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\begin{document}

\noindent
\vfil \noindent
\large
\hfil Math 126 C - Winter 2006 \hfil \pp
\hfil Mid-Term Exam Number Two\hfil  \pp
\hfil Solutions \hfil \pp
\hfil February 16, 2006 \hfil \pp
\normalsize

\begin{enumerate}

\item {\textit{Eliminate the parameter in the following parametric equation pair
to get a Cartesian equation for the curve that involves no trigonometric functions.
}}
\[
x = \cos t, y = \sin t - \cos t
\]


There are many different ways to solve this.  Here's one:

We know 
\[
\sin^2 t + \cos^2 t = 1
\]
and
\[
\cos t = x
\]
and 
\[
\sin t = y + \cos t = y + x
\]
so that
\[
(y+x)^2 +x^2 = 1
\]
and we're done.

Here's another way: notice that we can write
\[
\sin t = \pm \sqrt{1-\cos^2 t} = \pm \sqrt{1- x^2}
\]
so that 
\[
y = \pm \sqrt{1-x^2} - x.
\]



\item {\textit{Consider the curve defined parametrically by the parametric equations}}
\[
x = \ln \ln t, y = \ln t - (\ln t)^2.
\]
{\textit{
Find the equation of the tangent line to the curve at the
point $t=e$.}}

If $t=e$ then $x=0$, and $y=0$. 
We have
\[
\frac{dx}{dt} = \frac{1}{t \ln t} = \frac{1}{e}
\]
when $t=e$,
and
\[
\frac{dy}{dt} = \frac{1}{t} - 2 \left( \ln t \right) \frac{1}{t} = \frac{1}{e} - \frac{2}{e} = -\frac{1}{e}
\]
when $t=e$.

Thus, when $t=e$,
\[
\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = -1
\]
so the tangent line has equation $y=-x$.

%\item Find all points of intersection between the curve defined by the
%polar equation 
%\[
%r = 4 \csc \theta
%\]
%and the line $y=x$.  


%\newp

%\item Find the unit tangent vector $\vec{T}(t)$ at the point given by
%$\dsps t=\frac{\pi}{2}$ on the curve defined by the vector function
%\[
%\vec{r}(t) = 3 \sin t \vec{i} + 5 \cos t \vec{j} + 7 \sin^2 t \vec{k}
%\]

%\vfil

\item {\textit{Find the parametric equations for the tangent line to the
curve defined by
\[
x = t^3-t, y = t^6+t^2+1, z = \frac{1}{2}t^2+5t
\]
at the point $(0,1,0)$.}}

We find 
\[
\frac{dx}{dt} = 3t^2-1
\]
\[
\frac{dy}{dt} = 6t^5+2t
\]
\[
\frac{dz}{dt} = t+5
\]

The point $(0,1,0)$ corresponds to $t=0$.  To see this, note that we need
\[
t^3-t=0
\]
which tells us
\[
t(t^2-1)=0
\]
so we know $t=0$, $t=1$ or $t=-1$.  Checking these values with
$y=t^6+t^2+1$, we find that only $t=0$ works.

Plugging $t=0$ into the derivatives above, and using the point 
$(0,1,0)$, we have the tangent line equations
\[
x = -t, y=1, z = 5t.
\]

\item {\textit{At what point does the curve $y=e^x$ have maximum curvature?}}

We have a formula to find the curvature function $\kappa(x)$ for the graph of 
a given 
function $f(x)=e^x$:
\[
\kappa(x) = \frac{|f''(x)|}{(1+(f'(x))^2)^{3/2}}= \frac{e^x}{(1+e^{2x})^{3/2}}
\]
Note that we have used the fact that $e^x>0$ for all $x$ to remove the
absolute value symbol.

Now we want to find out how large this function $\kappa(x)$ can get.
We can start our search for maxima of this function be studying the function's 
first derivative.

\[
\kappa'(x) = 
\frac{e^x (1+e^{2x})^{3/2} - e^x \frac{3}{2}(1+e^{2x})^{1/2}2x}{(1+e^{2x})^3}
\]
\[
=\frac{e^x(1+e^{2x})^{1/2}\left( (1+e^{2x}) - 3e^{2x} \right)}{(1+e^{2x})^3}
=\frac{e^x(1+e^{2x})^{1/2}(1-2e^{2x})}{(1+e^{2x})^3}
\]
We note that this is defined for all $x$, so the only critical points will occur
where this is zero.  If this is zero, then
\[
1-2e^{2x}=0
\]
since $e^x>0$ for all $x$, and $(1+e^{2x})^{1/2}>0$ for all $x$.

Hence, the only critical point is at
\[
x = \frac{1}{2}\ln \frac{1}{2}.
\]
Since $2e^{2x}$ is a strictly increasing function, we can see that 
that $\kappa'(x)$ is going to be negative for $x$ greater than the value
we just found, and it will be positive for $x$ less than that value.
In other words, we can conclude that this value of $x$ gives us the
maximum value of $\kappa(x)$.

The point on the curve where curvature is maximum is thus
\[
\left( \frac{1}{2}\ln \frac{1}{2}, \frac{1}{\sqrt{2}} \right)
\]

\item Find the length of the curve defined by
\[
\vec{r}(t) = \left\langle \frac{2\sqrt{2}}{3} t^{3/2},t, \frac{1}{2}t^2 \right\rangle,
0 \le t \le 4
\]

Conveniently,
\[
\left(
\frac{d}{dt} 
\frac{2\sqrt{2}}{3} t^{3/2}
\right)^2
+
\left(
\frac{d}{dt}
t
\right)^2
+
\left(
\frac{d}{dt}
\frac{1}{2}t^2
\right)^2
= (t+1)^2
\]
so the arc length is just
\[
\int_0^4 (t+1) \, dt = 12.
\]


\item {\textit{Find the curvature of the curve defined by
\[
\vec{r}(t) = \left\langle \frac{1}{2}t^2-2t,t^2-t,t^2+t \right\rangle
\]
at the point $t=0$.}}

We have two useful formulas for finding the curvature of a 
3D curve.  One is
\[
\kappa(t) = \frac{|\vec{T}'(t)|}{|r'(t)|}
\]
and the other is
\[
\kappa(t) = \frac{|\vec{r}\,'(t) \times \vec{r}\,''(t)}{|\vec{r}\,'(t)|^3}
\]
For this problem, the second equation is {\textbf{much}} easier to use.
I really can't think of a time when it would be preferable to use the first one,
expect in that rare occasion where you are given $\vec{T}'(t)$ and
don't have to derive it from $\vec{r}(t)$.

So we use the second formula.  Since we are interested in $\kappa(0)$,
we need only find $\vec{r}\,'(0)$ and $\vec{r}\,''(0)$ and plug them into
the formula:
\[
\vec{r}\,'(t) = \langle t-2,2t-1,2t+1 \rangle
\]
\[
\vec{r}\,'(0) = \langle -2,-1,1 \rangle
\]
\[
\vec{r}\,''(t) = \langle 1,2,2 \rangle
\]
\[
\vec{r}\,''(0) = \langle 1,2,2 \rangle
\]
Plugging these into our formula gives us
\[
\kappa(0) = \frac{\sqrt{50}}{(\sqrt{6})^3} = \frac{5}{6 \sqrt{3}}.
\]













\end{enumerate}




\end{document}


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