\input homework_macros.tex

%\distiller
\printer


\def\dd#1#2{\frac{d#1}{d#2}}
\def\sinh{\mathop{\rm sinh}\nolimits}
\def\cosh{\mathop{\rm cosh}\nolimits}
\def\tanh{\mathop{\rm tanh}\nolimits}
\def\coth{\mathop{\rm coth}\nolimits}
\def\sech{\mathop{\rm sech}\nolimits}
\def\csch{\mathop{\rm csch}\nolimits}

\def\arc#1{\mathop{\rm arc}\hskip-1.5pt#1}

\SN{\jobname}
\pn3.If $\cosh x=\frac{17}{15}$, $x>0$ find the values of the
remaining hyperbolic trig. functions.
By definition $\sech x=\frac{15}{17}$.
From $\cosh^2 x - \sinh^2 x=1$ we see
$\sinh^2 x=\br2{\frac{17}{15}}^2-1=\frac{289}{225}-1=
\frac{64}{225}$ so $\sinh x=\frac{8}{15}$.
By definition $\csch x=\frac{15}{8}$;
$\tanh x=\frac{\frac{8}{15}}{\frac{17}{15}}=\frac{8}{17}$; and
$\coth x=\frac{17}{8}$.

\pn15.If $y=2\sqrt t\ \tanh\sqrt t$, find $\dd yt$.
By the Product Rule, 
$\dd yt=\br1{\dd{2\sqrt t}t}\tanh\sqrt t+\br1{2\sqrt t}
\dd{\tanh\sqrt t}t$.
From the Power Rule,
$\dd{2\sqrt t}t=2\frac 12 t^{-1/2}=\frac{1}{\sqrt t}$.
From the formula for the derivative of $\tanh$ and the Chain Rule,
$\dd{\tanh\sqrt t}t=\br2{\sech^2(\sqrt t)}\dd{\sqrt t}t=
\br1{\sech^2\sqrt t}\br1{\frac 12 t^{-1/2}}=
\frac{\sech^2\sqrt t}{2\sqrt t}$.
Hence $\dd yt=\frac{\tanh\sqrt t}{\sqrt t}+\sech^2\sqrt t$.

\pn19.If $y=\sech\theta\ (1-\ln\sech\theta)$ find $\dd y\theta$.
From the Product Rule, it follows that
$\dd y\theta=\br1{\dd{\sech\theta}\theta} \br1{1-\ln\sech\theta}+
\br1{\sech\theta}\br1{\dd{(1-\ln\sech\theta)}\theta}$.
From the book, 
$\dd{\sech\theta}\theta=-\sech\theta \tanh\theta$.
From the Sum Rule, the Chain Rule applied to the $\ln$,
$\dd{(1-\ln\sech\theta)}\theta=0-
\br2{\frac{\dd{\sech\theta}\theta}{\sech\theta}}=
-\frac{-\sech\theta\tanh\theta}{\sech\theta}=\tanh\theta$.
Plugging back in,
$\dd y\theta=
\br1{-\sech\theta\tanh\theta} \br1{1-\ln\sech\theta}+
\sech\theta\tanh\theta=
(\sech\theta)(\tanh\theta)(\ln\sech\theta)$.


\pn31.If  $y=\arc\cos x-x\arc\sech x$, find $\dd yx$.
From the Sum Rule and the Product Rule,
$\dd yx=\dd{\arc\cos x}x-\br2{\dd xx\arc\sech x+
x\dd{\arc\sech x}x}$.
From formulae in the book,
$\dd yx=\frac{-1}{\sqrt{1-x^2\ }}-\br2{
\arc\sech x + x\frac{-1}{x\sqrt{1-x^2\ }}}=-\arc\sech x$


\pn33.If $y=\arc\csch\br2{\br1{\frac 12}^\theta}$
find $\dd y\theta$.
By the Chain Rule and a formula in the book,
$\dd y\theta=\frac{-\dd{\br1{\frac 12}^\theta}\theta}
{\br1{\frac 12}^\theta\sqrt{1+\br2{\br1{\frac 12}^\theta}^2\ }}=
\frac{\ln(\frac12)\br1{\frac 12}^\theta}
{\br1{\frac 12}^\theta\sqrt{1+\br2{\br1{\frac 12}^\theta}^2\ }}=
\frac{-\ln2}{\sqrt{1+\br2{\br1{\frac 12}^\theta}^2\ }}=
\frac{-2^\theta\ln2}{\sqrt{2^{2\theta}+1}}
$

\pn39.Verify
$\int x\arc\coth x\  dx=\frac{x^2-1}2\arc\coth x+\frac x2+C$.
Compute $\dd yx$ where $y=\frac{x^2-1}2\arc\coth x+\frac x2$.
By the Sum Rule and the Product Rule,
$\dd yx=\dd{\frac{x^2-1}2}x\arc\coth x+\frac{x^2-1}2\dd{\arc\coth x}x+
\frac 12$.
Plugging in the answers for the two derivatives (one by inspection 
and the other from the book)
$\dd yx=x\arc\coth x+\frac{x^2-1}2\frac{1}{1-x^2}+\frac12=
x\arc\coth x+\frac{-1}2+\frac12$.
The integral formula is verified.
One can {\sl derive\/} formulae like this using Integration by Parts (\S7.2).

\pn45.$\int \tanh\frac x7\ dx$.
The $\frac x7$ is annoying so substitute $u=\frac x7$ ($du=\frac{dx}7$).
$\int \tanh\frac x7\ dx=7\int\tanh u\>du$.
We concentrate on $\int\tanh u\>du$.
By analogy with the ordinary trig. functions, try the substitution
$w=\cosh u$. 
Then $dw=\sinh u\>du$ and
$\int\tanh u\>du=\int\frac{\cosh u}{\sinh u}\>du=
\int\frac{dw}w=\ln\vert w\vert+C=\ln\vert\cosh u\vert+C$.
By definition, $\cosh u\geq 1$ so
$\int\tanh u\>du=\ln\cosh u+C$ and
$\int \tanh\frac x7\ dx=7\ln\cosh(\frac x7)+C$.

\pn49.$\int\frac{\sech\sqrt t\tanh\sqrt t\ dt}{\sqrt t}$.
The $\sqrt t$ is annoying so substitute $u=\sqrt t$ 
($du=\frac{dt}{2\sqrt t}$).
$\int\frac{\sech\sqrt t\tanh\sqrt t\ dt}{\sqrt t}=
2\int \sech u \tanh u\ du$.
From the book, $\int \sech u \tanh u\ du=-\sech u\ +C$, so
$\int\frac{\sech\sqrt t\tanh\sqrt t\ dt}{\sqrt t}=
-2\sech \sqrt t\ +C$.

\pn53.$\int_{-\ln 4}^{-\ln 2}2e^\theta\cosh \theta\ d\theta$.
One could do this by Integration by Parts in analogy with 
$\int e^x\cos x\>dx$, but at this point in the book we have
only substitution.
If we recall that $\cosh \theta=\frac{e^\theta+e^{-\theta}}{2}$,
the substitution $u=e^\theta$ looks promising.
Well $du=e^\theta d\theta$ so
$\int_{-\ln 4}^{-\ln 2}2e^\theta\cosh \theta\ d\theta=
2\int_{\frac14}^{\frac12} \frac{u+u^{-1}}2\ du=
\int_{\frac14}^{\frac12}  u+u^{-1}\>du=\frac{u^2}2+\ln\vert u\vert
\Bigg\vert^{\frac12}_{\frac14}=
\br2{\frac{(\frac12)^2}2+\ln\frac12}-
\br2{\frac{(\frac14)^2}2+\ln\frac14}=
\br2{\frac18-\ln2}-\br2{\frac1{32}-\ln4}=
\frac3{32}+2\ln2-\ln2=\frac3{32}+\ln2
$

\pn77.{\bf Part a).}
With the verbiage removed, we are being asked to verify that
$v=\sqrt{\frac{mg}k}\tanh\br2{\br2{\sqrt{\frac{gk}m}}t}$ satisfies the
differential equation $m\dd vt=mg-kv^2$ with initial condition
$v(0)=0$.
In this problem, $m$, $g$ and $k$ are constants.
To simplify the writing (typing) let $a=\sqrt{\frac{gk}m}$ 
and check $v=\frac ga\tanh(at)$.
$v(0)=\frac ga\tanh(a\cdot0)=\frac ga\tanh 0=
\frac ga\frac{e^0-e^{-0}}{e^0+e^{-0}}=
\frac ga\cdot 0=0$, which verifies the initial condition.
Compute $\dd vt=\frac ga a \sech^2(at)=g\sech^2(at)$.
We are being asked whether
$m\dd vt$ is $mg-kv^2$.
Well, $m\dd vt=mg\sech^2(at)$ and
$mg-kv^2=mg-k\br1{\frac ga\tanh(ah)}^2=
mg-\frac{kg^2}{a^2}\tanh^2(at)=mg-mg\tanh^2(at)$.
Since $\sech^2 u=1-\tanh^2 u$ the two quantities are equal.
\item{}{\bf Part b).}
We are being asked to compute
$\displaystyle\lim_{t\to\infty}v(t)=
\displaystyle\lim_{t\to\infty}\frac ga\tanh(at)$.
Since $a>0$, $\displaystyle\lim_{t\to\infty}e^{at}=\infty$ and
$\displaystyle\lim_{t\to\infty}e^{-at}=0$.
Hence $\displaystyle\lim_{t\to\infty}\cosh(at)=
\displaystyle\lim_{t\to\infty}\sinh(at)=\infty$ so are
limit is of the form $\frac\infty\infty$ and hence a candidate for
L'H\^opital's Rule.
$\displaystyle\lim_{t\to\infty}\frac ga\tanh(at)=
\displaystyle\lim_{t\to\infty}\frac ga\frac{\sinh(at)}{\cosh(at}=
\frac ga\displaystyle\lim_{t\to\infty}\frac{a\cosh(at)}{a\sinh(at)}=
\frac ga\displaystyle\lim_{t\to\infty}\frac ga\coth(at)$.
This looks like no improvement so some other method must be
employed.
Recall $\tanh x=\frac{e^x-e^{-x}}{e^x+e^{-x}}$ so we may rewrite
$\tanh x=\frac{ 1-e^{-2x}}{1+e^{-2x}}$.
Now our limit becomes
$\displaystyle\lim_{t\to\infty}\frac ga\tanh(at)=
\frac ga \displaystyle\lim_{t\to\infty}
\frac{ 1-e^{-2atx}}{1+e^{-2at}}=\frac ga\frac{1-0}{1+0}=
\frac ga=\sqrt{\frac{mg}k}$.
\item{}{\bf Part c).}
We are being asked to numerically evaluate
$\sqrt{\frac{mg}k}$ when $mg=160$ and $k=0.005$ or to find
$\sqrt{32000}=\sqrt{3.2\cdot 10^4}=\sqrt{3.2}\cdot10^2=
1.78885\cdot 10^2$.
\end

