\begin{homework}\
\label{homework:6}

\medskip

\exercise %1.
Find $d/dx$ of

\begin{center}
\begin{tabular}{ll}
 
(a) $x^4-3x^3+0.5x^2+7x-\pi$,  &
(b) $x^3-2x^2+4\sqrt{x}$,\\ 
\multicolumn{2}{l}{(c) $(x^2+1)^3$ (expand out the cube first),}\\
(d) $(x^2+1)^3$ (use the chain rule), 
&
(e) $x\sqrt{169-x^2}$,\\
(f) $(x^2-4x+5)\sqrt{25-x^2}$,  
&
(g) $(x^2+x+1)/(1-x)$,\\
(h) $\sqrt{25-x^2}/x$, 
&
(i) $\sqrt{(169/x)-x}$ (use the chain rule), \\
\multicolumn{2}{l}{(j) $\sqrt{(169/x)-x}$ (factor out the $1/x$ and use
the product or quotient rule),}\\
(k) $\sqrt{x^3-x^2-(1/x)}$,
&
(l) $100/(100-x^2)^{3/2}$,  \\
(m) ${\root 3 \of{x+x^3}}$,  
&
(n) $\sqrt{(x^2+1)^2+\sqrt{1+(x^2+1)^2}}$.
\end{tabular}\\

\end{center}

\medskip

\exercise %2.
Find the first derivative $d/dt$ and also the second derivative $d^2/dt^2$
of the following functions of $t$ (here $a$, $b$, and $c$ are constants):
(a) $-4.9t^2+20t+50$, (b) $at+b$, (c) $\sqrt{at+b}$, (d)
$\sqrt{at^2+bt+c}$.

\medskip

\exercise %3.
Suppose that the \index{ bicycle} bicycle in Example 6.8 has \index{ distance function} distance function 
$x(t)=50-5t+t^2$.  Find $v(t)=\dot{x}(t)$ and $a(t)=\ddot{x}(t)$.  Graph
$x(t)$ and $v(t)$, and explain in words what the bicycle is doing.

\medskip

\exercise %4.
If $f(x)=\sqrt{3+x^n}$, then $f(1)=2$.  Use the \index{ tangent line approximation} tangent line approximation 
to find $f(1+h)$.  Express your answer in terms of $n$ and $\Delta
x$.

\medskip

\exercise %5.
You want a \index{ ladder} ladder to reach a window which is about 24 feet above the
ground.  You place the bottom of the ladder exactly 7 feet away from the
side of the building.  If the window were exactly 24 feet high, you compute
that you would need 25 feet of ladder to reach it.  Suppose that the error
in your value of 24 for the height of the window is $\pm h$.  Find
the error in the figure of 25 feet for the length of ladder.


%\newpage

\medskip

\exercise %6.
In the situation of Problem 6 of the last homework, suppose that you want
to fit the triangle in the smallest possible {\it sphere} (rather than
circle).  In that case the triangle's hypotenuse becomes a diameter of the
sphere.  Find the volume of that smallest sphere, and include an error
estimate obtained using the tangent line approximation.

\medskip

\exercise %7.
Suppose that an object falling from a height of 80 meters (see Problem 4 of
Chapter 3) is viewed from a height of 2
meters by a person standing 10 meters from the side of the building where
the object is falling.  (a) Find a formula in terms of $t$ for the distance
from the person's eye to the object.  (b) Find a formula in terms of $t$
for the rate at which that distance is changing.

\end{homework}



