(Most) Final Exam Solutions

\#2b)

Most people did chain rule:

$f^{\prime}(x)=\frac{1}{(x^{2}+1)\sin x}\left(2x\cdot\sin x+(x^{2}+1)\cos x\right)$

Most errors resulted in not having parentheses, not knowing how to
do chain rule, or perhaps as simple as not knowing the derivative
of $\sin x$ is $\cos x$.

A simpler way to do the problem is by writing

$f(x)=\ln((x^{2}+1)\sin x)=\ln(x^{2}+1)+\ln(\sin x)$

So that

$f^{\prime}(x)=\frac{2x}{x^{2}+1}+\frac{1}{\sin x}\cos x=\frac{2x}{x^{2}+1}+\cot x$.

Note that doing it this way part of the problem comes for free provided
you did \#2a correctly.

\#2c)

This is standard from last exam. Most people took the natural log
of both sides. But many seem to forget that when you take the derivative
of $\ln y$ you should obtain $\frac{y^{\prime}}{y}$. Thus $\frac{y^{\prime}}{y}=(\cos x\ln(\sin x))^{\prime}=-\sin x\ln(\sin x)+\cos x\left(\frac{1}{\sin x}\cos x\right)$

Thus $y^{\prime}=y\left(-\sin x\ln(\sin x)+\cos x\left(\frac{1}{\sin x}\cos x\right)\right)$
$=\left(\sin x\right)^{\cos x}\left(-\sin x\ln(\sin x)+\cos x\left(\frac{1}{\sin x}\cos x\right)\right)$

Just like \#2b, many errors were from not knowing the derivative of
$\sin x$ is $\cos x$ and the derivative of $\cos x$ is $-\sin x$.

Not putting necessary parentheses or forgetting to substitute $\left(\sin x\right)^{\cos x}$
back for $y$ or even forgetting about $y$ altogether was a reason
to lose points.

Acceptible simplifications include:

$\left(\sin x\right)^{\cos x}\left(-\sin x\ln(\sin x)+\frac{\cos^{2}x}{\sin x}\right)$

$\left(\sin x\right)^{\cos x}\left(-\sin x\ln(\sin x)+\cos x\cdot\cot x\right)$

$-\left(\sin x\right)^{\cos x+1}\ln\sin x+\left(\sin x\right)^{\cos x-1}\cos^{2}x$

$\left(\sin x\right)^{\cos x+1}\left(-\ln\sin x+\cot^{2}x\right)$

\#2d)

This problem seemed to involve too many important ideas at once...

Also, it seems many students have trouble distinguishing a function
and it's derivative.

$f(x)=\int_{x}^{x^{2}}e^{\sin t}dt=\int_{0}^{x^{2}}e^{\sin t}dt-\int_{0}^{x}e^{\sin t}dt$

$f^{\prime}(x)=e^{\sin x^{2}}\cdot2x+e^{\sin x}$

There were many errors in the chain rule here. Unfortunately my posts
on the subject either went unseen or were incomprehensible.

Neither did anybody have a problem with the similar practice final
exam problem... and yet...

\#2e)

This was hard for many students to differentiate, as it requires not
only knowing how to use chain rule and product rule like the back
of your hand, but also algebraic manipulation.

$1+x=\sin(xy^{2})$

Differentiate:

$0+1=\cos(xy^{2})\left(y^{2}+x(2y\cdot y^{\prime})\right)$

From here there are two ways to solve for $y^{\prime}$.

I did what I thought was simpler:

$\left(\frac{1}{\cos(xy^{2})}-y^{2}\right)\frac{1}{2xy}=y^{\prime}$.

It was typical to see the arguably simpler: $\frac{1}{2xy\cos(xy^{2})}-\frac{y}{2x}=y^{\prime}$.

If you would like to see this done in more steps let me know.

Another way to solve for $y^{\prime}$ begins by distributing the
$\cos(xy^{2})$

$1=y^{2}\cos(xy^{2})+2xy\cos(xy^{2})y^{\prime}$

So $\frac{1-y^{2}\cos(xy^{2})}{2xy\cos(xy^{2})}=y^{\prime}$.

One can see this answer and the answer above are the same, simply
whether or not you choose combine the two terms over a common denominator.

\#3b)

You should accomplish this integral by realizing it is an odd function.

\#3c)

This integral is easiest to compute when realizing it is the area
of a quarter circle or radius $a$.

\#3d)

First, split up the integrals, you can treat each one separately.

$\int_{-1}^{2}|x+4|dx-3\int_{-1}^{2}|x-1|dx$.

You can choose to integrate these functions by treating them as piecewise
functions, or draw their graphs and do the problem geometrically.

I present the first method here:

$|x+4|=\begin{cases}
x+4 & \mbox{if }x+4\geq0\\
-x-4 & \mbox{if }x+4<0\end{cases}=\begin{cases} x+4 & \mbox{if }x\geq-4\\ -x-4 & \mbox{if }x<-4\end{cases}$ $|x-1|=\begin{cases} x-1 & \mbox{if }x-1\geq0\\ -x+1 & \mbox{if }x-1<0\end{cases}=\begin{cases} x-1 & \mbox{if }x\geq1\\ -x+1 & \mbox{if }x<1\end{cases}$

Of course you can do this in your head.

In any case,

$\int_{-1}^{2}\left(x+4\right)dx-3\int_{-1}^{1}\left(-x+1\right)dx-3\int_{1}^{2}\left(x-1\right)dx$

In the end you should get the answer is 6.

If you drew the graph,

For $x+4$ you get a trapezoid which you compute the area to be $\frac{3+6}{2}\cdot3=\frac{27}{2}$

For $|x-1|$ you get the triangle to the left of $x=1$ to be $-\frac{6\cdot2}{2}=-\frac{12}{2}$
and the triangle to the right of $x=1$ to be $-\frac{3\cdot1}{2}=-\frac{3}{2}$

Thus in total we have $\frac{12}{2}=6$. Same answer as by integrating.

\#4a)

This problem was difficult because you have to realize that you should
take the difference of functions:

Let $h(x)=g(x)-f(x)$.

Then $h$ is continuous on $[0,2]$ and differentiable on $(0,2)$
with $h(0)=0$ and $h^{\prime}(x)=g^{\prime}(x)-f^{\prime}(x)>0$
for all $0<x<2$. Thus $h(2)>0$. Otherwise, suppose $h(2)\leq0$. By the mean value
theorem $\exists c\in(0,2)$ such that $h^{\prime}(c)=\frac{h(2)-h(0)}{2-0}\leq0$.
Contradiction.

$g(2)-f(2)=h(2)>0$ implies $g(2)>f(2)$.

\#4b)

There are many examples to show the result is not true if $f$ is
not continuous at $0$ or $2$.

One example is

$f(x)=\begin{cases}
x & \mbox{if }x\in[0,2)\\
3 & \mbox{if }x=2\end{cases}$

$g(x)=2x$

\#5)

First differentiate to obtain:

$y^{2}+x2yy^{\prime}+2xy+x^{2}y^{\prime}=0$ .

Then we want to find when $y^{\prime}=0$ so let $y^{\prime}=0$.

We get $y^{2}+2xy=0$.

Thus $y(y+2x)=0$.

Then $y=0$ or $y\neq0$, $y=-2x$

Plugging $y=0$in the original equation we get $0=2$. Ridiculous.

Try the other solution: $x(4x^{2})+x^{2}(-2x)=2$

This simplifies to $2x^{3}=2$. We conclude $x=1$ and so $y=-2$.

Just to check with the derivative we get $4-4y^{\prime}-4+y^{\prime}=0$.
So $-3y^{\prime}=0$.

We conclude at the point $(1,-2)$ we have $y^{\prime}=0$.

\#6a)

$g(x)=\int_{0}^{x^{2}}\sin tdt$

$g^{\prime}(x)=2x\sin x^{2}$

$g^{\prime}(2x)=4x\sin(4x^{2})$

$g^{\prime}(2x)/x=4\sin(4x^{2})$

Because $\sin(4x^{2})$ has a maximum value of $1$ then $g^{\prime}(2x)/x$
has a maximum value of $4$.

There was difficulty with many students in plugging in $2x$.

COMMON ERROR: $g^{\prime}(2x)=2x\sin(4x^{2})$

ANOTHER COMMON ERROR: Not noticing to plug in $2x$.

THIRD COMMON ERROR: Confusing the derivative with the original function.

One way to go about the problem without using the fundamental theorem
of calculus would be to evalute the integral.

$g(x)=\int_{0}^{x^{2}}\sin tdt=-\cos t|_{0}^{x^{2}}=\left[-\cos x^{2}-\left(-\cos0\right)\right]=1-\cos x^{2}$

COMMON ERROR: Not knowing the antiderivative of $\sin t$ is $-\cos t$.

ANOTHER COMMON ERROR: Thinking that $\cos0=0$.

THIRD COMMON ERROR: Forgetting the minus sign, either from the $-\cos t$
or from the Fundamental Theorem of Calculus which says $\int_{a}^{b}f(t)dt=F(b)-F(a)$
where $F$ is an antiderivative of $f$.

In any case, then $g^{\prime}(x)=\sin x^{2}\cdot2x=2x\cdot\sin x^{2}$
by using the chain rule.

Note that in general this is how we get the chain rule for integrals.

$g(x)=\int_{v(x)}^{u(x)}f(t)dt=F(u(x))-F(v(x))$. Where $F^{\prime}(t)=f(t)$.

Thus, $g^{\prime}(x)=F^{\prime}(u(x))\cdot u^{\prime}(x)-F^{\prime}(v(x))\cdot v^{\prime}(x)=f(u(x))\cdot u^{\prime}(x)-f(v(x))\cdot v^{\prime}(x)$

For the special case $u(x)=x$ and $v(x)=a$, a constant, that is
$g(x)=\int_{a}^{x}f(t)dt$

We have $g^{\prime}(x)=f(x)$.

\#6b)

$g^{\prime}(x)=2x\sin x^{2}=0$ implies $x=0$ or $\sin x^{2}=0$.
Then second gives $x^{2}=k\pi$ for $k\in\mathbb{Z}$.

But $x\in[0,\sqrt{\pi/2}]$ implies that the only zeros of $g^{\prime}(x)$
are $x=0$.

This coincides with an endpoint.

$g(0)=0$.

$g(\sqrt{\pi/2})=\int_{0}^{\pi/2}\sin tdt=-\cos t|_{0}^{\pi/2}=1$

ERROR: To not look for critical points but only consider endpoints.

ERROR: To only look for critical points and forget to consider endpoints.

Various other errors overlap with errors made from \#6a).

One unfortunate error is even needing to evaluate $g(0)=\int_{0}^{0}\sin tdt$

This error indicates a lack of understanding. Intuitively, the integral
$\int_{a}^{b}f(x)dx$ is the area under the curve $f(x)$ from $a$
to $b$.

If we look at the are under the curve from $0$ to $0$ we wouldn't
have any area. So the integral would have to be $0$.

I do suppose it's more obvious from my point of view, but you should
learn to make it obvious from your point of view as well.

INTEGRAL=AREA UNDER THE CURVE.

\#7a)

NOTE: It's really not important what a definition of a curve is, but
the mathematical definition of a curve is different from the normal
use of the term. The mathematical definition doesn't require that
a curve bends.

If the two curves enclose a region, then they should intersect at
two points.

For a given $m$, the intersections between $mx$ and $\frac{x}{x^{2}+1}$
are found by solving the equation: $mx=\frac{x}{x^{2}+1}$

Then $m=\frac{1}{x^{2}+1}$ , $x\neq0$ or $x=0$.

Then $x=\pm\sqrt{\frac{1}{m}-1}$, $x\neq0$ or $x=0$.

Note that for the first to actually be a solution, it is necessary
that $\frac{1}{m}-1>0$. That is, $1>m>0$.

On the other hand, if $m$ is in this range, then the two curves intersect.

\#7b)

Choose any $m$. $m=\frac{1}{2}$ is a simple choice.

\#7c)

There are actually two enclosed regions, one for $x<0$ and one for $x>0$ but by symmetry, you the areas are equal.

Thus,

\begin{align*}
\mbox{Area of the enclosed region} & =2\int_{0}^{\sqrt{\frac{1}{m}-1}}\left(\frac{x}{x^{2}+1}-mx\right)dx\\
& =\left[\ln(x^{2}+1)-mx^{2}\right]_{0}^{\sqrt{\frac{1}{m}-1}}\\
& =\left[\ln\left(\frac{1}{m}\right)-m\left(\frac{1}{m}-1\right)\right]-\left[\ln(1)-0\right]\\
& =m-1-\ln(m)\end{align*}


\#8

We should choose the method of integration by the functions involved.
Though in fact when I solved 8a I used the harder method.

Easier method for

\#8a)

The height of a cylinder will be given by $(8-x^{2})-x^{2}=8-2x^{2}$.
We solve to know the range on which to integrate $x$. $8-x^{2}=x^{2}$
implies $x^{2}=4$ and so $x=\pm2$. The radius of a cylinder at a
point $x$ is $3-x$. This checks out intuitively, if we plug in $x=3$
we get radius 0, if we plug in $x=0$ we get radius $3$ and if we
plug in $x=-1$ we get radius $4$. Great.

$\mbox{Volume}=2\pi\int_{-2}^{2}(3-x)(8-2x^{2})dx=2\pi\int_{-2}^{2}24-8x-6x^{2}+2x^{3}dx$

I know the odd terms disappear so

$=2\pi\int_{-2}^{2}24-6x^{2}dx=4\pi\int_{0}^{2}24-6x^{2}dx=4\pi\left[24x-2x^{3}\right]_{0}^{2}=4\pi\left[48-16\right]=4\pi\cdot32=128\pi$

Harder method for

\#8a)

In the other direction I use symmetry to save work and integrate from
0 to 4 but double up my answer

Thus the bottom half is bounded by $-\sqrt{y}$ and $\sqrt{y}$.

Thus we have $3-(-\sqrt{y})=3+\sqrt{y}$ and $3-\sqrt{y}$.

$\mbox{Half Volume}=\pi\int_{0}^{4}(3+\sqrt{y})^{2}-(3-\sqrt{y})^{2}dy=\pi\int_{0}^{4}\left(9+6\sqrt{y}+y\right)-\left(9-6\sqrt{y}+y\right)dy$

$=\pi\int_{0}^{4}12\sqrt{y}dy=12\pi\left[\frac{2}{3}y^{3/2}\right]_{0}^{4}=8\pi\left[y^{3/2}\right]_{0}^{4}=64\pi$

Thus $\mbox{Volume}=128\pi$.

The expected method for

\#8b)

$\mbox{Volume}=\pi\int_{0}^{4}(2-\frac{x}{2})^{2}-(2-\sqrt{x})^{2}dx=\pi\int_{0}^{4}(4-2x+\frac{x^{2}}{4})-(4-4\sqrt{x}+x)dx=\pi\int_{0}^{4}(-3x+4\sqrt{x}+\frac{x^{2}}{4})dx$

$=\pi\left[\frac{-3}{2}x^{2}+\frac{4\cdot2}{3}x^{3/2}+\frac{1}{12}x^{3}\right]_{0}^{4}=\pi\left[-24+\frac{64}{3}+\frac{16}{3}\right]=\frac{8}{3}\pi$