\input homework_macros.tex

%\distiller
\printer

\def\dd#1#2{\frac{d#1}{d#2}}

\SN{6.3}
\pn1.\hskip8true pt a) $e^{\ln 7.2}=7.2$: since  ($e^{\ln x}=x$).
\itemitem{b)} $e^{-\ln\>x^2}=e^{-\ln(x^2)}=\frac1{e^{\ln(x^2)}}=
\frac 1{x^2}=x^{-2}$.
\itemitem{c)} $e^{\ln x-\ln y}=\frac{e^{\ln x}}{e^{\ln y}}=\frac xy$.

\pn3.\hskip8truept a) $2\ln\sqrt e=2\ln(e^{1/2})=
2\cdot\frac 12\cdot\ln e=
2\frac 12\cdot 1=1$.
\itemitem{b)} $\ln(\ln e^e)=\ln( e\ln e)=\ln(e\cdot 1)=\ln e=1$.
\itemitem{c)} $\ln(e^{-x^2-y^2})=-x^2-y^2$: ( since $\ln e^z=z$).

\pn5.Solve $\ln y=2t+4$, so $y=e^{\ln y}=e^{2t+4}$.

\pn9. $\ln(y-1)-\ln 2 = x+\ln x$, so
$\ln(y-1)=x+\ln x+\ln 2$ and
$y-1=e^{\ln(y-1)}=e^{x+\ln x+\ln2}=e^x\cdot e^{\ln x}\cdot e^{\ln 2}=
2xe^x$, so $y=1+2xe^x$.

\pn13.\hskip8true pt a) Solve $e^{-0.3t}=27$: $(-0.3)t=\ln(e^{-0.3t})=\ln(27)=
\ln(3^3)=3\ln3$.
Hence $t=\hfill$\par\hskip\parindent$-\frac{3\ln 3}{0.3}=-10\ln3$.
\itemitem{b)} Solve $e^{kt}=\frac 12$: 
$kt=\ln(e^{kt})=\ln(\frac 12)=-\ln 2$, so
$t=-\frac {\ln2}k$.
\itemitem{c)} Solve $e^{(\ln 0.2)t}=0.4$:
$(\ln0.2)t=\ln\br1{e^{(\ln 0.2)t}}=\ln(0.4)$, so
$t=\frac{\ln 0.4} {\ln0.2}$.

\pn19. $y=e^{5-7x}$: find $\dd yx$. $\dd yx= e^{5-7x}\dd {(5-7x)}x=
e^{5-7x}(-7)=-7e^{5-7x}$.

\pn23. $y=(x^2-2x+2)e^x$: find $\dd yx$. 
$\dd yx=\br1{\dd {(x^2-2x+2)}x} e^x+\br0{x^2-2x+2}\dd{(e^x)}x=
(2x-2)e^x+\br0{x^2-2x+2} e^x=x^2\>e^x$.

\pn27. $y=\cos\br2{e^{-\theta^2}}$: find $\dd y\theta$.
$\dd y\theta=-\br3{\sin\br2{e^{-\theta^2}}}
\dd{\br2{e^{-\theta^2}}}\theta=
-\br3{\sin\br2{e^{-\theta^2}}}\br3{e^{-\theta^2}}\dd{(-\theta^2)}{\theta}=
-\br3{\sin\br2{e^{-\theta^2}}}\br3{e^{-\theta^2}}(-2\theta)=
2\theta e^{-\theta^2}\sin\br2{e^{-\theta^2}}$.

\pn37. $\ln y=e^y\sin x$: find $\dd yx$.
This uses implicit differentiation, or more fundamentally, the Chain Rule.
Differentiate both sides with respect to $x$, treating $y$ as a function 
of $x$.
$\frac{y^\prime}{y}=\dd{(e^y)}x\sin x+e^y\dd{(\sin x)}x=
e^yy^\prime\sin x+e^y\cos x$.
Now solve for $y^\prime$ in terms of $x$ and $y$:
$y^\prime=y^\prime ye^y\sin x + ye^y\cos x$, or
$y^\prime-y^\prime ye^y\sin x=ye^y\cos x$ or
$y^\prime=\frac{ye^y\cos x}{1-ye^y\sin x}$.


\pn41. $\int \br 1{e^{3x}+5e^{-x}}\> dx=
\int e^{3x}\> dx+5\int e^{-x}\> dx=
\frac{e^{3x}}3+5\frac{e^{-x}}{-1}+C=\frac{e^{3x}}3-5e^{-x}+C$.

\pn49. $\int \frac{e^{\sqrt r}}{\sqrt r} dr$. Substitute $t=\sqrt r$:
$dt=\frac 12 r^{-1/2}dr=\frac 1{2\sqrt r} dr$, or
$\frac{dr}{\sqrt r}=2dt$ and our integral becomes
$\int e^t(2dt)=2e^t+C=2e^{\sqrt r}+C$.

\pn55. $\int_0^{\frac{\pi}4} \br1{1+e^{\tan\theta}}\sec^2\theta\> d\theta$.
Substitute $s=\tan\theta$: $ds =\sec^2\theta d\theta$.
Hence our integral becomes
$\int_0^1 (1+e^s)\> ds=(s+e^s)\Big\vert_0^1=
(1+e)-(0+e^0)=1+e-1=e$.

\pn63. Solve $\frac{dy}{dt}=e^t\sin\br1{e^t-2}$, $y(\ln2)=0$.
First integrate to find all functions satisfying the differential
equation:
$y=\int e^t\sin\br1{e^t-2}\> dt$: substitute $s=e^t$: $ds=e^t dt$
so $y=\int \sin(s-2)\> ds=-\cos(s-2)+C=
-\cos(e^t-2)+C$.
Now $y(\ln2)=-\cos(e^{\ln 2}-2)+C=-\cos(2-2)+C=-\cos 0+C=C-1$.
But $y(\ln2)$ is also $0$, so $C=1$ and
$y=1-\cos(e^t-2)$.

\pn65. $\frac{d^2y}{dx^2}=2e^{-x}$, $y(0)=1$, $y^\prime(0)=0$.
Integrate twice.
$\dd yx=\int 2e^{-x}=2\frac{e^{-x}}{=1}+C=-2e^{-x}+C$.
When $x=0$ $y^\prime=0$ and also $-2e^{-0}+C=C-2e^0=C-2$.
Hence $C=2$ and $y^\prime=2-2e^{-x}$.
But then $y=\int (2-2e^{-x})\> dx=2x-(-2e^{-x})+C=2x+2e^{-x}+C$.
Since $y(0)=1$, $1=2\cdot 0+2e^{-0}+C=2+C$ so $C=1$ and
$y=1+2x+2e^{-x}$.

\pn69. Find absolute minimum value of $f(x)=x^2\ln \br1{\frac 1x}$.
Locate critical points: solve $0=f^\prime(x)=
2x\ln\br1{\frac 1x}+x^2\br2{
\frac{\frac{d(\frac 1x)}{dx}}{\frac 1x}}=
-2x\ln x+x^2\br1{-x^{-2}\cdot x}=
-2x\ln x-x=(-x)(1+2\ln x)$.
This product vanishes if $x=0$ or if $1+2\ln x=0$.
The point $x=0$ is not in the domain of $f$ so it is not a critical point.
Hence the only critical point occurs when $1+2\ln x=0$, or
$\ln x=-\frac 12$ and $x=e^{\ln x}=e^{-1/2}=\frac 1{\sqrt e}$.
The value of $f$ at $x=\frac 1{\sqrt e}$ is 
$(\frac 1{\sqrt e})^2\ln\br2{\frac{1}{\frac 1{\sqrt e}}}=-\frac 1{e}
\cdot\ln\sqrt e=-\frac 1e\cdot\ln\br1{e^{1/2}}=
-\frac 1e\cdot(\frac 12)=-\frac1{2e}$.
We still have to see that $f$ has a relative maximum.
Compute $f^\prime(1)=(-1)\br1{1-2\ln1}=-(1-2\cdot0)=-1<0$, 
so $f$ is idecreasing to the right of $\frac 1{\sqrt e}$ (note $1<\sqrt e$).
Compute $f^\prime(e^{-2})=
-e^{-2}\br1{3+2\ln(e^{-2})}=-e^{-2}\cdot(3-4)=e^{-2}>0$ so $f$ is
increasing to the left of $\frac 1{\sqrt e}$.
As an alternative to show $f$ has a local minimum at $\sqrt e$ apply
the 2nd Derivative Test.
$\frac{d^2f}{dx^2}=-\frac{d\br2{x(1+2\ln x)}}{dx}=
-\br 2{1\cdot(1+2\ln x)+x\cdot(\frac 2x)}=-(1+2\ln x+2)=-(3+2\ln x)$.
Hence $f^{\prime\prime}(\frac 1{\sqrt e})=
-\br1{3+2\ln(\frac 1{\sqrt e})}=-(3-1)=-2<0$,
so the 2nd Derivative Test shows $f$ has a local maximum at 
$x=\frac1{\sqrt e}$.
Since $f$ has only the one critical point and is continuous, 
the local maximum must be an absolute (or global) maximum.
%%%
\pn71.
Draw a graph and determine that the region is bounded on the left
by the vertical line $x=0$ (since this is where the two graphs intersect);
on the left by $x=\ln 3$ (we were told this); above by $y=e^{2x}$ and
below by $y=e^x$.
The area is given by $\int_0^{\ln 3}(e^{2x}-e^x)\> dx$ and this is the
formula which should be familiar to you (using calculus to find the area
of planar regions \S5.1).

\pn72. Same ideas as 71.  The curves still intersect at $x=0$, so the area
is $\int_0^{2\ln 2}(e^{x/2}-e^{-x/2})\> dx$.

\pn73. The length of the graph of $y=f(x)$ between the point 
$\bigl(a,f(a)\bigr)$ and the point $\bigl(b,f(b)\bigr)$ is
$\int_a^b\sqrt{1+{\br1{f^\prime(x)}}^2\ } dx$.
Once you recall this formula (\S5.5) and stare at the problem, you
see you are being asked to find a function $f(x)$ such that
$f^\prime(x)=\sqrt{\frac 14 e^x}=\frac{e^{x/2}}2$.
$f(x)=e^{x/2}$ is such a function.
Any function of the form $e^{x/2}+C$ is also such a function, so
the problem has many answers.

\pn74. From \S5.6, we recall that the surface area of the surface of
revolution obtained by revolving the graph $s=f(t)$ about the $t$ axis
for $a\leq t\leq b$ is given by
$$2\pi\int_a^b f(t)\sqrt{1+{\br1{f^\prime(t)}}^2\ } dt\ .$$
In our case $s=x$; $y=t$; $f(t)=\frac{e^t+e^{-t}}2$; $a=0$; $b=\ln2$.
$f^\prime(t)=\frac{e^t-e^{-t}}2$;
${\br1{f^\prime(t)}}^2=\frac{e^{2t}-2+e^{-2t}}4$ so
${1+{\br1{f^\prime(t)}}^2\ }=\frac{e^{2t}+2+e^{-2t}}4=
{\br3{\frac{e^t+e{-t}}2}}^2$, so our integral is
$\int_0^{\ln2} \frac{e^t+e{-t}}2\sqrt{\br3{\frac{e^t+e{-t}}2}^2\ } dt=
\int_0^{\ln2} {\br2{\frac{e^t+e^{-t}}2}}^2 dt=\int_0^{\ln2}
\frac{e^{2t}+2+e^{-2t}}4 dt=\cdots$.

\end