## 175 – Final exam

December 15, 2009

Here is the final exam, and here are the solutions.

December 5, 2009

Here is quiz 8.

## 175 – Quiz 7

November 25, 2009

Happy Thanksgiving!

Here is quiz 7.

## 175 – Quiz 6

November 13, 2009

Here is quiz 6.

## 175 – Self test and Extra credit problems

November 12, 2009

Here are the two parts of the test (part 1, part 2), in case you want a blank copy. As I mentioned, they do not include all topics; they mostly cover material related to chapter 7 of the book, and even then, they are not comprehensive (for example, the second part does not include parts or word problems), but I hope you found them helpful in identifying topics that may require further study or review.

Here are two extra credit problems. Please let me know if you need me to clarify either one.

November 6, 2009

Here is quiz 5.

## 175 – Midterm 2

November 1, 2009

Here is the second midterm.

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.

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