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\newp 11.
If we coordinatize so that the ladder is sliding down the $y${--}axis
and to the right along the $x${--}axis, then we are being asked
to compute $\frac{dy}{dt}$ when $x=10$ and $\frac{dx}{dt}=5$. 

The relation between $x$ and $y$ is that $x^2 + y^2 = 20^2$.
Differentiating with respect to $t$ we get
$2 x \frac{dx}{dt} + 2 y \frac{dy }{dt} = 0$, or
$\frac{dy}{dt} = - \frac{x}{y} \frac{dx}{dt}$.

When $x=10$, $10^2 + y^2 = 20^2$ or $y^2 = 300$ or $y=\pm10\sqrt{3}$ and
since we are on the positive side of the $y${--}axis, $y=10\sqrt 3$.
Hence
$\frac{dy}{dt} = - \frac{10}{10\sqrt{3}} \cdot 5= - \frac{5}{\sqrt{3}}$.
We also accepted $\frac{5}{\sqrt{3}}$ as the rate the top of the ladder
was sliding {\sl down}.

\endp
\newp 12.
$L(x)=f(a) + f^\prime(a)(x-a)$ with $f(x)=x^{1/3}$ and $x=27.3$
and $a=27$. $f^\prime(x)=\frac{1}{3}x^{-2/3}, \ f(a)=(27)^{1/3}=3,
\ f^\prime(a)=\frac{1}{3}\frac{1}{(27)^{2/3}} = \frac{1}{27}.$
Thus, $L(27.3)= 3 + \frac{1}{27}(27.3-27)= 3 + \frac{0.3}{27}.$

\endp
\newp 13.
Differentiate implicitly to get
$$15x^2y'+30xy+10=2(x^2+y^2)(2x+2yy')$$

Set $x=1$ and $y=2$ and solve for $y'$.

$$15y'+60+10=2(1+4)(2+4y')$$
$$50=25y'$$
$$y'=2$$

The tangent line is the line through $(1,2)$ with slope 2.
$$y=2+2(x-1)$$
$$y=2x$$
\endp
\newp 14.
Since $f(x) = x^3 - 12x$ is continuous on the interval $[-3,5]$, it
must have an absolute minimum and an absolute maximum and these must
occur at critical points or at endpoints.

The critical points are given by setting $f'(x) = 3x^2 - 12 = 0$ and
solving for $x$.  Since $3x^2 - 12 = 3(x-2)(x+2)= 0$, the critical points
are $x = \pm 2$.

Checking the values of $f(x)$ at $x = -3, -2, 2, 5$, we see that
$f(-3) = 9$,  $f(-2) = 16$, $f(2) = -16$, and $f(5) = 65$.

Thus on the interval $[-3,5]$, $f(x)$ has an absolute minimum of $-16$ at
$x = 2$ and an absolute maximum of $65$ at $x = 5$.

\endp
\newp 15.
$f(x)=\frac{1}{x}$ is differentiable for all real numbers $\neq 0$
since $f^\prime(x)=\frac{-1}{x^2}$.
Therefore $f$ is continuous for all real numbers $\neq 0$ since
a function differentiable at a point is continuous at that point.
But then $f$ is continuous on the subinterval $[1,4]$ and
differentiable on the subinterval $(1,4)$, so the Mean Value
Theorem applies to this function on this interval.

The equation that the MVT asserts has a solution is the equation
$$f^\prime(c)=\frac{f(4)-f(1)}{4-1}\hskip10pt{\rm and}\hskip 8pt 
c\in(1,4)$$
For us, $f^\prime(c)=\frac{-1}{c^2}$, $f(4)=\frac{1}{4}$, $f(1)=1$,
so $\frac{-1}{c^2}=\frac{\frac{1}{4}-1}{3}=\frac{-\frac{3}{4}}{3}=
-\frac{1}{4}$, so $c^2=4$.
This equation has solutions $c=\pm 2$, but the only solution
satisfying our additional condition is $c=2$.

\endp
\end
