(\mathrm{es})$. If we let $\Delta t$ approach 0 , then we get the differential equation
\[
\boxed{\frac{d x}{d t}=k x-a} .
\]

Answer: \frac{d x}{d t}=k x-a


Subproblem 0: If the complex number $z$ is given by $z = 1+\sqrt{3} i$, what is the magnitude of $z^2$?

Solution: $z^{2}$ has argument $2 \pi / 3$ and radius 4, so by Euler's formula, $z^{2}=4 e^{i 2 \pi / 3}$. Thus $A=4, \theta=\frac{2\pi}{3}$, so our answer is $\boxed{4}$.

Answer: 4


Subproblem 0: In the polar representation $(r, \theta)$ of the complex number $z=1+\sqrt{3} i$, what is $r$?

Solution: For z, $r=2$ and $\theta=\pi / 3$, so its polar coordinates are $\left(2, \frac{\pi}{3}\right)$. So $r=\boxed{2}$.

Answer: 2


Preamble: In the following problems, take $a = \ln 2$ and $b = \pi / 3$.

Subproblem 0: Given $a = \ln 2$ and $b = \pi / 3$, rewrite $e^{a+b i}$ in the form $x + yi$, where $x, y$ are real numbers.

Solution: Using Euler's formula, we find that the answer is $\boxed{1+\sqrt{3} i}$.

Answer: 1+\sqrt{3} i


Subproblem 0: Find the general solution of the differential equation $y^{\prime}=x-2 y$ analytically using integrating factors, solving for $y$. Note that a function $u(t)$ such that $u \dot{x}+u p x=\frac{d}{d t}(u x)$ is an integrating factor. Additionally, note that a general solution to a differential equation has the form $x=x_{p}+c x_{h}$ where $x_{h}$ is a nonzero solution of the homogeneous equation $\dot{x}+p x=0$.

Solution: In standard form, $y^{\prime}+2 y=x$, so $u=C e^{2 x}$. Then $y=u^{-1} \int u x d x=e^{-2 x} \int x e^{2 x} d x$. Integrating by parts yields $\int x e^{2 x} d x=$ $\frac{x}{2} e^{2 x}-\frac{1}{2} \int e^{2 x} d x=\frac{x}{2} e^{2 x}-\frac{1}{4} e^{2 x}+c$. Therefore, $y=\boxed{x / 2-1 / 4+c e^{-2 x}}$.

Final answer: The final answer is x / 2-1 / 4+c e^{-2 x}. I hope it is correct.

Subproblem 1: For what value of $c$ does the straight line solution occur?

Solution: The straight line solution occurs when $c=\boxed{0}$.

Answer: 0


Preamble: The following subproblems relate to applying Euler's Method (a first-order numerical procedure for solving ordinary differential equations with a given initial value) onto $y^{\prime}=y^{2}-x^{2}=F(x, y)$ at $y(0)=-1$, with $h=0.5$. Recall the notation \[x_{0}=0, y_{0}=-1, x_{n+1}=x_{h}+h, y_{n+1}=y_{n}+m_{n} h, m_{n}=F\left(x_{n}, y_{n}\right)\].

Subproblem 0: Use Euler's method to estimate the value at $x=1.5$.

Solution: $y_3 = \boxed{-0.875}$

Answer: -0.875


Subproblem 0: Rewrite the function $f(t) = \cos (2 t)+\sin (2 t)$ in the form $A \cos (\omega t-\phi)$. It may help to begin by drawing a right triangle with sides $a$ and $b$.

Solution: Here, our right triangle has hypotenuse $\sqrt{2}$, so $A=\sqrt{2}$. Both summands have "circular frequency" 2, so $\omega=2 . \phi$ is the argument of the hypotenuse, which is $\pi / 4$, so $f(t)=\boxed{\sqrt{2} \cos (2 t-\pi / 4)}$.

Answer: \sqrt{2} \cos (2 t-\pi / 4)


Subproblem 0: Given the ordinary differential equation $\ddot{x}-a^{2} x=0$, where $a$ is a nonzero real-valued constant, find a solution $x(t)$ to this equation such that $x(0) = 0$ and $\dot{x}(0)=1$.

Solution: First, notice that both $x(t)=e^{a t}$ and $x(t)=e^{-a t}$ are solutions to $\ddot{x}-a^{2} x=0$. Then for any constants $c_{1}$ and $c_{2}$, $x(t)=c_{1} e^{a t}+c_{2} e^{-a t}$ are also solutions to $\ddot{x}-a^{2} x=0$. Moreover, $x(0)=c_{1}+c_{2}$, and $\dot{x}(0)=a\left(c_{1}-c_{2}\right)$. Assuming $a \neq 0$, to satisfy the given conditions, we need $c_{1}+c_{2}=0$ and $a\left(c_{1}-c_{2}\right)=1$, which implies $c_{1}=-c_{2}=\frac{1}{2 a}$. So $x(t)=\boxed{\frac{1}{2a}(\exp{a*t} - \exp{-a*t})}$.

Answer: \frac{1}{2a}(\exp{a*t} - \exp{-a*t})


Subproblem 0: Find a solution to the differential equation $\ddot{x}+\omega^{2} x=0$ satisfying the initial conditions $x(0)=x_{0}$ and $\dot{x}(0)=\dot{x}_{0}$.

Solution: Suppose \[x(t)=a \cos (\omega t)+b \sin (\omega t)\] $x(0)=a$, therefore $a=x_{0}$. Then \[x^{\prime}(0)=-a \omega \sin 0+b \omega \cos 0=b \omega=\dot{x}_{0}\] Then $b=\dot{x}_{0} / \omega$. The solution is then $x=\boxed{x_{0} \cos (\omega t)+$ $\dot{x}_{0} \sin (\omega t) / \omega}$.

Answer: x_{0} \cos (\omega t)+$ $\dot{x}_{0} \sin (\omega t) / \omega


Subproblem 0: Find the complex number $a+b i$ with the smallest possible positive $b$ such that $e^{a+b i}=1+\sqrt{3} i$.

Solution: $1+\sqrt{3} i$ has modulus 2 and argument $\pi / 3+2 k \pi$ for all integers k, so $1+\sqrt{3} i$ can be expressed as a complex exponential of the form $2 e^{i(\pi / 3+2 k \pi)}$. Taking logs gives us the equation $a+b i=\ln 2+i(\pi / 3+2 k \pi)$. The smallest positive value of $b$ is $\pi / 3$. Thus we have $\boxed{\ln 2 + i\pi / 3}$

Answer: \ln 2 + i\pi / 3


Subproblem 0: Find the general solution of the differential equation $\dot{x}+2 x=e^{t}$, using $c$ for the arbitrary constant of integration which will occur.

Solution: We can use integrating factors to get $(u x)^{\prime}=u e^{t}$ for $u=e^{2 t}$. Integrating yields $e^{2 t} x=e^{3 t} / 3+c$, or $x=\boxed{\frac{e^{t}} {3}+c e^{-2 t}}$.

Final answer: The final answer is \frac{e^{t}} {3}+c e^{-2 t}. I hope it is correct.

Subproblem 1: Find a solution of the differential equation $\dot{x}+2 x=e^{t}$ of the form $w e^{t}$, where $w$ is a constant (which you should find).

Solution: When $c=0, x=\boxed{e^{t} / 3}$ is the solution of the required form.

Answer: e^{t} / 3


Subproblem 0: For $\omega \geq 0$, find $A$ such that $A \cos (\omega t)$ is a solution of $\ddot{x}+4 x=\cos (\omega t)$.

Solution: If $x=A \cos (\omega t)$, then taking derivatives gives us $\ddot{x}=-\omega^{2} A \cos (\omega t)$, and $\ddot{x}+4 x=\left(4-\omega^{2}\right) A \cos (\omega t)$. Then $A=\boxed{\frac{1}{4-\omega^{2}}}$.

Final answer: The final answer is \frac{1}{4-\omega^{2}}. I hope it is correct.

Subproblem 1: For what value of $\omega$ does resonance occur?

Solution: Resonance occurs when $\omega=\boxed{2}$.

Answer: 2


Subproblem 0: Find a purely sinusoidal solution of $\frac{d^{4} x}{d t^{4}}-x=\cos (2 t)$.

Solution: We choose an exponential input function whose real part is $\cos (2 t)$, namely $e^{2 i t}$. Since $p(s)=s^{4}-1$ and $p(2 i)=15 \neq 0$, the exponential response formula yields the solution $\frac{e^{2 i t}}{15}$. A sinusoidal solution to the original equation is given by the real part: $\boxed{\frac{\cos (2 t)}{15}}$.

Final answer: The final answer is \frac{\cos (2 t)}{15}. I hope it is correct.

Subproblem 1: Find the general solution to $\frac{d^{4} x}{d t^{4}}-x=\cos (2 t)$, denoting constants as $C_{1}, C_{2}, C_{3}, C_{4}$.

Solution:
To get the general solution, we take the sum of the general solution to the homogeneous equation and the particular solution to the original equation. The homogeneous equation is $\frac{d^{4} x}{d t^{4}}-x=0$. The characteristic polynomial $p(s)=s^{4}-1$ has 4 roots: $\pm 1, \pm i$. So the general solution to $\frac{d^{4} x}{d t^{4}}-x=0$ is given by $C_{1} e^{t}+C_{2} e^{-t}+C_{3} \cos (t)+C_{4} \sin (t)$ for arbitrary real constants $C_{1}, C_{2}, C_{3}, C_{4}$.
The solution to the equation is $\boxed{\frac{\cos (2 t)}{15}+C_{1} e^{t}+C_{2} e^{-t}+C_{3} \cos (t)+C_{4} \sin (t)}$.

Answer: \frac{\cos (2 t)}{15}+C_{1} e^{t}+C_{2} e^{-t}+C_{3} \cos (t)+C_{4} \sin (t)


Subproblem 0: For $\omega \geq 0$, find $A$ such that $A \cos (\omega t)$ is a solution of $\ddot{x}+4 x=\cos (\omega t)$.

Solution: If $x=A \cos (\omega t)$, then taking derivatives gives us $\ddot{x}=-\omega^{2} A \cos (\omega t)$, and $\ddot{x}+4 x=\left(4-\omega^{2}\right) A \cos (\omega t)$. Then $A=\boxed{\frac{1}{4-\omega^{2}}}$.

Answer: \frac{1}{4-\omega^{2}}


Subproblem 0: Find a solution to $\dot{x}+2 x=\cos (2 t)$ in the form $k_0\left[f(k_1t) + g(k_2t)\right]$, where $f, g$ are trigonometric functions.  Do not include homogeneous solutions to this ODE in your solution.

Solution: $\cos (2 t)=\operatorname{Re}\left(e^{2 i t}\right)$, so $x$ can be the real part of any solution $z$ to $\dot{z}+2 z=e^{2 i t}$. One solution is given by $x=\operatorname{Re}\left(e^{2 i t} /(2+2 i)\right)=\boxed{\frac{\cos (2 t)+\sin (2 t)}{4}}$.

Answer: \frac{\cos (2 t)+\sin (2 t)}{4}


Preamble: The following subproblems refer to the differential equation. $\ddot{x}+4 x=\sin (3 t)$

Subproblem 0: Find $A$ so that $A \sin (3 t)$ is a solution of $\ddot{x}+4 x=\sin (3 t)$.

Solution: We can find this by brute force. If $x=A \sin (3 t)$, then $\ddot{x}=-9 A \sin (3 t)$, so $\ddot{x}+4 x=-5 A \sin (3 t)$. Therefore, when $A=\boxed{-0.2}, x_{p}(t)=-\sin (3 t) / 5$ is a solution of the given equation.

Answer: -0.2


Subproblem 0: Find the general solution of the differential equation $y^{\prime}=x-2 y$ analytically using integrating factors, solving for $y$. Note that a function $u(t)$ such that $u \dot{x}+u p x=\frac{d}{d t}(u x)$ is an integrating factor. Additionally, note that a general solution to a differential equation has the form $x=x_{p}+c x_{h}$ where $x_{h}$ is a nonzero solution of the homogeneous equation $\dot{x}+p x=0$.

Solution: In standard form, $y^{\prime}+2 y=x$, so $u=C e^{2 x}$. Then $y=u^{-1} \int u x d x=e^{-2 x} \int x e^{2 x} d x$. Integrating by parts yields $\int x e^{2 x} d x=$ $\frac{x}{2} e^{2 x}-\frac{1}{2} \int e^{2 x} d x=\frac{x}{2} e^{2 x}-\frac{1}{4} e^{2 x}+c$. Therefore, $y=\boxed{x / 2-1 / 4+c e^{-2 x}}$.

Answer: x / 2-1 / 4+c e^{-2 x}


Subproblem 0: Find a purely exponential solution of $\frac{d^{4} x}{d t^{4}}-x=e^{-2 t}$.

Solution: The characteristic polynomial of the homogeneous equation is given by $p(s)=$ $s^{4}-1$. Since $p(-2)=15 \neq 0$, the exponential response formula gives the solution $\frac{e^{-2 t}}{p(-2)}=\boxed{\frac{e^{-2 t}}{15}}$.

Final answer: The final answer is \frac{e^{-2 t}}{15}. I hope it is correct.

Subproblem 1: Find the general solution to $\frac{d^{4} x}{d t^{4}}-x=e^{-2 t}$, denoting constants as $C_{1}, C_{2}, C_{3}, C_{4}$.

Solution:
To get the general solution, we take the sum of the general solution to the homogeneous equation and the particular solution to the original equation. The homogeneous equation is $\frac{d^{4} x}{d t^{4}}-x=0$. The characteristic polynomial $p(s)=s^{4}-1$ has 4 roots: $\pm 1, \pm i$. So the general solution to $\frac{d^{4} x}{d t^{4}}-x=0$ is given by $C_{1} e^{t}+C_{2} e^{-t}+C_{3} \cos (t)+C_{4} \sin (t)$ for arbitrary real constants $C_{1}, C_{2}, C_{3}, C_{4}$.
Therefore, the general solution to the equation is $\boxed{\frac{e^{-2 t}}{15}+C_{1} e^{t}+C_{2} e^{-t}+ C_{3} \cos (t)+C_{4} \sin (t)}$.

Answer: \frac{e^{-2 t}}{15}+C_{1} e^{t}+C_{2} e^{-t}+ C_{3} \cos (t)+C_{4} \sin (t)


Preamble: Consider the differential equation $\ddot{x}+\omega^{2} x=0$. \\

Subproblem 0: A differential equation $m \ddot{x}+b \dot{x}+k x=0$ (where $m, b$, and $k$ are real constants, and $m \neq 0$ ) has corresponding characteristic polynomial $p(s)=m s^{2}+b s+k$.\\
What is the characteristic polynomial $p(s)$ of $\ddot{x}+\omega^{2} x=0$?

Solution: The characteristic polynomial $p(s)$ is $p(s)=\boxed{s^{2}+\omega^{2}}$.

Answer: s^{2}+\omega^{2}


Subproblem 0: Rewrite the function $\cos (\pi t)-\sqrt{3} \sin (\pi t)$ in the form $A \cos (\omega t-\phi)$. It may help to begin by drawing a right triangle with sides $a$ and $b$.

Solution: The right triangle has hypotenuse of length $\sqrt{1^{2}+(-\sqrt{3})^{2}}=2$. The circular frequency of both summands is $\pi$, so $\omega=\pi$. The argument of the hypotenuse is $-\pi / 3$, so $f(t)=\boxed{2 \cos (\pi t+\pi / 3)}$.

Answer: 2 \cos (\pi t+\pi / 3)


Preamble: The following subproblems refer to the damped sinusoid $x(t)=A e^{-a t} \cos (\omega t)$.

Subproblem 0: What is the spacing between successive maxima of $x(t)$? Assume that $\omega \neq 0$.

Solution: The extrema of $x(t)=A e^{-a t} \cos (\omega t)$ occur when $\dot{x}(t)=0$, i.e., $-a \cos (\omega t)=\omega \sin (\omega t)$. When $\omega \neq 0$, the extrema are achieved at $t$ where $\tan (\omega t)=-a / \omega$. Since minima and maxima of $x(t)$ are alternating, the maxima occur at every other $t \operatorname{such}$ that $\tan (\omega t)=-a / \omega$. If $t_{0}$ and $t_{1}$ are successive maxima, then $t_{1}-t_{0}=$ twice the period of $\tan (\omega t)=\boxed{2 \pi / \omega}$,

Answer: 2 \pi / \omega


Preamble: The following subproblems refer to a spring/mass/dashpot system driven through the spring modeled by the equation $m \ddot{x}+b \dot{x}+k x=k y$. Here $x$ measures the position of the mass, $y$ measures the position of the other end of the spring, and $x=y$ when the spring is relaxed.

Subproblem 0: In this system, regard $y(t)$ as the input signal and $x(t)$ as the system response. Take $m=1, b=3, k=4, y(t)=A \cos t$. Replace the input signal by a complex exponential $y_{c x}(t)$ of which it is the real part, and compute the exponential ("steady state") system response $z_p(t)$; leave your answer in terms of complex exponentials, i.e. do not take the real part.

Solution: The equation is $\ddot{x}+3 \dot{x}+4 x=4 A \cos t$, with the characteristic polynomial $p(s)=s^{2}+3 s+4$. The complex exponential corresponding to the input signal is $y_{c x}=A e^{i t}$ and $p(i)=3+3 i \neq 0$. By the Exponential Response Formula, $z_{p}=\frac{4 A}{p(i)} e^{i t}=\boxed{\frac{4 A}{3+3 i} e^{i t}}$.

Answer: \frac{4 A}{3+3 i} e^{i t}


Preamble: The following subproblems refer to a circuit with the following parameters. Denote by $I(t)$ the current (where the positive direction is, say, clockwise) in the circuit and by $V(t)$ the voltage increase across the voltage source, at time $t$. Denote by $R$ the resistance of the resistor and $C$ the capacitance of the capacitor (in units which we will not specify)-both positive numbers. Then
\[
R \dot{I}+\frac{1}{C} I=\dot{V}
\]

Subproblem 0: Suppose that $V$ is constant, $V(t)=V_{0}$. Solve for $I(t)$, with initial condition $I(0)$.

Solution:
When $V$ is constant, the equation becomes $R \dot{I}+\frac{1}{C} I=0$, which is separable. Solving gives us
\[
I(t)=\boxed{I(0) e^{-\frac{t}{R C}}
}\].

Answer: I(0) e^{-\frac{t}{R C}}


Subproblem 0: Find the general (complex-valued) solution of the differential equation $\dot{z}+2 z=e^{2 i t}$, using $C$ to stand for any complex-valued integration constants which may arise.

Solution: Using integrating factors, we get $e^{2 t} z=e^{(2+2 i) t} /(2+2 i)+C$, or $z=\boxed{\frac{e^{2 i t}}{(2+2 i)}+C e^{-2 t}}$, where $C$ is any complex number.

Final answer: The final answer is \frac{e^{2 i t}}{(2+2 i)}+C e^{-2 t}. I hope it is correct.

Subproblem 1: Find a solution of the differential equation $\dot{z}+2 z=e^{2 i t}$ in the form $w e^{t}$, where $w$ is a constant (which you should find).

Solution: When $C=0, z=\boxed{\frac{e^{2 i t}}{(2+2 i)}}$.

Answer: \frac{e^{2 i t}}{(2+2 i)}


Preamble: The following subproblems consider a second order mass/spring/dashpot system driven by a force $F_{\text {ext }}$ acting directly on the mass: $m \ddot{x}+b \dot{x}+k x=F_{\text {ext }}$. So the input signal is $F_{\text {ext }}$ and the system response is $x$. We're interested in sinusoidal input signal, $F_{\text {ext }}(t)=A \cos (\omega t)$, and in the steady state, sinusoidal system response, $x_{p}(t)=g A \cos (\omega t-\phi)$. Here $g$ is the gain of the system and $\phi$ is the phase lag. Both depend upon $\omega$, and we will consider how that is the case. \\
Take $A=1$, so the amplitude of the system response equals the gain, and take $m=1, b=\frac{1}{4}$, and $k=2$.\\

Subproblem 0: Compute the complex gain $H(\omega)$ of this system. (This means: make the complex replacement $F_{\mathrm{cx}}=e^{i \omega t}$, and express the exponential system response $z_{p}$ as a complex multiple of $F_{\mathrm{cx}}, i.e. z_{p}=H(\omega) F_{\mathrm{cx}}$).

Solution: Set $F_{\mathrm{cx}}=e^{i \omega t}$. The complex replacement of the equation is $\ddot{z}+\frac{1}{4} \dot{z}+2 z=e^{i \omega t}$, with the characteristic polynomial $p(s)=s^{2}+\frac{1}{4} s+2.$ Given that $p(i \omega)=-\omega^{2}+\frac{\omega}{4} i+2 \neq 0$, so by the exponential response formula, $z_{p}=e^{i \omega t} / p(i \omega)=F_{\mathrm{cx}} / p(i \omega)$, and $H(\omega)=z_{p} / F_{\mathrm{cx}}=1 / p(i \omega)=$ $\frac{2-\omega^{2}-\omega i / 4}{\left(2-\omega^{2}\right)^{2}+(\omega / 4)^{2}}=\boxed{\frac{2-\omega^{2}-\omega i / 4}{\omega^{4}-\frac{63}{16} \omega^{2}+4}}$.

Answer: \frac{2-\omega^{2}-\omega i / 4}{\omega^{4}-\frac{63}{16} \omega^{2}+4}


Preamble: The following subproblems refer to the following "mixing problem": A tank holds $V$ liters of salt water. Suppose that a saline solution with concentration of $c \mathrm{gm} /$ liter is added at the rate of $r$ liters/minute. A mixer keeps the salt essentially uniformly distributed in the tank. A pipe lets solution out of the tank at the same rate of $r$ liters/minute.

Subproblem 0: Write down the differential equation for the amount of salt in the tank in standard linear form. [Not the concentration!] Use the notation $x(t)$ for the number of grams of salt in the tank at time $t$.

Solution:
The concentration of salt at any given time is $x(t) / V \mathrm{gm} /$ liter, so for small $\Delta t$, we lose $r x(t) \Delta t / V$ gm from the exit pipe, and we gain $r c \Delta t \mathrm{gm}$ from the input pipe. The equation is $x^{\prime}(t)=r c-\frac{r x(t)}{V}$, and in standard linear form, it is
$\boxed{x^{\prime}+\frac{r}{V} x-r c=0}$.

Answer: x^{\prime}+\frac{r}{V} x-r c=0


Subproblem 0: Find the polynomial solution of $\ddot{x}-x=t^{2}+t+1$, solving for $x(t)$.

Solution: Since the constant term of the right-hand side is nonzero, the undetermined coefficients theorem asserts that there is a unique quadratic polynomial $a t^{2}+b t+c$ satisfying this equation. Substituting this form into the left side of the equation, we see that $a=-1,-b=1$, and $2 a-c=1$, so $b=-1$ and $c=-3$. Finally, $x(t) = \boxed{-t^2 - t - 3}$

Answer: -t^2 - t - 3


Preamble: In the following problems, take $a = \ln 2$ and $b = \pi / 3$.

Subproblem 0: Given $a = \ln 2$ and $b = \pi / 3$, rewrite $e^{a+b i}$ in the form $x + yi$, where $x, y$ are real numbers.

Solution: Using Euler's formula, we find that the answer is $\boxed{1+\sqrt{3} i}$.

Final answer: The final answer is 1+\sqrt{3} i. I hope it is correct.

Subproblem 1: Given $a = \ln 2$ and $b = \pi / 3$, rewrite $e^{2(a+b i)}$ in the form $x + yi$, where $x, y$ are real numbers.

Solution: $e^{n(a+b i)}=(1+\sqrt{3} i)^{n}$, so the answer is $\boxed{-2+2 \sqrt{3} i}$.

Final answer: The final answer is -2+2 \sqrt{3} i. I hope it is correct.

Subproblem 2: Rewrite $e^{3(a+b i)}$ in the form $x + yi$, where $x, y$ are real numbers.

Solution: $e^{n(a+b i)}=(1+\sqrt{3} i)^{n}$, so the answer is $\boxed{-8}$.

Answer: -8


Subproblem 0: Find a purely sinusoidal solution of $\frac{d^{4} x}{d t^{4}}-x=\cos (2 t)$.

Solution: We choose an exponential input function whose real part is $\cos (2 t)$, namely $e^{2 i t}$. Since $p(s)=s^{4}-1$ and $p(2 i)=15 \neq 0$, the exponential response formula yields the solution $\frac{e^{2 i t}}{15}$. A sinusoidal solution to the original equation is given by the real part: $\boxed{\frac{\cos (2 t)}{15}}$.

Answer: \frac{\cos (2 t)}{15}


Preamble: In the following problems, take $a = \ln 2$ and $b = \pi / 3$.

Subproblem 0: Given $a = \ln 2$ and $b = \pi / 3$, rewrite $e^{a+b i}$ in the form $x + yi$, where $x, y$ are real numbers.

Solution: Using Euler's formula, we find that the answer is $\boxed{1+\sqrt{3} i}$.

Final answer: The final answer is 1+\sqrt{3} i. I hope it is correct.

Subproblem 1: Given $a = \ln 2$ and $b = \pi / 3$, rewrite $e^{2(a+b i)}$ in the form $x + yi$, where $x, y$ are real numbers.

Solution: $e^{n(a+b i)}=(1+\sqrt{3} i)^{n}$, so the answer is $\boxed{-2+2 \sqrt{3} i}$.

Answer: -2+2 \sqrt{3} i


Subproblem 0: Find a solution of $\ddot{x}+4 x=\cos (2 t)$, solving for $x(t)$, by using the ERF on a complex replacement. The ERF (Exponential Response Formula) states that a solution to $p(D) x=A e^{r t}$ is given by $x_{p}=A \frac{e^{r t}}{p(r)}$, as long as $\left.p (r\right) \neq 0$). The ERF with resonance assumes that $p(r)=0$ and states that a solution to $p(D) x=A e^{r t}$ is given by $x_{p}=A \frac{t e^{r t}}{p^{\prime}(r)}$, as long as $\left.p^{\prime} ( r\right) \neq 0$.

Solution: The complex replacement of the equation is $\ddot{z}+4 z=e^{2 i t}$, with the characteristic polynomial $p(s)=s^{2}+4$. Because $p(2 i)=0$ and $p^{\prime}(2 i)=4 i \neq 0$, we need to use the Resonant ERF, which leads to $z_{p}=\frac{t e^{2 i t}}{4 i}$. A solution of the original equation is given by $x_{p}=\operatorname{Re}\left(z_{p}\right)=\boxed{\frac{t}{4} \sin (2 t)}$.

Answer: \frac{t}{4} \sin (2 t)


Subproblem 0: Given the ordinary differential equation $\ddot{x}-a^{2} x=0$, where $a$ is a nonzero real-valued constant, find a solution $x(t)$ to this equation such that $x(0) = 1$ and $\dot{x}(0)=0$.

Solution: First, notice that both $x(t)=e^{a t}$ and $x(t)=e^{-a t}$ are solutions to $\ddot{x}-a^{2} x=0$. Then for any constants $c_{1}$ and $c_{2}$, $x(t)=c_{1} e^{a t}+c_{2} e^{-a t}$ are also solutions to $\ddot{x}-a^{2} x=0$. Moreover, $x(0)=c_{1}+c_{2}$, and $\dot{x}(0)=a\left(c_{1}-c_{2}\right)$. Assuming $a \neq 0$, to satisfy the given conditions, we need $c_{1}+c_{2}=1$ and $a\left(c_{1}-c_{2}\right)=0$, which implies $c_{1}=c_{2}=1 / 2$. So $x(t)=\boxed{\frac{1}{2}(\exp{a*t} + \exp{-a*t})}$.

Answer: \frac{1}{2}(\exp{a*t} + \exp{-a*t})


Subproblem 0: Find the general solution of the differential equation $\dot{x}+2 x=e^{t}$, using $c$ for the arbitrary constant of integration which will occur.

Solution: We can use integrating factors to get $(u x)^{\prime}=u e^{t}$ for $u=e^{2 t}$. Integrating yields $e^{2 t} x=e^{3 t} / 3+c$, or $x=\boxed{\frac{e^{t}} {3}+c e^{-2 t}}$.

Answer: \frac{e^{t}} {3}+c e^{-2 t}


Subproblem 0: Find a solution of $\ddot{x}+3 \dot{x}+2 x=t e^{-t}$ in the form $x(t)=u(t) e^{-t}$ for some function $u(t)$.  Use $C$ for an arbitrary constant, should it arise.

Solution: $\dot{x}=\dot{u} e^{-t}-u e^{-t}$ and $\ddot{x}=\ddot{u} e^{-t}-2 \dot{u} e^{-t}+u e^{-t}$. Plugging into the equation leads to $e^{-t}(\ddot{u}+\dot{u})=t e^{-t}$. Cancelling off $e^{-t}$ from both sides, we get $\ddot{u}+\dot{u}=t$. To solve this equation for $u$, we use the undetermined coefficient method. However, the corresponding characteristic polynomial $p(s)=s^{2}+s$ has zero as its constant term. So set $w=\dot{u}$, then the equation can be rewritten as $\dot{w}+w=t$. This can be solved and one solution is $w=t-1$, and hence $\dot{u}=t-1$, and one solution for $u$ is $u=\frac{t^{2}}{2}-t+C$. Back to the original equation, one solution is given by $x=\boxed{\left(\frac{t^{2}}{2}-t+C\right) e^{-t}}$

Answer: \left(\frac{t^{2}}{2}-t+C\right) e^{-t}


Subproblem 0: If the complex number $z$ is given by $z = 1+\sqrt{3} i$, what is the real part of $z^2$?

Solution: $z^{2}$ has argument $2 \pi / 3$ and radius 4 , so by Euler's formula, $z^{2}=4 e^{i 2 \pi / 3}=-2+2 \sqrt{3} i$. Thus $a = -2, b = 2\sqrt 3$, so our answer is \boxed{-2}.

Answer: -2


Subproblem 0: Find a purely exponential solution of $\frac{d^{4} x}{d t^{4}}-x=e^{-2 t}$.

Solution: The characteristic polynomial of the homogeneous equation is given by $p(s)=$ $s^{4}-1$. Since $p(-2)=15 \neq 0$, the exponential response formula gives the solution $\frac{e^{-2 t}}{p(-2)}=\boxed{\frac{e^{-2 t}}{15}}$.

Answer: \frac{e^{-2 t}}{15}


Preamble: The following subproblems refer to the exponential function $e^{-t / 2} \cos (3 t)$, which we will assume is a solution of the differential equation $m \ddot{x}+b \dot{x}+k x=0$.

Subproblem 0: What is $b$ in terms of $m$? Write $b$ as a constant times a function of $m$.

Solution: We can write $e^{-t / 2} \cos (3 t)=\operatorname{Re} e^{(-1 / 2 \pm 3 i) t}$, so $p(s)=m s^{2}+b s+k$ has solutions $-\frac{1}{2} \pm 3 i$. This means $p(s)=m(s+1 / 2-3 i)(s+1 / 2+3 i)=m\left(s^{2}+s+\frac{37}{4}\right)$. Then $b=\boxed{m}$,

Answer: m


Preamble: The following subproblems refer to the differential equation. $\ddot{x}+4 x=\sin (3 t)$

Subproblem 0: Find $A$ so that $A \sin (3 t)$ is a solution of $\ddot{x}+4 x=\sin (3 t)$.

Solution: We can find this by brute force. If $x=A \sin (3 t)$, then $\ddot{x}=-9 A \sin (3 t)$, so $\ddot{x}+4 x=-5 A \sin (3 t)$. Therefore, when $A=\boxed{-0.2}, x_{p}(t)=-\sin (3 t) / 5$ is a solution of the given equation.

Final answer: The final answer is -0.2. I hope it is correct.

Subproblem 1: What is the general solution, in the form $f_0(t) + C_1f_1(t) + C_2f_2(t)$, where $C_1, C_2$ denote arbitrary constants?

Solution: To find the general solution, we add to $x_{p}$ the general solution to the homogeneous equation $\ddot{x}+4 x=0$. The characteristic polynomial is $p(s)=s^{2}+4$, with roots $\pm 2 i$, so the general solution to $\ddot{x}+4 x=0$ is $C_{1} \sin (2 t)+C_{2} \cos (2 t)$. Therefore, the general solution to $\ddot{x}+4 x=\sin (3 t)$ is given by $\boxed{-\sin (3 t) / 5+ C_{1} \sin (2 t)+C_{2} \cos (2 t)}$.

Answer: -\sin (3 t) / 5+ C_{1} \sin (2 t)+C_{2} \cos (2 t)


Subproblem 0: What is the smallest possible positive $k$ such that all functions $x(t)=A \cos (\omega t-\phi)$---where $\phi$ is an odd multiple of $k$---satisfy $x(0)=0$? \\

Solution: $x(0)=A \cos \phi$. When $A=0$, then $x(t)=0$ for every $t$; when $A \neq 0$, $x(0)=0$ implies $\cos \phi=0$, and hence $\phi$ can be any odd multiple of $\pi / 2$, i.e., $\phi=\pm \pi / 2, \pm 3 \pi / 2, \pm 5 \pi / 2, \ldots$ this means $k=\boxed{\frac{\pi}{2}}$

Answer: \frac{\pi}{2}


Preamble: The following subproblems refer to the differential equation $\ddot{x}+b \dot{x}+x=0$.\\

Subproblem 0: What is the characteristic polynomial $p(s)$ of $\ddot{x}+b \dot{x}+x=0$?

Solution: The characteristic polynomial is $p(s)=\boxed{s^{2}+b s+1}$.

Answer: s^{2}+b s+1


Preamble: The following subproblems refer to the exponential function $e^{-t / 2} \cos (3 t)$, which we will assume is a solution of the differential equation $m \ddot{x}+b \dot{x}+k x=0$.

Subproblem 0: What is $b$ in terms of $m$? Write $b$ as a constant times a function of $m$.

Solution: We can write $e^{-t / 2} \cos (3 t)=\operatorname{Re} e^{(-1 / 2 \pm 3 i) t}$, so $p(s)=m s^{2}+b s+k$ has solutions $-\frac{1}{2} \pm 3 i$. This means $p(s)=m(s+1 / 2-3 i)(s+1 / 2+3 i)=m\left(s^{2}+s+\frac{37}{4}\right)$. Then $b=\boxed{m}$,

Final answer: The final answer is m. I hope it is correct.

Subproblem 1: What is $k$ in terms of $m$? Write $k$ as a constant times a function of $m$.

Solution: Having found that $p(s)=m(s+1 / 2-3 i)(s+1 / 2+3 i)=m\left(s^{2}+s+\frac{37}{4}\right)$ in the previous subproblem, $k=\boxed{\frac{37}{4} m}$.

Answer: \frac{37}{4} m


Preamble: In the following problems, take $a = \ln 2$ and $b = \pi / 3$.

Subproblem 0: Given $a = \ln 2$ and $b = \pi / 3$, rewrite $e^{a+b i}$ in the form $x + yi$, where $x, y$ are real numbers.

Solution: Using Euler's formula, we find that the answer is $\boxed{1+\sqrt{3} i}$.

Final answer: The final answer is 1+\sqrt{3} i. I hope it is correct.

Subproblem 1: Given $a = \ln 2$ and $b = \pi / 3$, rewrite $e^{2(a+b i)}$ in the form $x + yi$, where $x, y$ are real numbers.

Solution: $e^{n(a+b i)}=(1+\sqrt{3} i)^{n}$, so the answer is $\boxed{-2+2 \sqrt{3} i}$.

Final answer: The final answer is -2+2 \sqrt{3} i. I hope it is correct.

Subproblem 2: Rewrite $e^{3(a+b i)}$ in the form $x + yi$, where $x, y$ are real numbers.

Solution: $e^{n(a+b i)}=(1+\sqrt{3} i)^{n}$, so the answer is $\boxed{-8}$.

Final answer: The final answer is -8. I hope it is correct.

Subproblem 3: Rewrite $e^{4(a+b i)}$ in the form $x + yi$, where $x, y$ are real numbers.

Solution: $e^{n(a+b i)}=(1+\sqrt{3} i)^{n}$, so the answer is $\boxed{-8-8 \sqrt{3} i}$.

Answer: -8-8 \sqrt{3} i


Subproblem 0: Rewrite the function $\operatorname{Re} \frac{e^{i t}}{2+2 i}$ in the form $A \cos (\omega t-\phi)$. It may help to begin by drawing a right triangle with sides $a$ and $b$.

Solution: $e^{i t}=\cos (t)+i \sin (t)$, and $\frac{1}{2+2 i}=\frac{1-i}{4}$. the real part is then $\frac{1}{4} \cos (t)+$ $\frac{1}{4} \sin (t)$. The right triangle here has hypotenuse $\frac{\sqrt{2}}{4}$ and argument $\pi / 4$, so $f(t)=\boxed{\frac{\sqrt{2}}{4} \cos (t-\pi / 4)}$.

Answer: \frac{\sqrt{2}}{4} \cos (t-\pi / 4)


Preamble: The following subproblems refer to the differential equation $\ddot{x}+b \dot{x}+x=0$.\\

Subproblem 0: What is the characteristic polynomial $p(s)$ of $\ddot{x}+b \dot{x}+x=0$?

Solution: The characteristic polynomial is $p(s)=\boxed{s^{2}+b s+1}$.

Final answer: The final answer is s^{2}+b s+1. I hope it is correct.

Subproblem 1: For what value of $b$ does $\ddot{x}+b \dot{x}+x=0$ exhibit critical damping?

Solution: To exhibit critical damping, the characteristic polynomial $s^{2}+b s+1$ must be a square, i.e., $(s-k)^{2}$ for some $k$. Multiplying and comparing yields $-2 k=b$ and $k^{2}=1$. Therefore, $b$ could be either one of $=-2, 2$. When $b=-2, e^{t}$ is a solution, and it exhibits exponential growth instead of damping, so we reject that value of $b$. Therefore, the value of $b$ for which $\ddot{x}+b \dot{x}+x=0$ exhibits critical damping is $b=\boxed{2}$

Answer: 2


Subproblem 0: Find the general (complex-valued) solution of the differential equation $\dot{z}+2 z=e^{2 i t}$, using $C$ to stand for any complex-valued integration constants which may arise.

Solution:
Using integrating factors, we get $e^{2 t} z=e^{(2+2 i) t} /(2+2 i)+C$, or $z=\boxed{\frac{e^{2 i t}}{(2+2 i)}+C e^{-2 t}}$, where $C$ is any complex number.

\section{Dynamics and Control (2.003 Spring 2005)}

Answer: \frac{e^{2 i t}}{(2+2 i)}+C e^{-2 t}


Preamble: Consider the first-order system
\[
\tau \dot{y}+y=u
\]
driven with a unit step from zero initial conditions. The input to this system is \(u\) and the output is \(y\).

Subproblem 0: Derive and expression for the settling time \(t_{s}\), where the settling is to within an error \(\pm \Delta\) from the final value of 1.

Solution:
Rise and Settling Times.  We are given the first-order transfer function
\[
H(s)=\frac{1}{\tau s+1}
\]
The response to a unit step with zero initial conditions will be \(y(t)=1-e^{-t / \tau}\). To determine the amount of time it take \(y\) to settle to within \(\Delta\) of its final value, we want to find the time \(t_{s}\) such that \(y\left(t_{s}\right)=1-\Delta\). Thus, we obtain
\[
\begin{aligned}
&\Delta=e^{-t_{s} / \tau} \\
&t_{s}=\boxed{-\tau \ln \Delta}
\end{aligned}
\]

Answer: -\tau \ln \Delta


Preamble: Consider the first-order system
\[
\tau \dot{y}+y=u
\]
driven with a unit step from zero initial conditions. The input to this system is \(u\) and the output is \(y\).

Subproblem 0: Derive and expression for the settling time \(t_{s}\), where the settling is to within an error \(\pm \Delta\) from the final value of 1.

Solution: Rise and Settling Times.  We are given the first-order transfer function
\[
H(s)=\frac{1}{\tau s+1}
\]
The response to a unit step with zero initial conditions will be \(y(t)=1-e^{-t / \tau}\). To determine the amount of time it take \(y\) to settle to within \(\Delta\) of its final value, we want to find the time \(t_{s}\) such that \(y\left(t_{s}\right)=1-\Delta\). Thus, we obtain
\[
\begin{aligned}
&\Delta=e^{-t_{s} / \tau} \\
&t_{s}=\boxed{-\tau \ln \Delta}
\end{aligned}
\]

Final answer: The final answer is -\tau \ln \Delta. I hope it is correct.

Subproblem 1: Derive an expression for the \(10-90 \%\) rise time \(t_{r}\) in terms of $\tau$.

Solution:
The \(10-90 \%\) rise time \(t_{r}\) may be thought of as the difference between the \(90 \%\) settling time \((\Delta=0.1)\) and the \(10 \%\) settling time \((\Delta=0.9)\).
\[
t_{r}=t_{\Delta=0.1}-t_{\Delta=0.9}
\]
Therefore, we find \(t_{r}=\boxed{2.2 \tau}\).

Answer: 2.2 \tau


Preamble: For each of the functions $y(t)$, find the Laplace Transform $Y(s)$ :

Subproblem 0: $y(t)=e^{-a t}$

Solution:
This function is one of the most widely used in dynamic systems, so we memorize its transform!
\[
Y(s)=\boxed{\frac{1}{s+a}}
\]

Answer: \frac{1}{s+a}


Preamble: For each Laplace Transform \(Y(s)\), find the function \(y(t)\) :

Subproblem 0: \[
Y(s)=\boxed{\frac{1}{(s+a)(s+b)}}
\]

Solution: We can simplify with partial fractions:
\[
Y(s)=\frac{1}{(s+a)(s+b)}=\frac{C}{s+a}+\frac{D}{s+b}
\]
find the constants \(C\) and \(D\) by setting \(s=-a\) and \(s=-b\)
\[
\begin{aligned}
\frac{1}{(s+a)(s+b)} &=\frac{C}{s+a}+\frac{D}{s+b} \\
1 &=C(s+b)+D(s+a) \\
C &=\frac{1}{b-a} \\
D &=\frac{1}{a-b}
\end{aligned}
\]
therefore
\[
Y(s)=\frac{1}{b-a} \frac{1}{s+a}-\frac{1}{b-a} \frac{1}{s+b}
\]
By looking up the inverse Laplace Transform of \(\frac{1}{s+b}\), we find the total solution \(y(t)\)
\[
y(t)=\boxed{\frac{1}{b-a}\left(e^{-a t}-e^{-b t}\right)}
\]

Final answer: The final answer is \frac{1}{b-a}\left(e^{-a t}-e^{-b t}\right). I hope it is correct.

Subproblem 1: \[
Y(s)=\frac{s}{\frac{s^{2}}{\omega_{n}^{2}}+\frac{2 \zeta}{\omega_{n}} s+1}
\]
You may assume that $\zeta < 1$.

Solution:
First, note that the transform is
\[
\begin{aligned}
Y(s) &=\frac{s}{\frac{s^{2}}{\omega_{n}^{2}}+\frac{2 \zeta}{\omega_{n}} s+1} \\
&=s \cdot \frac{\omega_{n}^{2}}{s^{2}+2 \zeta \omega_{n} s+\omega_{n}^{2}}
\end{aligned}
\]
We will solve this problem using the property
\[
\frac{d f}{d t}=s F(s)-f(0)
\]
therefore
\[
\begin{aligned}
y(t) &=\frac{d}{d t}\left(\frac{\omega_{n}}{\sqrt{1-\zeta^{2}}} e^{-\zeta \omega_{n} t} \sin \left(\omega_{n} \sqrt{1-\zeta^{2}} t\right)\right) \\
&=\boxed{\omega_{n}^{2} e^{-\zeta \omega_{n} t} \cos \left(\omega_{n} \sqrt{1-\zeta^{2}} t\right)-\frac{\zeta \omega_{n}^{2}}{\sqrt{1-\zeta^{2}}} e^{-\zeta \omega_{n} t} \sin \left(\omega_{n} \sqrt{1-\zeta^{2}} t\right)}
\end{aligned}
\]
remember that for this form to be correct, \(\zeta\) must be less than 1 .

Answer: \omega_{n}^{2} e^{-\zeta \omega_{n} t} \cos \left(\omega_{n} \sqrt{1-\zeta^{2}} t\right)-\frac{\zeta \omega_{n}^{2}}{\sqrt{1-\zeta^{2}}} e^{-\zeta \omega_{n} t} \sin \left(\omega_{n} \sqrt{1-\zeta^{2}} t\right)


Subproblem 0: A signal \(x(t)\) is given by
\[
x(t)=\left(e^{-t}-e^{-1}\right)\left(u_{s}(t)-u_{s}(t-1)\right)
\]
Calculate its Laplace transform \(X(s)\). Make sure to clearly show the steps in your calculation.

Solution:
Simplify the expression in to a sum of terms,
\[
x(t)=e^{-t} u_{s}(t)-e^{-1} u_{s}(t)-e^{-t} u_{s}(t-1)+e^{-1} u_{s}(t-1)
\]
Now take the Laplace transform of the first, second and fourth terms,
\[
X(s)=\frac{1}{s+1}-\frac{e^{-1}}{s}-\mathcal{L} e^{-t} u_{s}(t-1)+\frac{e^{-1} e^{-s}}{s}
\]
The third term requires some massaging to get it in a form available on the table. The term can be modified into the form of a time delay, by factoring out \(e^{-1}\).
\[
\mathcal{L}\left\{e^{-t} u_{s}(t-1)\right\}=e^{-1} \mathcal{L}\left\{e^{-(t-1)} u_{s}(t-1)\right\}
\]
Now applying the Laplace Transform for a time delay from the table
\[
e^{-1} \mathcal{L}\left\{e^{-(t-1)} u_{s}(t-1)\right\}=\frac{e^{-1} e^{-s}}{s+1}
\]
Substituting this piece back into the expression above gives the solution
\[
X(s)=\boxed{\frac{1}{s+1}-\frac{e^{-1}}{s}-\frac{e^{-1} e^{-s}}{s+1}+\frac{e^{-1} e^{-s}}{s}}
\]

Answer: \frac{1}{s+1}-\frac{e^{-1}}{s}-\frac{e^{-1} e^{-s}}{s+1}+\frac{e^{-1} e^{-s}}{s}


Preamble: You are given an equation of motion of the form:
\[
\dot{y}+5 y=10 u
\]

Subproblem 0: What is the time constant for this system?

Solution: We find the homogenous solution, solving:
\[
\dot{y}+5 y=0
\]
by trying a solution of the form $y=A \cdot e^{s, t}$.
Calculation:
\[
\dot{y}=A \cdot s \cdot e^{s \cdot t} \mid \Rightarrow A \cdot s \cdot e^{s t}+5 A \cdot e^{s t}=0
\]
yields that $s=-5$, meaning the solution is $y=A \cdot e^{-5 \cdot t}=A \cdot e^{-t / \tau}$, meaning $\tau = \boxed{0.2}$.

Final answer: The final answer is 0.2. I hope it is correct.

Subproblem 1: If \(u=10\), what is the final or steady-state value for \(y(t)\)?

Solution: Steady state implies $\dot{y} = 0$, so in the case when $u=10$, we get $y=\boxed{20}$.

Answer: 20


Subproblem 0: A signal \(w(t)\) is defined as
\[
w(t)=u_{s}(t)-u_{s}(t-T)
\]
where \(T\) is a fixed time in seconds and \(u_{s}(t)\) is the unit step. Compute the Laplace transform \(W(s)\) of \(w(t)\). Show your work.

Solution:
The Laplace Transform of \(x(t)\) is defined as
\[
\mathcal{L}[x(t)]=X(s)=\int_{0}^{\infty} x(t) e^{-s t} d t
\]
therefore
\[
\begin{aligned}
W(s) &=\int_{0}^{\infty} e^{-s t} d t-\left(\int_{0}^{T} 0 d t+\int_{T}^{\infty} e^{-s t} d t\right) \\
&=-\left.\frac{1}{s} e^{-s t}\right|_{0} ^{\infty}-\left(0+-\left.\frac{1}{s} e^{-s t}\right|_{T} ^{\infty}\right) \\
&=\boxed{\frac{1}{s}-\frac{1}{s} e^{-s T}}
\end{aligned}
\]

Answer: \frac{1}{s}-\frac{1}{s} e^{-s T}


Preamble: Assume that we apply a unit step in force separately to a mass \(m\), a dashpot \(c\), and a spring \(k\). The mass moves in inertial space. The spring and dashpot have one end connected to inertial space (reference velocity \(=0\) ), and the force is applied to the other end.  Assume zero initial velocity and position for the elements.
Recall that the unit step function \(u_{S}(t)\) is defined as \(u_{S}(t)=0 ; t<0\) and \(u_{S}(t)=1 ; t \geq 0\). We will also find it useful to introduce the unit impulse function \(\delta(t)\) which can be defined via
\[
u_{S}(t)=\int_{-\infty}^{t} \delta(\tau) d \tau
\]
This means that we can also view the unit impulse as the derivative of the unit step:
\[
\delta(t)=\frac{d u_{S}(t)}{d t}
\]

Subproblem 0: Solve for the resulting velocity of the mass.

Solution:
\[
\begin{aligned}
m \ddot{x}_{m} &=u_{s}(t) \\
\dot{x}_{m}=v_{m} &=\int_{-\infty}^{t} \frac{1}{m} u_{s}(t) d t=\boxed{\frac{1}{m} t} \\
\end{aligned}
\]

Answer: \frac{1}{m} t


Preamble: For each of the functions $y(t)$, find the Laplace Transform $Y(s)$ :

Subproblem 0: $y(t)=e^{-a t}$

Solution: This function is one of the most widely used in dynamic systems, so we memorize its transform!
\[
Y(s)=\boxed{\frac{1}{s+a}}
\]

Final answer: The final answer is \frac{1}{s+a}. I hope it is correct.

Subproblem 1: $y(t)=e^{-\sigma t} \sin \omega_{d} t$

Solution: \[
Y(s)=\boxed{\frac{\omega_{d}}{(s+\sigma)^{2}+\omega_{d}^{2}}}
\]

Final answer: The final answer is \frac{\omega_{d}}{(s+\sigma)^{2}+\omega_{d}^{2}}. I hope it is correct.

Subproblem 2: $y(t)=e^{-\sigma t} \cos \omega_{d} t$

Solution:
\[
Y(s)=\boxed{\frac{s+\sigma}{(s+\sigma)^{2}+\omega_{d}^{2}}}
\]

Answer: \frac{s+\sigma}{(s+\sigma)^{2}+\omega_{d}^{2}}


Preamble: For each of the functions $y(t)$, find the Laplace Transform $Y(s)$ :

Subproblem 0: $y(t)=e^{-a t}$

Solution: This function is one of the most widely used in dynamic systems, so we memorize its transform!
\[
Y(s)=\boxed{\frac{1}{s+a}}
\]

Final answer: The final answer is \frac{1}{s+a}. I hope it is correct.

Subproblem 1: $y(t)=e^{-\sigma t} \sin \omega_{d} t$

Solution:
\[
Y(s)=\boxed{\frac{\omega_{d}}{(s+\sigma)^{2}+\omega_{d}^{2}}}
\]

Answer: \frac{\omega_{d}}{(s+\sigma)^{2}+\omega_{d}^{2}}


Preamble: Consider the mass \(m\) sliding horizontally under the influence of the applied force \(f\) and a friction force which can be approximated by a linear friction element with coefficient \(b\).

Subproblem 0: Formulate the state-determined equation of motion for the velocity \(v\) as output and the force \(f\) as input.

Solution:
The equation of motion is
\[
\boxed{m \frac{d v}{d t}+b v=f} \quad \text { or } \quad \frac{d v}{d t}=-\frac{b}{m} v+\frac{1}{m} f
\]

Answer: m \frac{d v}{d t}+b v=f


Preamble: Consider the rotor with moment of inertia \(I\) rotating under the influence of an applied torque \(T\) and the frictional torques from two bearings, each of which can be approximated by a linear frictional element with coefficient \(B\).

Subproblem 0: Formulate the state-determined equation of motion for the angular velocity $\omega$ as output and the torque $T$ as input.

Solution: The equation of motion is
\[
\boxed{I \frac{d \omega}{d t}+2 B \omega=T} \quad \text { or } \quad \frac{d \omega}{d t}=-\frac{2 B}{I} \omega+\frac{1}{I} T
\]

Final answer: The final answer is I \frac{d \omega}{d t}+2 B \omega=T. I hope it is correct.

Subproblem 1: Consider the case where:
\[
\begin{aligned}
I &=0.001 \mathrm{~kg}-\mathrm{m}^{2} \\
B &=0.005 \mathrm{~N}-\mathrm{m} / \mathrm{r} / \mathrm{s}
\end{aligned}
\]
What is the steady-state velocity \(\omega_{s s}\), in radians per second, when the input is a constant torque of 10 Newton-meters?

Solution:
The steady-state angular velocity, when \(T=10\) Newton-meters, and \(I=0.001 \mathrm{~kg}-\mathrm{m}^{2}\), and \(B=0.005 \mathrm{~N}-\mathrm{m} / \mathrm{r} / \mathrm{s}\) is
\[
\omega_{s s}=\frac{T}{2 B}=\frac{10}{2(0.005)}=\boxed{1000} \mathrm{r} / \mathrm{s}
\]

Answer: 1000


Preamble: Consider the mass \(m\) sliding horizontally under the influence of the applied force \(f\) and a friction force which can be approximated by a linear friction element with coefficient \(b\).

Subproblem 0: Formulate the state-determined equation of motion for the velocity \(v\) as output and the force \(f\) as input.

Solution: The equation of motion is
\[
\boxed{m \frac{d v}{d t}+b v=f} \quad \text { or } \quad \frac{d v}{d t}=-\frac{b}{m} v+\frac{1}{m} f
\]

Final answer: The final answer is m \frac{d v}{d t}+b v=f. I hope it is correct.

Subproblem 1: Consider the case where:
\[
\begin{aligned}
m &=1000 \mathrm{~kg} \\
b &=100 \mathrm{~N} / \mathrm{m} / \mathrm{s}
\end{aligned}
\]
What is the steady-state velocity \(v_{s s}\) when the input is a constant force of 10 Newtons? Answer in meters per second.

Solution:
The steady-state velocity, when \(f=10\) Newtons, and \(m=1000 \mathrm{~kg}\), and \(b=100 \mathrm{~N} / \mathrm{m} / \mathrm{s}\) is
\[
v_{s s}=\frac{f}{b}=\frac{10}{100}=\boxed{0.10} \mathrm{~m} / \mathrm{s}
\]

Answer: 0.10


Subproblem 0: Obtain the inverse Laplace transform of the following frequency-domain expression: $F(s) = -\frac{(4 s-10)}{s(s+2)(s+5)}$.
Use $u(t)$ to denote the unit step function.

Solution:
Using partial fraction expansion, the above can be rewritten as
\[
F(s) = \frac{1}{s} - \frac{3}{s+2} + \frac{2}{s+5}
\]
Apply the inverse Laplace transform, then we end up with
\[
f(t) = \boxed{(1 - 3e^{-2t} + 2e^{-5t}) u(t)}
\]

Answer: (1 - 3e^{-2t} + 2e^{-5t}) u(t)


Subproblem 0: A signal has a Laplace transform
\[
X(s)=b+\frac{a}{s(s+a)}
\]
where \(a, b>0\), and with a region of convergence of \(|s|>0\). Find \(x(t), t>0\).

Solution:
Each term of \(X(s)\) can be evaluated directly using a table of Laplace Transforms:
\[
\mathcal{L}^{-1}\{b\}=b \delta(t)
\]
and
\[
\mathcal{L}^{-1}\left\{\frac{a}{s(s+a)}\right\}=1-e^{-a t}
\]
The final result is then
\[
\mathcal{L}^{-1}\{X(s)\}=\boxed{b \delta(t)+1-e^{-a t}}
\]

Answer: b \delta(t)+1-e^{-a t}


Preamble: For each Laplace Transform \(Y(s)\), find the function \(y(t)\) :

Subproblem 0: \[
Y(s)=\boxed{\frac{1}{(s+a)(s+b)}}
\]

Solution:
We can simplify with partial fractions:
\[
Y(s)=\frac{1}{(s+a)(s+b)}=\frac{C}{s+a}+\frac{D}{s+b}
\]
find the constants \(C\) and \(D\) by setting \(s=-a\) and \(s=-b\)
\[
\begin{aligned}
\frac{1}{(s+a)(s+b)} &=\frac{C}{s+a}+\frac{D}{s+b} \\
1 &=C(s+b)+D(s+a) \\
C &=\frac{1}{b-a} \\
D &=\frac{1}{a-b}
\end{aligned}
\]
therefore
\[
Y(s)=\frac{1}{b-a} \frac{1}{s+a}-\frac{1}{b-a} \frac{1}{s+b}
\]
By looking up the inverse Laplace Transform of \(\frac{1}{s+b}\), we find the total solution \(y(t)\)
\[
y(t)=\boxed{\frac{1}{b-a}\left(e^{-a t}-e^{-b t}\right)}
\]

Answer: \frac{1}{b-a}\left(e^{-a t}-e^{-b t}\right)


Preamble: Consider the rotor with moment of inertia \(I\) rotating under the influence of an applied torque \(T\) and the frictional torques from two bearings, each of which can be approximated by a linear frictional element with coefficient \(B\).

Subproblem 0: Formulate the state-determined equation of motion for the angular velocity $\omega$ as output and the torque $T$ as input.

Solution:
The equation of motion is
\[
\boxed{I \frac{d \omega}{d t}+2 B \omega=T} \quad \text { or } \quad \frac{d \omega}{d t}=-\frac{2 B}{I} \omega+\frac{1}{I} T
\]

Answer: I \frac{d \omega}{d t}+2 B \omega=T


Subproblem 0: Obtain the inverse Laplace transform of the following frequency-domain expression: $F(s) = \frac{4}{s^2(s^2+4)}$.
Use $u(t)$ to denote the unit step function.

Solution:
Since $F(s) = \frac{1}{s^2} + \frac{-1}{s^2+4}$, its inverse Laplace transform is
\[
f(t) = \boxed{(t + \frac{1}{2} \sin{2t}) u(t)}
\]

Answer: (t + \frac{1}{2} \sin{2t}) u(t)


Preamble: This problem considers the simple RLC circuit, in which a voltage source $v_{i}$ is in series with a resistor $R$, inductor $L$, and capacitor $C$.  We measure the voltage $v_{o}$ across the capacitor.  $v_{i}$ and $v_{o}$ share a ground reference.

Subproblem 0: Calculate the transfer function \(V_{o}(s) / V_{i}(s)\).

Solution:
Using the voltage divider relationship:
\[
\begin{aligned}
V_{o}(s) &=\frac{Z_{e q}}{Z_{\text {total }}}V_{i}(s)=\frac{\frac{1}{C s}}{R+L s+\frac{1}{C s}} V_{i}(s) \\
\frac{V_{o}(s)}{V_{i}(s)} &=\boxed{\frac{1}{L C s^{2}+R C s+1}}
\end{aligned}
\]

Answer: \frac{1}{L C s^{2}+R C s+1}


Preamble: You are given an equation of motion of the form:
\[
\dot{y}+5 y=10 u
\]

Subproblem 0: What is the time constant for this system?

Solution:
We find the homogenous solution, solving:
\[
\dot{y}+5 y=0
\]
by trying a solution of the form $y=A \cdot e^{s, t}$.
Calculation:
\[
\dot{y}=A \cdot s \cdot e^{s \cdot t} \mid \Rightarrow A \cdot s \cdot e^{s t}+5 A \cdot e^{s t}=0
\]
yields that $s=-5$, meaning the solution is $y=A \cdot e^{-5 \cdot t}=A \cdot e^{-t / \tau}$, meaning $\tau = \boxed{0.2}$.

Answer: 0.2


Preamble: This problem considers the simple RLC circuit, in which a voltage source $v_{i}$ is in series with a resistor $R$, inductor $L$, and capacitor $C$.  We measure the voltage $v_{o}$ across the capacitor.  $v_{i}$ and $v_{o}$ share a ground reference.

Subproblem 0: Calculate the transfer function \(V_{o}(s) / V_{i}(s)\).

Solution: Using the voltage divider relationship:
\[
\begin{aligned}
V_{o}(s) &=\frac{Z_{e q}}{Z_{\text {total }}}V_{i}(s)=\frac{\frac{1}{C s}}{R+L s+\frac{1}{C s}} V_{i}(s) \\
\frac{V_{o}(s)}{V_{i}(s)} &=\boxed{\frac{1}{L C s^{2}+R C s+1}}
\end{aligned}
\]

Final answer: The final answer is \frac{1}{L C s^{2}+R C s+1}. I hope it is correct.

Subproblem 1: Let \(L=0.01 \mathrm{H}\). Choose the value of $C$ such that \(\omega_{n}=10^{5}\) and \(\zeta=0.05\).  Give your answer in Farads.

Solution: $C=\frac{1}{\omega_{n}^{2}L}=\boxed{1e-8}[\mathrm{~F}]$

Answer: 1e-8


Preamble: Here we consider a system described by the differential equation
\[
\ddot{y}+10 \dot{y}+10000 y=0 .
\]

Subproblem 0: What is the value of the natural frequency \(\omega_{n}\) in radians per second?

Solution:
$\omega_{n}=\sqrt{\frac{k}{m}}$
So
$\omega_{n} =\boxed{100} \mathrm{rad} / \mathrm{s}$

Answer: 100