October 18, 2009

Here is quiz 4.

## 175 – Integrating products of secants and tangents

October 9, 2009

(In what follows, I will write ${\tan^n x}$ for ${(\tan x)^n}$ and ${\sec^m x}$ for ${(\sec x)^m.}$)

Recall that

$\displaystyle \tan x=\frac{\sin x}{\cos x}\quad\mbox{ and }\quad\sec x=\frac1{\cos x},$

that

$\displaystyle (\tan x)'=\sec^2x\quad\mbox{ and }\quad(\sec x)'=\sec x\cdot\tan x,$

and that

$\displaystyle \sec^2x=\tan^2x+1.$

The formulas below make use of these identities repeatedly.

We want a series of methods and reduction formulas that allow us to evaluate any expression of the form

$\displaystyle \int \sec^m x\cdot \tan^n x\,dx,$

for ${m}$ and ${n}$ integers, ${m,n\ge0.}$

## 175 – Quiz 3

October 2, 2009

Here is quiz 3.

Problem 1 is Exercise 9.3.16 from the book. Here is a graph showing the cardioid and the circle. The easiest way to compute the required area is by subtracting the area of the circle from that of the cardioid. (Note that the circle is completely contained in the cardioid as, for each ${ \theta\in[0,\pi/2],}$ we have that ${0<\cos\theta< 1+\cos\theta.}$ This shows that in the portion of the circle in the first quadrant is within the portion of the cardiod there. The same holds in the fourth quadrant by symmetry. Of course, all of this also follows directly from the graph.)

The area of the cardioid is given by ${ \displaystyle\int_0^{2\pi}\frac12(1+\cos\theta)^2\,d\theta}$ and the area of the circle is ${ \displaystyle\int_0^\pi \frac12(\cos\theta)^2\,d\theta}$ (note that as ${ \theta}$ varies from ${ 0}$ to ${ \pi}$ the whole circle is traveled once).

By symmetry, ${ \displaystyle\int_0^{2\pi}\frac12(1+\cos\theta)^2\,d\theta= \int_0^{\pi}(1+\cos\theta)^2\,d\theta,}$ so the area we want is given by

$\displaystyle \int_0^{\pi}\left((1+\cos\theta)^2-\frac12(\cos\theta)^2\right)\,d\theta$ $\displaystyle =\int_0^\pi\left(\frac{\cos^2\theta}2+2\cos\theta+1\right)\,d\theta.$

Recall that ${ \displaystyle\cos^2\theta=\frac{1+\cos2\theta}2,}$ so the integral is ${ \displaystyle\int_0^\pi\left(\frac{1+\cos2\theta}4+2\cos\theta+1\right)\,d\theta.}$

Now note from the Cartesian graph of ${ \cos\theta}$ that ${ \displaystyle\int_0^\pi\cos2\theta\,d\theta=\int_0^\pi\cos\theta\,d\theta=0,}$ so the area we want is just ${ \displaystyle\int_0^\pi\frac14+1\,d\theta=\frac54\pi.}$

Problem 2 is Exercise 9.5.50 from the book. Recall that the standard polar equation of a line not going through the origin is given by

$\displaystyle r = \frac{r_0}{\cos(\theta-\theta_0)},$

where ${ r_0}$ is the distance from the line to the origin, and ${ \theta_0}$ is the angle of the point in the line that realizes this distance, i.e., ${ (r_0,\theta_0)}$ are the polar coordinates of the point on the line closest to the origin.

There are at least two ways we can proceed:

1) We can directly find the distance from the line to the origin. To do this, we recall that the line that goes through the origin and is perpendicular to ${ y=mx+b,}$ where ${ m\ne0,}$ is given by ${ \displaystyle y=-\frac1m x.}$ The point in ${ y=mx+b}$ closest to the origin is the intersection of these two lines. In the case that interests us, this is the intersection of ${ y={\sqrt3}x-1}$ and ${ \displaystyle y=-\frac1{\sqrt3} x,}$ so ${ \displaystyle{\sqrt3} x -1=-\frac1{\sqrt3} x}$ or ${ \displaystyle\left({\sqrt3}+\frac1{\sqrt3}\right)x=1,}$ or ${ \displaystyle x=\frac{\sqrt3}4,}$ so ${ \displaystyle y=-\frac1{\sqrt3} x=-\frac14.}$ We have found that the point ${ \displaystyle\left(\frac{\sqrt3}4,-\frac{1}4\right)}$ is closest in ${ y=\sqrt3-1}$ to the origin. Its distance is ${ r_0=\sqrt{x^2+y^2}=2/4=1/2.}$ Its angle ${ \theta_0}$ is on the fourth quadrant and satisfies ${ \tan\theta_0=1/\sqrt3,}$ i.e., ${ \theta_0=-\pi/6.}$

Putting this together, ${ \displaystyle r=\frac{1/2}{\cos(\theta+\pi/6)}}$ is the desired equation.

2) The other method we saw in lecture is based in the fact that

$\displaystyle \cos(\theta-\theta_0)=\cos\theta_0\cos\theta+\sin\theta_0\sin\theta,$

and consists in first writing the Cartesian equation of the line directly in polar coordinates, and then using the resulting expression to find ${ r_0}$ and ${ \theta_0:}$ We have ${ y={\sqrt 3} x-1,}$ so ${ r\sin\theta=\sqrt 3 r\cos\theta-1,}$ or ${ r(\sqrt 3\cos\theta-\sin\theta)=1.}$ This gives us

$\displaystyle r=\frac 1{\sqrt3 \cos\theta-\sin\theta}.$

We find a constant ${ k}$ such that ${ k\sqrt3}$ and ${ -k}$ are the cosine and the sine of some angle, precisely ${ \theta_0.}$ We then have ${ k\sqrt3\cos\theta-k\sin\theta=\cos\theta_0\cos\theta+\sin\theta_0\sin\theta=\cos(\theta-\theta_0).}$

To find this ${ k,}$ we use that ${ \cos^2\alpha+\sin^2\alpha=1}$ for any ${ \alpha,}$ so ${ k}$ must satisfy ${ {k^2}3+k^2=1,}$ or ${ k^2=1/4,}$ so ${ k=1/2.}$ The value of ${ \theta_0}$ such that ${ \cos\theta_0={\sqrt3}/2}$ and ${ \sin\theta_0=-1/2}$ is ${ \theta_0=-\pi/6.}$ This means that the equation we are looking for is

$\displaystyle r=\frac k{k\sqrt3 \cos\theta-k\sin\theta}=\frac{1/2}{\cos(\theta+\pi/6)},$

just as with the previous method.

Typeset using LaTeX2WP.

## 175 – Midterm 1

September 25, 2009

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)

## 175 – Dandelin spheres

September 22, 2009

Here is a link to the Wikipedia page on Dandelin spheres, giving us the “modern” proof of the equivalence between the definition of conic sections as regions of intersection of planes and cones, with the standard definition in terms of distance to foci. The links on the Wikipedia page provide further explanations and nice graphics illustrating the argument.

## 175 – Quiz 2

September 19, 2009

Here is quiz 2.

Problem 1 is (simplification of part of) exercise 6.6.16 from the book.

To solve the question, use coordinates as in the accompanying figure in the book, so the origin is at ground level, and $y$ increases downwards. The units of $y$ are feet. For a fix $y$ with $0\le y\le 20,$ the thin slice of water in the tank at depth $y$ and of tickness $dy$ has volume $dV=10\times 12\times dy$ and weighs $dF=62.4\times dV=7488\,dy.$ This is a constant force, so the work required to remove it to ground level is just $dW={\rm distance}\times dF,$ where ${\rm distance}$ is the depth at which the slice is located, i.e., $y.$ Hence, $dW=7488 y\,dy.$ The total work is obtained by adding all these contributions, i.e., $W=\int_0^{20} 7488 y\,dy=7488\times 200=1497600$ ft-lb.

Problem 2 is exercise 9.2.16 from the book.

The curve is $r^2=-\cos\theta.$ Since $\cos\theta=\cos(-\theta),$ the graph is symmetric about the $x$-axis (because whenever $(r,\theta)$ is in the graph, then so is $(r,-\theta)$).

Since $(-r)^2=r^2,$ the graph is symmetric about the origin (because whenever $(r,\theta)$ is in the graph, then so is $(-r,\theta)$).

Since the graph is symmetric about both the origin and the $x$-axis, it is also symmetric about the $y$-axis.

To sketch the curve, look first at $0\le \theta< \pi/2.$ Here $\cos\theta>0,$ so $r^2<0,$ which is impossible, so there is nothing to graph here. Consider now what happens when $\pi/2\le\theta\le \pi.$ As $\theta$ increases, $-\cos\theta$ increases, from $0$ to $1.$ So the same occurs with $r^2.$ This means that $r$ increases from $0$ to $1$, and $-r$ decreases from $0$ to $-1.$ The part with $r$ gives us a curve in the second quadrant, and the part with $-r$ gives us its reflection about the origin. This part of the curve is in the fourth quadrant. Their reflections on the $x$-axis complete the curve, which can be seen here.

Note that $\tan(\pi/2)$ is undefined. This corresponds to the fact that at the origin the tangent to the curve is the $y$-axis, as can be seen from the graph.

## 175 – Quiz 1

September 4, 2009

Here is quiz 1.

Problem 1 is exercise 6.1.16 from the book. (The graph in the book has a typo. The maximum height of the graph is $1/2$ rather than 1.)

Using the disk method, the volume can be expressed as $\int_0^{\pi/2}\pi y^2\,dx.$ Here, $y^2=\sin^2 x\cdot \cos^2 x.$ If this expression were not squared, to find the integral would be easier, by a direct substitution. Being squared, we need to work harder. One possible approach is to use some trigonometric identities. For example:

• $\sin x\cdot \cos x=\displaystyle\frac{\sin 2x}2,$ so $y^2=\displaystyle\frac{\sin^2 2x}4.$ Again, the fact that the sine expression is squared makes things difficult. We use another trigonometric identity:
• $\sin^2\theta=\displaystyle\frac{1-\cos2\theta}2,$ so $y^2=\displaystyle\frac{1-\cos 4x}8.$ Now we can proceed to integrate:

$\displaystyle\int_0^{\pi/2}\pi y^2\,dx=\displaystyle\frac{\pi}8\int_0^{\pi/2} (1-\cos 4x)\,dx=\displaystyle\frac\pi8\left.\bigl(x+\frac{\sin 4x}4\bigr)\right|_0^{\pi/2},$ which simplifies to $\displaystyle\frac{\pi^2}{16}.$

There are other approaches. For example, other trigonometric identities could be used as well. Also, the book includes a formula for the integral of powers of sine and cosine; we will study this formula later. Trying to use the shell method leads to rather messy expressions, I didn’t work out the full details. I believe that the argument above is perhaps the most efficient.

Problem 2 is exercise 6.3.14 from the book. Since $x$ is given as a function of $y,$ the most efficient route seems to be to use $y$ as the parameter, so the expression for the length of the curve takes the form $\int_2^3\sqrt{(x')^2+1}\,dy,$ where the derivative of $x$ is with respect to $y.$

We have $x'=\displaystyle\frac{y^2}2-\frac1{2y^2},$ so $(x')^2+1=\displaystyle \frac{y^4}4-\frac12+\frac1{4y^4}+1=\displaystyle\frac{y^4}4+\frac12+\frac1{4y^4}=\displaystyle\bigl(\frac{y^2}2+\frac1{2y^2}\bigr)^2.$

The expression for the length then reduces to $\displaystyle\int_2^3\bigl(\frac{y^2}2+\frac1{2y^2}\bigr)\,dy=\displaystyle\left.\bigl(\frac{y^3}6-\frac1{2y}\bigr)\right|_2^3$ $=\displaystyle\bigl(\frac{27}6-\frac16\bigr)-\bigl(\frac86-\frac14\bigr)=\displaystyle3+\frac14=3.25.$