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\begin{center}
{\Large Key: TEST 1}
\end{center}

\begin{enumerate}

\item 
  \begin{enumerate}
  \item  Write a parametric equation of the line containing the points $(1,2,3)$ and $(3, -1, 4)$.
  \\
  \sol: We find the direction of the line:
  \[
  \mathbf{v} = (1-3, 2-(-1), 3-4) = (-2, 3, -1)
  \]
  The parametric equation is:
  \[
  \mathbf{r}(t) = (1,2,3) + t(-2,3,-1) = (1-2t, 2+3t, 3-t)
  \]

  \item  Find the distance between the line above and the point $(1,4,8)$.
  \\
  \sol:  There are a couple ways to do this.
  The method that I choose to use is use vector projection (this is not necessarily the easiest way).
  We know that the point $(1,2,3)$ is on the line.
  The vector between our point and $(1,2,3)$ is $\mathbf{u} = (0,2,5)$.
  We first project the vector $\mathbf{u}$ to the line (which means we project to the vector $\mathbf{v}$:
  \[
  \mathrm{Pr}_\mathbf{v} \mathbf{u} = \frac{\mathbf{u} \dotprod \mathbf{v}}{|\mathbf{v}|^2} v
     = \frac{1}{14} (-2,3,-1) = \left( -\frac{1}{7}, \frac{3}{14}, -\frac{1}{14} \right)
  \]
  Now, notice (draw the picture) that the vector $\mathbf{w}$ defined by 
     $\mathbf{u} - \mathrm{Pr}_\mathbf{v}\mathbf{u}$ is orthogonal to $\mathbf{v}$ and gives the distance
     from the line to the point.
  Therefore, the distance is equal to 

	\[
	\mathrm{Distance} = \left| \mathbf{u} - \mathrm{Pr}_\mathbf{v}\mathbf{u} \right| = 
	\left| \left( \frac{1}{7}, \frac{25}{14}, \frac{71}{14} \right)\right| = \sqrt{\frac{405}{14}}
	\]

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	Another way to do this is to use first semester calculus and minimize 
	the distance between the point and a point on the line:
	\[
	D(t) = \left|(1,4,8) - (1-2t, 2+3t, 3-t) \right|
	\]
	Minimize this as a function of $t$.
  \end{enumerate}


\item  \begin{enumerate}
	\item  Compute the vector projection of the vector $\mbf{u}=(a,b,c)$ onto the vector $\mbf{i}=(1,0,0)$:
		  $\mathrm{Pr}_\mbf{i} \mbf{v}$.  Explain your answer graphically.
	\\
	\sol:
	We just compute the projection:
	\[
	\mathrm{Pr}_\mbf{i} \mbf{v} = \frac{\mathbf{v} \dotprod \mathbf{i}}{|\mathbf{i}|^2} \mathbf{i}
		= \frac{a}{1}(1,0,0) = (a,0,0)
	\]
	\item  Generalize this and explain what vector projection is (graphically) when projecting to the vectors
		$\mathbf{j}=(0,1,0)$ and $\mathbf{k}=(0,0,1)$.
	\\
	\sol:
	\[
	\mathrm{Pr}_\mbf{j} \mbf{v} = b\mathbf{j} \qquad \mathrm{Pr}_\mbf{k} \mbf{v} = c\mathbf{k}
	\]
	\end{enumerate}

\item  Consider the line given by the parametric equation:
  \[
  \mbf{r}(t) = \left(3-2t, -5+t, 7+3t \right)
  \]
  Find the equation of the plane that contains the line $\mbf{r}(t)$ and the point $(10, -9, -2)$.
	\\
	\sol:
We need to find the direction normal to the plane.
We know that the direction tangent to the line is: $\mathbf{v}=(-2,1,3)$.
A point on the line is $(3,-5,7)$ (just make $t=0$ in the equation of the line).
Using this point and the point $(10,-9,-2)$, we form the vector: $\mathbf{u} = (3,-5,7)-(10,-9,-2) = (-7,4,9)$.
We now take the cross product to get a normal vector:
\[
\mathbf{v} \times \mathbf{u} = (-3,-3,-1)
\]
So, a vector normal to the plane is $(3,3,1)$.
Therefore, the equation of the plane is $3x+3y+z=D$.
We use the point $(10,-9,-2)$ to solve for $D$: $3(10)+3(-9)+(-2) = D = 1$:
\[
3x+3y+z=1
\]

\item  Consider the curve $y=\ln x$.
  \begin{enumerate}
  \item  Write the curve as a parametric equation.
	\\
	\sol:
	\[
	\mathbf{r}(t) = (t, \ln t)
	\]
  \item  For your parametrization, find an expression for speed.
	\\ \sol:
	\begin{eqnarray*}
	\mathbf{v}(t) & = & \mathbf{r}'(t) = (1, \frac{1}{t}) \\
	\mathrm{Speed} & = & |\mathbf{v}(t)| = \sqrt{1+\frac{1}{t^2}} = \frac{\sqrt{t^2+1}}{t}
	\end{eqnarray*}
  \item  Find an expression for the curvature at an arbitrary point.
	\\ \sol:
	We use the curvature formula:
	\begin{eqnarray*}
	\kappa & = & \frac{\left|x'y'' - x''y' \right|}{\left((x')^2 + (y')^2 \right)^{(3/2)}} 
		 = \frac{\left|1(-1/t^2) - (0)(1/t) \right|}{\left((1)^2 + (1/t)^2 \right)^{(3/2)}} \\
		& = & \frac{\left|1(-1/t^2) - (0)(1/t) \right|}{\left((1)^2 + (1/t)^2 \right)^{(3/2)}} 
		= \frac{1/t^2}{\left(1 + (1/t)^2 \right)^{(3/2)}} \\
		& = & \frac{t}{(t^2+1)^{(3/2)}}
	\end{eqnarray*}
  \item  Find a general expression for the unit tangent vector and the unit normal vector.
	\\ \sol:
	$\mathbf{N}$ can be found by looking at the graph and finding an orthogonal vector to $\mathbf{T}$.
	It is important to note that $(-1,t)$ is orthogonal to $T$ but is in the opposite direction to $\mathbf{N}$, 
	so you have to be a bit careful.
	\begin{eqnarray*}
	\mathbf{T} & = & \frac{\mathbf{v}}{|\mathbf{v}|} = \frac{1}{\sqrt{t^2+1}} \left(t, 1\right) \\
	\mathbf{N} & = & \frac{1}{\sqrt{t^2+1}} \left(1, -t \right)
	\end{eqnarray*}
	Another option is to compute $\mathbf{N}$ from the definition of $\mathbf{N}$.
  \item  Find a general expression for the tangential acceleration and the normal acceleration.
	\\\sol:
	\begin{eqnarray*}
	a_T & = & \frac{d^2s}{dt^2} = |\mathbf{v}|' = -\frac{1}{t^2\sqrt{1+t^2}} \\
	a_N & = & \left(\frac{ds}{dt}\right)^2 \kappa = |\mathbf{v}|\kappa =
		\left(\frac{\sqrt{t^2+1}}{t}\right)^2 \left( \frac{t}{(t^2+1)^{(3/2)}} \right) \\
		& = & \frac{1}{t\sqrt{t^2+1}}
	\end{eqnarray*}
  \end{enumerate}





\end{enumerate}

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