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\begin{document}

\begin{center}
{\Large TEST 1 - KEY

Math 2210 - 2}
\end{center}

\hrule

\begin{enumerate}

\item (8 points) Draw a plane curve that satisfies the conditions:
(so you will draw 4 curves, one for each set of conditions)
\begin{enumerate}
\item  Simple and closed
\item  Simple but not closed
\item  Not simple, but closed
\item  Not simple, not closed
\end{enumerate}
\answer: See the book for this, there are lots of pictures and explainations there.

\item (10 points) Consider the parametric curve defined by the parametric equations:
\begin{eqnarray*}
x(t) & = & 2t-1 \\
y(t) & = & t-4
\end{eqnarray*}
\begin{enumerate}
\item  Eliminate the parameter $t$ to get an equation in $x$ and $y$.

\answer: You can see that $t=(x+1)/2$, so we plug this in to the equation with $y$ and get:
\begin{eqnarray*}
y & = & (x+1)/2-4 = \frac{1}{2}x - \frac{7}{2}
\end{eqnarray*}

\item  Graph the curve for $0 \leq t \leq 5$.  Indicate the initial and end points on your graph.

\answer:  The graph is a line.  You should be able to graph the line.
The only tricky thing is to use the interval $0 \leq t \leq 5$.
On this interval, $y$ starts at $-4$ and goes to $1$, so this is the section of the line to graph.
\end{enumerate}


\item (10+2 points) Consider the parametric curve given by the equations below:
\begin{eqnarray*}
x(t) & = & e^{2t}-5t \\
y(t) & = & \sin(t)+2
\end{eqnarray*}
\begin{enumerate}
\item  Find the slope of a line tangent to the curve when $t=0$.

\answer:  We just compute:
\begin{eqnarray*}
\frac{dy}{dx} & = & \frac{dy/dt}{dx/dt} = \frac{\cos t}{2e^{2t}-5}
\end{eqnarray*}
To find the slope when $t=0$, we just plug in $0$ for $t$:
\[
\frac{dy}{dx}|_{t=0} = \frac{\cos(0)}{2-5} = -\frac{1}{3}
\]

\item  Are there any points on the curve where the tangent line is horizontal?  Justify your answer.

\answer:  A horizontal tangent line corresponds to $dy/dx = 0$, and this happens if $\cos t=0$.  There are
lots of ways to make $\cos t=0$ (for example $t=\pi/2$), so yes there are points with horizontal tangent line.

\item  Are there any points on the curve where the tangent line is vertical?  Justify your answer.

\answer: we look for points where $dy/dx$ is undefined, or when $2e^{-2t}-5=0$, and this has a solution.
So there is at least one (there is actually exactly one) point with vertical tangent line.

\item  (Extra Credit)  Is there an interval $a \leq t \leq b$ so that the curve is not a simple curve?
	Justify your answer!

\answer:
First, I meant to put $y=\sin t +2t$.  In this case the answer to the extra credit would be easy because
then $dy/dt=\cos t +2$ which is always positive.  In other words, the $y$ coordinate is always increasing and
there could never be two $t$ values with the same $y$-value.

But, this was not the problem.
So, the easy solution is to graph it and see that in fact, there are points where the curve crosses itself.
But, there were no calculators allowed on the test.
I will explain some reasoning in class that would lead you to believe that the curve does cross itself, or you 
can come and we can discuss it together.
\end{enumerate}

\item (10 points)  Consider the vectors below:
\begin{eqnarray*}
\mbf{v} & = & \mbf{i} + 3\mbf{j} \\
\mbf{u} & = & \sqrt{3}\mbf{i} - \frac{\sqrt{3}}{6}\mbf{j}
\end{eqnarray*}
Find the cosine of the angle between $\mbf{v}$ and $\mbf{u}$.

\answer:
To do this, we first compute dot products and norms:
\begin{eqnarray*}
u \dotprod v & = & \sqrt{3} - \frac{3 \sqrt{3}}{6} = \frac{\sqrt{3}}{2} \\
|u| & = & \sqrt{1^2+3^2} = \sqrt{10} \\
|v| & = & \sqrt{3+(3/36)} = \sqrt{\frac{37}{12}} = \frac{\sqrt{111}}{6}
\end{eqnarray*}
We now just plug these into the formula for cosine of the angle:
\begin{eqnarray*}
\cos \theta & = & \frac{u \dotprod v}{|u| |v|} 
	= \frac{\sqrt{3}/2}{\sqrt{10}\sqrt{111}/6} \\
	& = & \frac{3\sqrt{370}}{370} = \frac{3}{\sqrt{370}}
\end{eqnarray*}

\item (10 points) Consider a particle moving in the plane with position vector given by
\[
\mbf{r}(t) = e^{-3t}\mbf{i} - (2t+1)^3 \mbf{j}
\]
\begin{enumerate}
\item  Find the position of the particle at time $t=0$.

\answer:  We just plug in $t=0$ and get $\mbf{r}=(1,-1)$.

\item  Find the velocity of the particle at time $t=0$.

\answer:  We take the derivative:
\begin{eqnarray*}
\mbf{v}(t) & = & \mbf{r}'(t) = (-3e^{-3t}, -6(2t+1)^2)
\end{eqnarray*}
We plug in $0$ and see that $\mbf{v}(0)=(-3,-6)$.

\item  Find the speed of the particle at time $t=0$.

\answer:  Here we just compute the magnitude of the velocity vector (thats what speed is).
\[
|v(0)| = \sqrt{(-3)^2+(-6)^2} = \sqrt{45}=3\sqrt{5}
\]
\item  Find the acceleration of the particle at time $t=0$.

\answer:
We take the derivative again:
\begin{eqnarray*}
\mbf{a}(t) & = & \mbf{v}'(t) = \mbf{r}''(t) = (9e^{-3t}, -24(2t+1))
\end{eqnarray*}
And, we plug in $0$ and get $\mbf{a}(0)=(9,-24)$.

\item  Is the particle moving at constant speed?  (Justify your answer!)

\answer:
There are several incomplete ways that people looked at this problem.
Here are the common ones and why they are wrong.
The basic counter example to the reasoning is circular motion: $x=\cos t$, $t=\sin t$.
\begin{enumerate}
\item  Agrument: The velocity is clearly not a constant vector, therefore speed is not constant.
	Why arguemtn is wrong:
	For the circular motion above, you can see (by computing) that $\mbf{v}$ is not 
	a constant vector, but speed is always equal to $1$.
\item  Arguemment: The acceleration is not zero so speed must be changing.
	Reason this is wrong:  Again, acceleration is a vector and it tells you how the velocity vector is 
	changing.  It does not immediately tell you how the speed is changing.
\end{enumerate}

Looking at the velocity formula, we can easily see that the velocity is not constant, but
that does not mean that the speed is not constant 
(although, in all likelyhood speed will not be constant if velocity is not constant).
So, we need to check that speed is not constant.
Remember that speed$=|v(t)|$.
So, we compute:
\[
\textrm{speed}= |v(t)| = \sqrt{9e^{-6t}+36(2t+1)^4}
\]
which is not constant.
\end{enumerate}

\item  (12 points)  Consider the vector-valued functions below:
\begin{eqnarray*}
\mbf{F}(t) & = & -t^3 \mbf{i} + e^{3t}\mbf{j} \\
\mbf{G}(t) & = & t \mbf{i} + e^{-t} \mbf{j}
\end{eqnarray*}
Find $(\mbf{F} \bullet \mbf{G})'(0)$ in two different ways:
\begin{enumerate}
\item  By first computing $\mbf{F}(t) \bullet \mbf{G}(t)$ and then taking the derivative.

\answer:
\begin{eqnarray*}
\mbf{F}(t) \bullet \mbf{G}(t) & = & (-t^3, e^{3t}) \dotprod (t, e^{-t})
	= -t^4+e^{2t}
\end{eqnarray*}
We can now take the derivative and get
\[
(\mbf{F} \bullet \mbf{G})' (t) = -4t^3 + 2e^{2t}
\]
And finally we plug in $t=0$ and get:
\[
(\mbf{F} \bullet \mbf{G})' (0) = 0 + 2 = 2
\]

\item  By using the product rule for dot products.

\answer:
The product rule for dot product is:
\[
(\mbf{F} \dotprod \mbf{G})' = (\mbf{F}' \dotprod \mbf{G}) + (\mbf{F} \dotprod \mbf{G}')
\]
So, we plug in using $\mbf{F}'(t)=(-3t^2,3e^{3t})$ and $\mbf{G}'(t)=(1,-e^{-t})$:
\begin{eqnarray*}
(\mbf{F} \dotprod \mbf{G})' & = & [(-3t^2,3e^{3t}) \dotprod (t, e^{-t})] + [ (-t^3, e^{3t}) \dotprod (1,-e^{-t})] \\
 	& = & -4t^3+2e^{2t}
\end{eqnarray*}
Plugging in $t=0$, we again (as expected) get $(\mbf{F} \bullet \mbf{G})' (0) = 0 + 2 = 2$.

\end{enumerate}


\end{enumerate}

\end{document}








