PAT 2021

1. What is the next number in the sequence $1,32,243,1024,3125$?

This is obviously not an arithmetic sequence. Next choice is a geometric sequence that does not work. However we notice that $1=1^5$, $32=2^5$, $243=3^5$, $1024=4^5$ and $3125=5^5$ so the sequence is $a_n=n^5$ with the next element $6^5=7776$.

2. What is the effective spring constant of the combination of springs shown in the diagram, if each spring has spring constant $k$?

3. Evaluate $\sum_{n=1}^{10} \qty(2-\frac{n}{2}+2^n)$.

Sum of arithmetic series: $\sum_{n=1}^N n=\frac{1}{2}N(N+1)$; sum of geometric series $\sum_{n=1}^N r^n=\frac{r\qty(1-r^n)}{1-r}$. $$ \begin{align*} \sum_{n=1}^{10} 2 & = 20 \\ \sum_{n=1}^{10} \frac{n}{2} & = \frac{1}{4} 10 \times 11 = \frac{55}{2} \\ \sum_{n=1}^{10} 2^n & = 2\qty(2^{10}-1) \end{align*} $$ Therefore $$ \sum_{n=1}^{10} \qty(2-\frac{n}{2}+2^n)=2^{11}-\frac{19}{2}. $$

4. Five different ions are accelerated from rest by the same potential difference. Which will have the smallest final velocity?

$$ \begin{array}{|c|c|c|c|c|} \hline {\bf A} & {\bf B} & {\bf C} & {\bf D} & {\bf E} \\ \hline {}^6_3{\rm Li}^{2+} & {}^7_3{\rm Li}^{2+} & {}^6_3{\rm Li}^{3+} & {}^9_4{\rm Li}^{3+} & {}^9_4{\rm Li}^{4+} \\ \hline m=6 & m=7 & m=6 & m=9 & m=9 \\ \hline q=2e & q=2e & q=3e & q=3e & q=4e \\ \hline a \propto \frac{1}{3} & a \propto \frac{2}{7} & a \propto \frac{1}{2} & a \propto \frac{1}{3} & a \propto \frac{4}{9} \\ \hline & ✔ & & &\\ \hline \end{array} $$

5. Gravity on the Moon satisfies $g_{\rm Moon}=\frac{1}{6} g_{\rm Earth}$. A ball dropped on Earth from a height $h$ takes a time $t$ to reach the ground. From which height should it be dropped on the Moon so that it take the same time $t$ to reach the surface? You can neglect all effects of air resistance.

We assume that the ball is dropped at rest; then $$ \frac{1}{2}g_{\rm Earth}t^2=h. $$ For the Moon we want $$ \frac{1}{2}g_{\rm Moon}t^2=h'. $$ We conclude that $$ h' = h \frac{g_{\rm Moon}}{g_{\rm Earth}} =\frac{h}{6}. $$

6. Two unbiased dice are rolled and the numbers obtained are added. If the probability of getting the sum $S$ is $\Pr(S)$, which of the following statements are true?

1. $\Pr(10)+\Pr(11)=\Pr(6)$
2. $\Pr(6)>\Pr(8)$
3. $\Pr(2)+\Pr(3)+\Pr(4)>\Pr(7)$
4. $\Pr(7)=\frac{3}{2}\Pr(5)$
5. $\Pr(11)=\Pr(3)$

The following table may help: $$ \begin{array}{|c|c|c|c|c|c|c|} \hline & {\bf 1} & {\bf 2} &{\bf 3} &{\bf 4} &{\bf 5} &{\bf 6} \\ \hline {\bf 1} & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline {\bf 2} & 3 & 4 & 5 & 6 & 7 & 8 \\ \hline {\bf 3} & 4 & 5 & 6 & 7 & 8 & 9 \\ \hline {\bf 4} & 5 & 6 & 7 & 8 & 9 & 10 \\ \hline {\bf 5} & 6 & 7 & 8 & 9 & 10 & 11 \\ \hline {\bf 6} & 7 & 8 & 9 & 10 & 11 & 12 \\ \hline \end{array} $$ Since each event has the same probability $p=1/36$ we have

1. $\Pr(10)+\Pr(11)=3p+2p=5p=\Pr(6)$; true.
2. $\Pr(6)=5p,\Pr(8)=5p$; false.
3. $\Pr(2)+\Pr(3)+\Pr(4)=p+2p+3p=6p,\Pr(7)=6p$; false.
4. $\Pr(7)=6p,\Pr(5)=4p,\frac{3}{2}\Pr(5)=6p=\Pr(7)$; true.
5. $\Pr(11)=2p,\Pr(3)=2p$; true.

7. A light ray follows a path through three media separated by plane boundaries as shown in the diagram, with refractive indices $n_1$, $n_2$ and $n_3$.

Since $\theta_1>\theta_2$, $\sin\theta_1>\sin\theta_2$. From Snell's law $$ n_1\sin\theta_1=n_2\sin\theta_2, \; \sin\theta_1=\frac{n_2}{n_1}\sin\theta_2, \; \frac{n_2}{n_1}\sin\theta_2>\sin\theta_2, \; n_2>n_1. $$ From the medium with refractive index $n_2$ to the medium with refractive index $n_3$ we have total internal reflection and so $$ \theta_2>\theta_c=\arcsin(\frac{n_3}{n_2}),\; 1 \geq \sin\theta_2>\frac{n_3}{n_2}, \; n_2>n_3. $$ The only answer that has $n_2>n_1$ and $n_2>n_3$ is (B).

8. If $f(x)=x^2$ and $g(x)=x+3$ find $\dv{y}{x}$ where $y=f(g(x))-g(f(x))$.

\begin{align*} y(x) & = (x+3)^2 - (x^2+3)=6(x+1) \\ \dv{y}{x} & = 6. \end{align*}

9. What is the current at the point P in the diagram?

10. Which of these represents a simpler form for $\cos\qty(\arcsin(x))$?

\begin{align*} x & = \sin(\theta) \\ \theta & = \arcsin(x) \\ \cos\theta & = \sqrt{1-\sin^2\theta} \\ \cos\theta & = \sqrt{1-x^2} \\ \cos(\arcsin(x)) & = \sqrt{1-x^2} \end{align*}

11. Consider the following five graphs.

1. Force ($y$-axis) against distance ($x$-axis).
2. Force ($y$-axis) against time ($x$-axis).
3. Velocity ($y$-axis) against time ($x$-axis).
4. Mass ($y$-axis) against velocity squared ($x$-axis).
5. Voltage ($y$-axis) against charge ($x$-axis).

For which graphs could the area under the graph potentially be a measurement of energy?

1. $I=\int F(x)dx$; since $F(x)dx$ measures work this could be a measurement of energy.
2. $I=\int F(t) dt$; $[F(t)dt]={\rm kg}\ {\rm m}\ {\rm s}^{-2}\ {\rm s}={\rm kg}\ {\rm m}\ {\rm s}^{-1}$; this measures momentum.
3. $I=\int v(t) dt$; $[v(t)dt]={\rm m}\ {\rm s}^{-1}\ {\rm s}={\rm m}$; this measures distance.
4. $I=\int m(v^2) dv^2$; since $m(v^2)dv^2$ measures kinetic energy this could be a measurement of energy.
5. $I=\int V(q) dq$; $[V(q)]={\rm J} \ {\rm C}^{-1}$, $[dq]={\rm C}$ so this could be a measurement of energy.

12. Which of the following integrals are equal to zero (you do not need to evaluate the integrals explicitly)?

1. $I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin(3x)dx$; $\sin(3x)$ is an odd function; limits symmetric around zero; integral equal to zero.
2. $I=\int_{-\sqrt{7}}^{\sqrt{7}} \qty(\frac{1}{9}x^5-\frac{1}{7}x^3+\frac{1}{21}x)dx$; integrand is an odd function; limits symmetric around zero; integral is equal to zero.
3. $I=\int_0^1 \cos x dx$; the period of $\cos$ is $2\pi$. This integral is not equal to zero.
4. $I=\int_{-2}^2 \qty(\frac{1}{4}x^4+\frac{1}{12}x^2)dx$; the integrand is an even function that only takes positive values; integral is not equal to zero.
5. $I=\int_{-90}^{90}\qty(\sin(5x)-\frac{1}{2}\sin x) dx$; the integrand is an odd function; limits symmetric around zero; integral is equal to zero.

13. A toilet roll is made up of an inner cardboard tube, diameter $x$ cm, with toilet paper wrapped around it to give an overall diameter of $y$ cm. The length of the cardboard tube is $z$ cm. Suppose the diameter of the inner tube is reduced from $x$ cm to $(x-1)$ cm, but the volume of the toiler paper is kept the same. What is the difference in the total volume of the roll?

The initial volume of the toilet paper is $\pi (y^2-x^2)z/4$; with the new diameter, $y_n$, of the toilet roll we want $$\frac{1}{4}\pi (y_n^2-(1-x)^2)z=\frac{1}{4}\pi(y^2-x^2)z \Rightarrow y_n^2-(1-x)^2 = y^2-x^2.$$ The difference in the total volume of the roll is $$ V_n - V = \frac{1}{4}\pi (y_n^2-y^2)z=\frac{1}{4}\pi \qty((1-x)^2-x^2)z=\frac{1}{4}\pi (1-2x)z. $$

14. A machine requires 10 kW of electrical power when operating at 100 V.

1. If power is delivered through a cable with resistance 100 $\Omega$, how much power is lost to the cable's resistance.
2. In order to reduce power loss, a transformer is used between the cable and the machine. What ratio of turns is required on the transformer in order to reduce power loss on the cable by a factor of $10^4$.

Since the machine requires 10 kW of power, if the cable had zero resistance we would require an electric current of $10^4 \ {\rm W}/ 100 \ {\rm V}=10^2 \ {\rm A}$. Given the resistance of the cable, the amount of power lost is $10^4 \ {\rm A}^2 \ 100 \ \Omega = 10^6 \ {\rm W}$.

The following simple schematic shows the transformer:

In order to reduce the power loss by a factor of $10^4$ the electric current in the cable must be $1 \ {\rm A}$. Power is conserved and so $V_p \times 1 \ {\rm A} = 100 \ {\rm V} \times 100 \ {\rm A}$, $V_p=10^4 \ {\rm V}$. Therefore $$ \frac{N_p}{N_s} = \frac{V_p}{V_s}=\frac{10^4}{10^2}=100. $$

15. The observed brightness of a Sun-like star shows periodic dips. The time period between these dips is $T=225 \ {\rm days}$.

1. If these dips are interpreted as being due to the transits of a planet in a circular orbit around the star, estimate the radius $R$ of that orbit.
   You may assume that the star has the same mass as the Sun and that the planet's mass is much smaller than the mass of the star. You can take the mean radius of the Earth's orbit around the Sun to be approximately $R_E=1.5 \times 10^{11} \ {\rm m}$.
2. A model of a transit and its effect on the star's brightness is shown in the figure below. Assuming that we observe the system in the plane of the planet's orbit, draw the relative positions of the planet and the star at the times $t_a$ to $t_d$ and calculate the radii of the planet, $R_p$ and the star, $R_s$.

The angular velocity of a planet of mass $m$ around a sun of mass $M$ is given by $$ G \frac{Mm}{R^2}=m \omega^2 R $$ where $G$ is the gravitational constant and $R$ is the orbital radius. Since $$ \omega =\sqrt{\frac{GM}{R^3}} $$ and $\omega=2\pi/T$, the orbital period is $$ T = 2 \pi \sqrt{\frac{R^3}{GM}}. $$ We know that $T=225 \ {\rm days}$; however we do not have the value for $M$ (it is the same as our sun but we don't remember it). Since the orbital period of the earth is 365 days we can write $$ T_E=2 \pi \sqrt{\frac{R_E^3}{GM}}. $$ Combining the last two equations we obtain $$ \frac{225}{365}=\qty(\frac{R}{R_E})^{\frac{3}{2}}, \; \frac{R}{R_E}=\qty(\frac{225}{365})^{\frac{2}{3}}, \; R = 1.086 \times 10^{11} \ {\rm m}. $$

The dip in brightness is caused by the partial obstruction of the distant sun by the planet crossing our line of sight. This phenomenon is known as the transit light curve. The following diagram helps to illustrate the process.

During the time interval $\Delta t_1$ the planet's motion covers an angle $\omega \Delta t_1=2\pi \Delta t_1/T=2 \pi \times \ 23 \ {\rm min}/225 \ {\rm days}$. The corresponding arc length is $R\omega \Delta t_1=1.086 \times 10^{11} \ {\rm m} \times 2 \pi \times 23 \ {\rm min}/225 \ {\rm days}=4.85 \times 10^7 \ {\rm m}$. Since this is equal to the diameter, $R_p=2.42 \times 10^7 \ {\rm m}$. It takes $2\Delta t_1 + \Delta t_2$ for the planet to traverse the sun in front of our line of sight. Therefore the diameter of the sun is given by $$R \omega (2\Delta t_1 + \Delta t_2)=1.086 \times 10^{11} \ {\rm m} \times 2 \pi \times (46 \ {\rm min} + 10.5 \ {\rm hr})/225 \ {\rm days}=1.42 \times 10^9 \ {\rm m}.$$ The radius is $R_s=7.12 \times 10^8 \ {\rm m}$. The following diagram shows the position of the planet at different times.

16. A vehicle travels in a fixed direction at a velocity $v(t)$ that varies with time as follows: \begin{align*} t & < t_1 & v(t) & = At^2 \\ t_1 & < t < t_2 & v(t) & = C-B(t-t_2)^2 \\ t & > t_2 & v(t) & = v_2. \end{align*} Here $A$, $B$, $C$ and $v_2$ are constants.

1. Find $A$, $B$, $C$ such that both the velocity and acceleration are continuous at $t=t_1$ and $t=t_2$.
2. Sketch the velocity and acceleration as a function of time.
3. Find the total distance travelled between $t=0$ and $t=t_3$, where $t_3>t_2$.

For continuity the velocity and its first derivative must agree at each boundary. The first set of equations is $$ \left.\dv{t}\qty(At^2)\right\vert_{t=t_1}=\left.\dv{t}\qty(C-B(t-t_2)^2)\right\vert_{t=t_1}, \; At_1^2=C-B(t_1-t_2)^2 $$ and the second set of equations is $$ \left.\dv{t}\qty(C-B(t-t_2)^2)\right\vert_{t=t_2}=\left.\dv{t}\qty(v_2)\right\vert_{t=t_2}, \; C = v_2 . $$ Evaluating the first set of equations we get $$ At_1=B(t_2-t_1), \; A t_1^2= v_2 - B (t_2-t_1)^2 $$ from which $A=v_2/(t_1t_2)$ and $B=v_2/((t_2-t_1)t_2)$. The second set of equations holds since $$ \left.\dv{t}\qty(C-B(t-t_2)^2)\right\vert_{t=t_2}=0=\left.\dv{t}\qty(v_2)\right\vert_{t=t_2}=0. $$

Given these equations the plot of the velocity and acceleration looks something like

To calculate the distance travelled, $d$, between $t=0$ and $t=t_3$ we must evaluate the following integral \begin{align*} d & = \int_0^{t_3} v(t) dt = \frac{v_2}{t_1t_2}\int_0^{t_1} t^2 dt + v_2\int_{t_1}^{t_2} \qty(1-\frac{(t-t_2)^2}{t_2(t_2-t_1)}) dt + v_2 \int_{t_2}^{t_3} dt \\ & = \frac{v_2}{t_1t_2} \left[ \frac{t^3}{3}\right]_0^{t_1} + v_2 \left[ t \right]_{t_1}^{t_2} - v_2 \frac{1}{t_2(t_2-t_1)}\left[\frac{s^3}{3}\right]_{t_1-t_2}^0 + v_2 \left[ t \right]_{t_2}^{t_3} \\ & = \frac{v_2t_1^2}{3t_2} + v_2 (t_2-t_1) - \frac{v_2(t_2-t_1)^2}{3t_2}+ v_2(t_3-t_2) \\ & = v_2\qty(t_3-\frac{t_1+t_2}{3}). \end{align*}

17. In Einstein's theory of gravity, light passing a star of mass $M$, at a distance $R$ from the centre of the star, is bent by an angle $\theta$ (masured in radians, $2\pi$ radians = $360^\circ$) as shown in the diagram.

1. Assume that $\theta$ depends on the gravitational constant $G$ and also depends on the mass of the star $M$, the distance $R$ and the speed of ligh $c=3 \times 10^8 \ {\rm m} \ {\rm s}^{-1}$, as $$ \theta=\lambda G M^\alpha R^\beta c^\gamma $$ where $\lambda$ is an undetermined, dimensionless constant. By considering the dimensions of these quantities determine $\alpha$, $\beta$ and $\gamma$.
2. In SI units the numerical value of $G$ is $6.67 \times 10^{-11}$.
   Neglecting any spatial extent to the star, and taking $\lambda=1$ and $M=2 \times 10^{30} \ {\rm kg}$, determine the distance $R$ such that light is bent so much that it falls directly int the star.

From Newton's law of universal gravitation between two masses $M,m$ separated by a distance $R$ $$ F = G \frac{Mm}{R^2} $$ where $G$ is the gravitational constant, and Newton's second law $$ F=ma $$ where $a$ is the acceleration we have \begin{align*} [F]=[m][a] & =[G] \frac{[M][m]}{[R]^2} \\ [a] & =[G] \frac{[M]}{[R]^2} \\ \frac{{\rm m}}{{\rm s}^2} & = [G] \frac{{\rm kg}}{{\rm m^2}} \\ [G] & = \frac{{\rm m}^3}{{\rm kg} \ {\rm s}^2}. \end{align*}

Since $[\theta]=[\lambda]=1$ we have \begin{align*} 1 & =[G] [M]^\alpha [R]^\beta [c]^\gamma \\ 1 & = \frac{{\rm m}^3}{{\rm kg} \ {\rm s}^2} {\rm kg}^\alpha {\rm m}^\beta \frac{{\rm m}^\gamma}{{\rm s}^\gamma}. \end{align*} It is immediately obvious that $\alpha=1$; next we must have $2+\gamma=0$, i.e. $\gamma=-2$. Finally since $3+\beta+\gamma=0$ we conclude that $\beta=-1$.

If we neglect the spatial extend of the star we must set $\theta=\pi/2$. If this is the case then $$ \frac{\pi}{2}=\lambda G M R^{-1} c^{-2} \Rightarrow R = \frac{2GM}{\pi c^2}=\frac{2 \times 6.67 \times 10^{-11} \times 2 \times 10^{30}}{\pi \times 9 \times 10^{16}} = 943.61 \ {\rm m}. $$

18. Find the area of the dark regions of this figure in terms of the overall radius of the outer circle (its radius is denoted as $R$).

$$ \left\{ \begin{aligned} \qty(\frac{R}{2}+r)^2 & = \qty(\frac{R}{2})^2 + x^2 \\ x + r & = R \end{aligned} \right\} \quad \left\{ \begin{aligned} \qty(\frac{R}{2}+r)^2 & = \qty(\frac{R}{2})^2 + (R-r)^2 \\ x + r & = R \end{aligned} \right\} $$ $$ \left\{ \begin{aligned} \qty(\frac{R}{2}+r)^2 -(R-r)^2 & = \qty(\frac{R}{2})^2 \\ x + r & = R \end{aligned} \right\} \quad \left\{ \begin{aligned} \frac{3R}{2}\qty(2r-\frac{R}{2}) & = \qty(\frac{R}{2})^2 \\ x + r & = R \end{aligned} \right\} \quad \left\{ \begin{aligned} r & = \frac{R}{3} \\ x & = \frac{2R}{3} \end{aligned} \right\} $$ $$ \text{ Shaded Area} = \pi R^2 - 2 \pi \qty(\frac{R}{2})^2-2 \pi \qty(\frac{R}{3})^2 = \pi \frac{5R^2}{18}. $$

19. Sand is poured at a constant rate onto a flat horizontal surface. It forms a pile in the shape of a cone with constant slope.

1. If $r$ is the radius of the base of the cone, find how $r$ varied with $t$, given that $r(0)=0$.
2. If it takes time $t_1$ for $r$ to reach $r_1$, how long does it take for the radius to reach $2r_1$ (in terms of $t_1$)?

The volume of sand is $V(t)=v t$ where $v$ is the constant rate. From the diagram above, since the slope of the sand cone is constant, $h=r \tan \alpha$. Since the volume of a cone is $$ V=\pi r^3 \frac{h}{3}=\pi\tan\alpha \frac{r^4}{3} $$ we have $$ V(t)=v t = \pi\tan\alpha \frac{r(t)^4}{3} \Rightarrow r(t)= \sqrt[4]{\frac{3vt}{\pi\tan\alpha}}. $$ Given that $$vt_1 = \pi\tan\alpha \frac{r_1^4}{3}$$ and $$vt_2 = \pi\tan\alpha \frac{(2r_1)^4}{3}$$ the time, $t_2$, it takes for the radius to reach $2r_1$ is $ t_2=16t_1 $.

20. Expand $(a+bx)^c$, where $c$ is a positive integer, as a polynomial in $x$ up to and including terms involving $x^2$.
If the coefficient of the $x^0$ term is equal to $\frac{1}{16}$, $b=\frac{1}{4}$ and the coefficient of the $x^2$ term is $\frac{3}{32}$, find $a$ and $c$.

$$ (a+bx)^c = a^c (bx)^0 + \binom{c}{1} a^{c-1}bx + \binom{c}{2} a^{c-2}(bx)^2+\cdots $$ From the given conditions we have $$ a^c =\frac{1}{16}, \; \binom{c}{2} a^{c-2}b^2 =\frac{ c (c-1)}{2} a^{c-2}b^2 = \frac{ c (c-1)}{32} a^{c-2}= \frac{3}{32}. $$ Since $c$ is an integer, from the first condition we can try $\qty(\frac{1}{2})^4=\frac{1}{16}$. With these values we can check the second condition $$ \frac{12}{32} \qty(\frac{1}{2})^2 = \frac{12}{32}\frac{1}{4}=\frac{3}{32}. $$

21. A pinhole camera consists of a box with a small aperture (of diameter $x$). A point light source $P$ is at a distance $u$ from the aperture outside the box. It projects an image on the screen, which is insie the box at a distance $D$ from the aperture.

1. If diffraction is ignored, what is the diameter $s_1$ of the spot projected on the screen?
2. If $D \gg x$, diffraction of light of wavelength $\lambda$ by the aperture produces a distribution of intensity that varies with angle $\theta$ (measured in radians from the $z$ axis marked on the diagram; $2\pi \ {\rm radians}=360^\circ$) approximately as shown below:
   
   What is the approximate diameter, $s_2$, of the central peak of the diffraction pattern projected on the screen?
3. Assume that the effects in parts (a) and (b) can be combined to give a total diameter $s_3$, where $$ s_3^2=s_1^2+s_2^2.$$ Find the pinhole size that produces the sharpest image of the light source on the screen.

For (a) start with the spherical waves emanating from $P$. If we ignore diffraction the picture looks like this:

Since $\triangle PAB \sim \triangle PSW$ we have $$ \frac{x/2}{u}=\frac{s_1/2}{u+D} \Rightarrow s_1 = \frac{u+D}{u}x. $$

For (b) it helps to go into a bit more detail following the analysis of Halliday-Resnik. The condition $D \gg x$ means that all points on the aperture are at the same distance from the centre of the screen. If we look at the right angled triangle formed by the rays arriving from points $A$, $C$

then $D^2+\qty(\frac{x}{2})^2=D'^2$ and so $1+\qty(\frac{x}{2D})^2=(\frac{D'}{D})^2$. From the condition $D\gg x$, $1+\qty(\frac{x}{2D})^2\approx 1$ and so $D' \approx D$. Therefore the rays arrive in phase generating constructive interference. A key feature of this condition is that $\angle BAC = \angle BCA \approx 90^\circ$.

We now consider the geometry for a point away from the centre like $E$. Looking at the rays emanating from points $A$, $C$, we can draw a circular arc $CY$ to determine their phase difference.

The cicular arc intersects segment $EA$ at point $Y$. The phase difference of rays $CE$ and $AE$ is $\Delta \phi = 2\pi |AY|/\lambda$. As in the case of $\angle BAC$, $\angle BCA$ we also have $\angle ECY = \angle EYC \approx 90^\circ$. Since $CY \perp AE$ and $CA \perp AB$ we have $\angle ACY = \angle BAE=\lambda/x$. From $\triangle CAY$ $$ \sin\qty(\frac{\lambda}{x})=\frac{2|AY|}{x}. $$ For destructive interference we must have $\Delta \phi=\pi$ and so $|AY|=\lambda/2$. Therefore $$ \sin\qty(\frac{\lambda}{x})=\frac{\lambda}{x} $$ which is the small angle approximation, $\sin\theta\approx\theta$. Since $\tan\qty(\frac{\lambda}{x})=s_2/2D$ and the small angle approximation of $\tan\theta$ is $\theta $ we conclude that $s_2=2D\lambda/x$.

For (c) we have $$ s_3^2=s_1^2+s_2^2 = \qty(\frac{u+D}{u})^2x^2 + \frac{\qty(2D\lambda)^2}{x^2}. $$ To produce a sharp image we must generate the smallest value of $s_3$. Since $$ 2 s_3 \dv{s_3}{x}=2\qty(\frac{u+D}{u})^2x-2\frac{\qty(2D\lambda)^2}{x^3} $$ if we set $\dv{s_3}{x}=0$ we obtain $$ x^4=\qty(\frac{2Du\lambda}{u+D})^2. $$ To check that this is a minimum write $$ 2 s_3 \dv[2]{s_3}{x} + 2 \qty(\dv{s_3}{x})^2 = 2\qty(\frac{u+D}{u})^2 + 12\frac{\qty(2D\lambda)^2}{x^4}. $$ Substituting the value for $x^4$ we obtain $$ 2 s_3 \dv[2]{s_3}{x} + 2 \qty(\dv{s_3}{x})^2 = 2\qty(\frac{u+D}{u})^2 + 12\qty(\frac{u+D}{u})^2=14\qty(\frac{u+D}{u})^2. $$ Since $\dv{s_3}{x}=0$, $s_3>0$ and $ 14\qty(\frac{u+D}{u})^2>0 $ we have $\dv[2]{s_3}{x}>0$. So for $x=\sqrt{2Du\lambda/(u+D)}$, we obtain the sharpest image.

22.

1. Sketch the functions $f(x)=x^2+2x-7$ and $g(x)=2/x$ on the same graph.
2. Determine the coordinates of the points of intersection of the two curves and label their exact values. You are not required to determine coordinates of intersections with $x$ or $y$ axes.
3. Evaluate the finite area enclosed between the two curves.

To determine the points of intersection of the two curves we must solve $$ x^2+2x-7=\frac{2}{x} $$ which is the cubic $$ x^3+2x^2-7x-2=0. $$ Since cubics are difficult, set $2=r-x$ to obtain $$ rx^2-7x-2=0. $$ This quadratic has solutions $$ x=\frac{7\pm\sqrt{49+8r}}{2r}. $$ For $r=4$, $49+32=81$ and so $$ x=\frac{7\pm9}{8}=2,-\frac{1}{4}. $$ From the condition $2=r-x$ the only admissible solution is $x=2$. Since $x-2$ is a factor of the quadratic write $$ (x-2)(x^2+ax+b)=x^2+ax^2+bx-2x^2-2ax-2b=x^3+2x^2-7x-2. $$ From the $x^0$ term $b=1$; from the $x$ term $b-2a=-7$ and so $a=4$. This value of $a$ produces the correct $x^2$ term. The quadratic $$ x^2+4x+1=0 $$ has solutions $x=-2\pm\sqrt{3}$. Therefore the points of intersection are $$ (2,1), \; (-2+\sqrt{3},-2(2+\sqrt{3})), \; (-2-\sqrt{3},-2(2-\sqrt{3})). $$ The area $A$ between the two curves is given by the integral $$ A=\int_{-(2+\sqrt{3})}^{-2+\sqrt{3}} \qty(\frac{2}{x}-\qty(x^2+2x-7)) dx = \qty[2 \ln x - \frac{x^3}{3}-x^2+7x]_{-(2+\sqrt{3})}^{-2+\sqrt{3}}. $$ Write the part with $\ln$ as $$ 2 \ln \frac{2-\sqrt{3}}{2+\sqrt{3}}=4 \ln (2-\sqrt{3}) $$ which since $4>3$ has a positive argument for the logarithm.

We must substract $$ \qty[\frac{x^2}{3}+x^2-7x]_b^a $$ to obtain the value of $A$. We can use $$ \frac{a^3-b^3}{3}=\frac{(a-b)(a^2+ab+b^2)}{3} $$ and $$ a^2-b^2=(a-b)(a+b) $$ to evaluate this expression. We have a common factor $ a-b=2\sqrt{3} $ ; then $a+b=-4$ and $$ \frac{a^2+ab+b^2}{3}=\frac{7-4\sqrt{3}+1+7+4\sqrt{3}}{3}=5. $$ So the value of the expression is $2\sqrt{3}(5-4-7)=-12\sqrt{3}$. Therefore $A=12\sqrt{3}+4 \ln (2-\sqrt{3})$.

23. The potential energy for a system of two atoms separated by a distance $x$ is $U(x)$. Sketch $-\dv*{U}{x}$. What does this new curve represent physically and what does the point $x_0$ represent?

The easiest way to determine what $-\dv{U}{x}$ represent is to look at the units. $U$ is potential energy measured in Joules; $x$ is distance measured in metres. ${\rm J}/{\rm m}$ has the units of a force. The negative sign is an adjustment that accounts for the fact that the force points to a direction opposite to the increase in potential energy (think of the weight of an object pointing in the opposite direction to the increase in gravitational potential). The point $x_0$ at which $\dv{U}{x}=0$ represents the distance for which the force is zero. The curve shown in the exercise is the Lennard-Jones potential that can be written as $$ U(x)=\frac{A}{x^{12}}-\frac{B}{x^6}. $$ with the first tem representing the repulsive potential and the second term representing the attractive potential. At very small distances the two atoms repel while at larger distances they attract each other. At very large distances there is close to zero force between the two atoms ($U(x)$ asymptotically approaches zero) while at a distance $x_0$ these two potentials are matched resulting in zero force between the two atoms.

24. If $f(x)$ is

and $g(x)$ is

then sketch $f(x-ct)+g(x+ct)$ for $t=a/2c$, $3a/4c$ and $a/c$. Each sketch should be done separately with its own set of axes.

To get an idea of each sketch it helps to write $$ f(x)=\max(x+2a,0)-2\max(x+a,0)+\max(x,0) $$ and $$ g(x)=-(\max(x,0)-2\max(x-a,0)+\max(x-2a,0)). $$ Then \begin{align*} f(x-ct)+g(x+ct) & = \max(x-ct+2a,0)-2\max(x-ct+a,0)+\max(x-ct,0) \\ &\; -\max(x+ct,0)+2\max(x+ct-a,0)-\max(x+ct-2a,0). \end{align*} For $t=a/2c$ we have $ct=a/2$ and $$ f\qty(x-\frac{a}{2})+g\qty(x+\frac{a}{2})=\max\qty(x+\frac{3a}{2},0)-3\max\qty(x+\frac{a}{2},0)+3\max\qty(x-\frac{a}{2},0)-\max\qty(x-\frac{3a}{2},0). $$

For $t=3a/4c$ we have $ct=3a/4$ \begin{align*} f\qty(x-\frac{3a}{4})+g\qty(x+\frac{3a}{4}) & =\max\qty(x+\frac{5a}{4},0)-\max\qty(x+\frac{3a}{4},0)-2\max\qty(x+\frac{a}{4},0) \\ & \; +2\max\qty(x-\frac{a}{4},0)+\max\qty(x-\frac{3a}{4},0)-\max\qty(x-\frac{5a}{4},0). \end{align*}

For $t=a/c$ we have $ct=a$. Lets follow a different approach here. From the information given we know that $g(x)=-f(x-2a)$. Therefore $$ f(x-a)+g(x+a)=f(x-a)-f(x+a-2a)=f(x-a)-f(x-a)=0. $$