\documentclass{article}

\setlength{\parindent}{0pt}
\everymath={\displaystyle}
\newcommand{\problemspace}{\bigskip\hrule\bigskip}

\begin{document}

\centerline{\textbf{Math 126: Calculus II, Exam I Solutions}}

\bigskip

1. a) Let $f(x) = x\ln x$. Find the value of $(f^{-1})'(2e^2)$.

$f'(x) = \ln(x) + x\cdot\frac{1}{x} = \ln(x) + 1$.
\newline Since $f(e^2) = 2e^2$, $(f^{-1})'(2e^2) = \frac{1}{f'(e^2)} =
\frac{1}{\ln(e^2) + 1} =\frac{1}{2+1} = \frac{1}{3}$.

b)  $\frac{d}{dx}\cos^{-1}(x^{-1}) = -\frac{1}{\sqrt{1 -
(x^{-1})^2}}\big(-x^{-2}\big) = \frac{1}{x^2\sqrt{1-(1/x^2)}} =
\frac{1}{|x|\sqrt{x^2 - 1}}$.

c) $\frac{d}{dx}x^{\cosh(x)} = \frac{d}{dx}e^{\cosh(x) \ln(x)}
= e^{\cosh(x) \ln(x)}(\sinh(x)\ln(x) + \cosh(x)(1/x))
\newline = x^{\cosh(x)}(\sinh(x)\ln(x) + \cosh(x)/x)$.

\problemspace

2. a) $\int e^{x}\sin(1 + e^{x})\,dx$.  Let $u = 1 + e^x$, $du = e^x\,dx$.
Then $\int e^{x}\sin(1 + e^{x})\,dx = \int \sin(u)\,du = -\cos(u) + C =
-\cos(1+e^x) + C$.

b) $\int_{0}^{1}\frac{x}{1+x^4}\,dx$. Let $u = x^2$, $du = 2x\,dx$,
or $x\,dx = (1/2)du$, and the new limits are from $u = 0^2$ to $u =
1^2$. Thus $\int_{0}^{1}\frac{x}{1+x^4}\,dx = \frac{1}{2}\int_0^1
\frac{1}{1+u^2}\,du = \frac{1}{2}\tan^{-1}(u)\Big|_0^1 =
\frac{1}{2}[\tan^{-1}(1) - \tan^{-1}(0)] = (1/2)[\pi/4 - 0] = \pi/8$.

c) $\int \frac{x}{x+1}\,dx$. First divide: $\frac{x}{x+1} = 1 -
\frac{1}{x+1}$. Then $\int \frac{x}{x+1}\,dx = \newline \int 1 -
\frac{1}{x+1}\,dx = x - \ln|x+1| + C$.

\problemspace

3. a) $\lim_{x\to 0} \frac{\sin^{-1}(x)}{x}$. The numerator and
denominator both approach 0, so by L'H\^opital's Rule the limit equals
$\lim_{x\to 0}\frac{1/\sqrt{1-x^2}}{1} = 1$.

b) $\lim _{x\to 0^+} e^{-1/x}\ln(x)$. Note that $e^{-1/x}\to 0$
and $\ln(x)\to \infty$. Rewrite the limit as $\lim _{x\to 0^+}
\frac{\ln(x)}{e^{1/x}}$ so both the numerator and denominator
approach infinity, and then apply L'H\^opital's Rule: $\lim _{x\to 0^+}
\frac{\ln(x)}{e^{1/x}} = \lim _{x\to 0^+} \frac{1/x}{e^{1/x}(-1/x^2)}
= \lim _{x\to 0^+} -\frac{x}{e^{1/x}}$. The numerator approaches 0 and
the denominator approaches infinity, so the limit is~0.

\problemspace

4. Decide whether $g(x)$ grows faster than, slower than, or at the same
rate as $f(x)$.

a) $f(x)=\frac{1}{x}$ and $g(x) = \sin(\tan^{-1}(x))$.
Since $\tan^{-1}(x)\to\pi/2$  and $1/x\to0$ as
$x\to\infty$, $\lim_{x\to\infty}\frac{f(x)}{g(x)} =
\lim_{x\to\infty}\frac{1/x}{\sin(\tan^{-1}(x))} = \frac{1}{\sin(\pi/2)}
=\frac{0}{1} = 0$.  (Note that L'H\^opital's Rule does not apply.)
Therefore, $g(x)$ grows faster than $f(x)$.

\bigskip

b) $f(x)=\sinh(\ln(x))$ and $g(x) = x$. Since $\sinh(\ln(x)) = (e^{\ln(x)}
- e^{-\ln(x)})/2 = (x - 1/x)/2$, $\lim_{x\to\infty}\frac{f(x)}{g(x)}
= \lim_{x\to\infty}\frac{(x - 1/x)/2}{x} =\lim_{x\to\infty}(1 -
1/x^2)/2 = 1/2$. (note that repeatedly applying L'H\^opital's Rule to
$\sinh(\ln(x))/x$ brings you back to the beginning.)  Therefore, $g(x)$
grows at the same rate as $f(x)$.

\problemspace

5. Warfarin is a drug used as an anticoagulant.  After
administration of the drug ends, the quantity remaining in a
patient's body decreases at a rate proportional to the quantity
remaining.  The half-life of Warfarin in the body is 37 hours.  How
many days does it take for the drug level in the body to be reduced
to 10\% of the original level.  

\bigskip

\emph{Solution.} Let $y(t)$ be the quantity remaining after~$t$
hours and let $y_0 = y(0)$ be the (unknown) initial amount. We must
solve for the time $t_1$ such that $y(t_1) = 0.1 y_0$. Since $y(t)$
decays exponentially, $y(t) = y_0 e^{kt}$.  The half life is 37 hours
which means that $(1/2)y_0 = y(37) = y_0e^{37k}$. Solving for $k$
gives $k = \ln(1/2)/37$.  Therefore, $y(t) = y_0 e^{\ln(1/2)t/37} =
y_0\big(\frac{1}{2}\big)^{t/37}$. We now solve for $t_1$: $0.1y_0 =
y(t_1) = y_0\big(\frac{1}{2}\big)^{t_1/37}$.  Taking logarithms gives
$\ln(0.1) = (t_1/37)\ln(0.5)$ so $t_1 = 37\ln(0.1))/\ln(0.5) = 122.9$
hours. The number of days is $t_1/24 = 122.9/24 = 5.1$.

\problemspace

6. a) Solve the differential equation\quad $x y' + 2 y = x$.

The standard form is $y' + \frac{2}{x}y = 1$, so $p(x) = \frac{2}{x}$ and
$q(x) = 1$. Multiplying the equation by $v(x) = e^{\int 2/x\,dx} =
e^{2\ln(x)} = x^2$ gives $x^2 y' + 2x y = x^2$. The left-hand side is the
derivative of $x^2y'$ so $\frac{d}{dx}\big(x^2 y\big) = x^2$.
Integrating gives, $x^2 y = \frac{1}{3}x^3 + C$, or
$y = \frac{x}{3} + \frac{C}{x^2}$.

\bigskip

b) Solve the initial value problem\quad $x^2 y' = y$,\quad $y(1) = 1$.

Separate the variables, $\frac{dy}{y} = \frac{dx}{x^2}$, and
integrate, $\int \frac{dy}{y} = \int \frac{dx}{x^2}$. This gives $\ln(y)
= -\frac{1}{x} + C$. Plug in the initial condition $y = 1$ when $x = 1$:
$\ln(1) = -1 + C$, so $C = 1$. Exponentiating gives $y = e^{1-1/x}$.

\end{document}


