## 175 – Midterm 1

Here is the midterm.

1. Hooke’s law says that the force required to stretch a spring a distance of $x$ units from its natural length is given by $F(x)=kx,$ where $k$ is a constant that only depends on the spring.

We are told that for the spring of problem 1 we have $F(2)=20,$ where units of distance are measured in meters, and forces in Newtons. This means that $k2=20,$ or $k=10.$ The work required to stretch the spring 10 m beyond its natural length is then $\displaystyle \int_0^{10}10x\,dx=10x^2/2|^{10}_0=500 \,N\cdot m.$

2. We are given a curve with equation $x=\sqrt{2y-1}$ and asked to rotate about the $y$-axis the segment where $1/2\le y\le 2.$ To find the area of the resulting surface, we can consider splitting the surface into little shells. A typical `thin’ shell, at height $y$ has surface  area $2\pi r\,ds$ where $r$ is the radius of the shell, in this case just the value of $x$ corresponding to $y,$ i.e., $\sqrt{2y-1}.$ Also, $ds$ is here the arc length element, given by $\sqrt{(x')^2+1}\,dy.$

We have: $\displaystyle x'=\frac12(2y-1)^{-1/2}2=\frac1{\sqrt{2y-1}},$ so $\displaystyle (x')^2+1=\frac1{2y-1}+1=\frac{2y}{2y-1}.$

Hence the requested area is given by $\displaystyle\int_{1/2}^2 2\pi\sqrt{2y-1}\sqrt{\frac{2y}{2y-1}}\,dy=\int_{1/2}^2 2\pi\sqrt{2y}\,dy.$

This expression reduces to $\displaystyle 2\pi\sqrt2\,\frac23\,y^{3/2}|^2_{1/2}=2\pi\sqrt2\frac23\left(2\sqrt2-\frac1{2\sqrt2}\right)=\frac{14\pi}3.$

3. We are given the region bounded by the curve $y=\sqrt x$ and the lines $y=2$ and $x=0.$ We rotate it about the $x$-axis, and are asked to find the volume of the resulting solid. A first attempt is to use the washer method. A typical thin washer consists of a slice of the solid for a fixed value of $x$ and with thickness $dx.$ Its volume is $\pi (r_1^2-r_2^2)\,dx.$ Here, $r_1$ is the exterior radius, which in this case is always 2, and $r_2$ is the interior radius, given by $y=\sqrt x.$ This means the washer contributes $\pi(4-x)\,dx$ of the full volume, and adding all of them together we obtain $\int_0^4 \pi(4-x)\,dx.$ The upper limit of 4 comes from observing that the curves $y=2$ and $y=\sqrt x$ cross precisely at the point $(4,2).$

Unfortunately, this integral is not one of the given expressions.

So we attempt a different approach, using now the shell method. Now a typical thin shell is obtained by fixing a height $y$ and rotating about the $x$-axis the slice of the figure of thickness $dy$ and height $y.$ Its volume is $2\pi r\,l\,dy,$ where $r$ is the radius of the shell, in this case $y;$ and $l$ is the length of the shell, in this case $x.$ We thus proceed to express $x$ in terms of $y:$ Since $y=\sqrt x,$ then $x=y^2.$ Hence the thin shell contributes $2\pi y\,y^2\,dy=2\pi y^3\,dy$ of the full volume, and the volume is obtained by adding all these contributions, obtaining $\int_0^2 2\pi y^3\,dy.$ This is expression (c).

4. The length of the curve $y=\sqrt{1-x^2}$ for $-1\le x\le 1$ is given by the integral $\displaystyle \int_{-1}^1\sqrt{1+(y')^2}\,dx.$ In this case, $\displaystyle y'=\frac12\,\frac{-2x}{\sqrt{1-x^2}}=-\frac x{\sqrt{1-x^2}},$ so $\displaystyle (y')^2=\frac{x^2}{1-x^2},$ and $\displaystyle 1+(y')^2=\frac1{1-x^2},$ so the integral reduces to $\displaystyle\int_{-1}^1\frac 1{\sqrt{1-x^2}}\,dx,$ which is expression (b).

[There is a small problem with this integral, though, because the denominator vanishes at both endpoints. Later, in Chapter 7, we will learn how to handle integrals of this kind. Note that if the question had been to actually find the length, there is an easier method: The curve is half a circumference of radius 1, so the length is $\pi.$]

5. (a) The differential equation $\displaystyle \frac{dy}{dx}=\frac{e^{3x+4y}}{e^{x^2-7y}}$ can be rewritten as $\displaystyle\frac{dy}{dx}=e^{3x-x^2}e^{11 y},$ which is clearly separable. (T)

(b) The area swept by $r=2f(\theta)$ for $\alpha\le\theta\le\beta,$ is four times the area swept by $r=f(\theta).$ Intuitively, we are making lengths twice as long, and areas carry two dimensions of length. Think of the area of a square of side 2, for example. Formally, we can check that this is the case: The area swept by the first curve is given by $\displaystyle\int_\alpha^\beta\frac12 (2f(\theta))^2\,d\theta=4\int_\alpha^\beta\frac12(f(\theta))^2\,d\theta,$ while the area swept by the second curve is given by $\displaystyle\int_\alpha^\beta\frac12 (f(\theta))^2\,d\theta.$ (T)

(c) The half life of a given element is 3 days. This means that 50% of the given amount of the element has decayed after 3 days. After 3 more days, another 25% would have decayed. After 3 more days, another 12.5%. After 3 more days, another 6.25%. Since $50+25+12.5+6.25>90,$ the given statement is clearly true.

A longer way of checking the same is by recalling that the total amount of the element that remains after time $t$ if originally the amount is $y_0$ is given by $y=y_0e^{-kt},$ where $k$ is a given constant. Measure $t$ in days. We are told that $y(3)=y_0/2,$ so $e^{3k}=2.$ Then $e^{27 k}=(e^{3k})^9=2^9=512,$ and $y(27)=y_0/512 (T)