\magnification=\magstep1
\centerline{Mathematics 241, section 1 -- Multivariable Calculus}
\centerline{Solutions for Exam 1}
\centerline{October 1, 2010}
 \bigskip
\noindent
I.  All parts of this question refer to the three points 
$P = (0,0,1)$, $Q = (1,2,-4)$, and $R = (1,-1,0)$ in ${\bf R}^3$.
\medskip
\item{A)}  (15) Find the equation of the plane containing $P,Q,R$.
\medskip
\item{} {\it Solution:}  The vector from $P$ to $Q$ is $Q - P = (1,2,-5)$
and the vector from $P$ to $R$ is $R - P = (1,-1,-1)$.  Hence 
$N = (Q-P) \times (R-P) = (-7,-4,-3)$ is a normal vector for the plane 
we want.  Using $N$ and the point $P$, equation is 
$(-7,-4,-3) \cdot (x - 0, y - 0, z - 1) = 0$, or 
$$7x + 4y + 3z = 3.$$
\item{B)}  (10) Give a parametrization of the line segment from $R$ to $Q$
(in that direction), including the proper range of $t$-values. 
\medskip
\item{} {\it Solution:}  The direction vector we want is $Q - R = (0,3,-4)$.
The line segment is 
$$(1,-1,0) + t(0,3,-4) = (1,-1+3t,-4t) \quad {\rm where}\ 0 \le t \le 1.$$
\item{C)}  (10) Which $t$-value gives the midpoint of the line segment 
from part B (the point equidistant from $R$
and $Q$) in your parametrization?
\medskip
\item{} {\it Solution:}  Since the line segment is traversed at
constant speed in this parametrization between $t = 0$ and $t = 1$, 
the midpoint will be reached at $t = {1\over 2}$.
The midpoint is $M = (1,{1\over 2},-2)$.  It can be checked that 
$\|M - Q\| = \|M - R\| = {5\over 2}$.
\medskip  
\item{D)}  (10)  Compute the angle between the vectors $\vec{PQ}$ and 
$\vec{PR}$.
\medskip
\item{} {\it Solution:}  $Q - P = (1,2,-5)$ and $R - P = (1,-1,-1)$.
The angle satisfies 
$$\cos(\theta) = {(1,2,-5)\cdot (1,-1,-1)\over \sqrt{30}\sqrt{3}} = {4\over 3\sqrt{10}} =
{4\sqrt{10}\over 30}.$$
The angle is 
$$\theta = \cos^{-1}\left({4\sqrt{10}\over 30}\right).  $$
\bigskip
\noindent
II. Let ${\bf u},{\bf v},{\bf w}$ be vectors in ${\bf R}^3$ with tails at the origin.
\item{A)} (10)  Show that 
$${\bf w} \times ({\bf u} + {\bf v}) = {\bf w}\times {\bf u} + {\bf w} \times {\bf v}.$$ 
\item{} {\it Solution:}  Let ${\bf u} = (u_1,u_2,u_3)$, ${\bf v} = (v_1,v_2,v_3)$, and
${\bf w} = (w_1,w_2,w_3)$.  Then 
$$\eqalign{
{\bf w} \times ({\bf u} + {\bf v}) &= (w_1,w_2,w_3) \times (u_1 + v_1,u_2 + v_2,u_3 + v_3)\cr 
&= (w_2(u_3 + v_3) - w_3(u_2 + v_2), \cr
&\qquad w_3(u_1 + v_1) - w_1(u_3 + v_3), w_1(u_2 + v_2) - w_2(u_1+v_1))\cr
&= (w_2u_3 - w_3u_2, w_3 u_1 - w_1 u_3, w_1u_2 - w_2u_1)\cr
&\qquad + (w_2v_3 - w_3v_2, w_3v_1 - w_1v_3, w_1v_2 - w_2v_1)\cr
&= {\bf w}\times {\bf u} + {\bf w}\times {\bf v}.\cr}$$
\item{B)} (5)  Show that if ${\bf w}$ is in the plane spanned by ${\bf u}$ and ${\bf v}$,
then $({\bf u} \times {\bf v}) \cdot {\bf w} = 0$.
\medskip
\item{} {\it Solution:}  If ${\bf w}$ is in the plane spanned by ${\bf u}$ and ${\bf v}$,
then there are scalars $s,t$ such that ${\bf w} = s{\bf u} + t{\bf v}$.  Then by another
vector identity,
$$({\bf u}\times {\bf v})\cdot {\bf w} = ({\bf u} \times {\bf v})\cdot (s{\bf u} + t{\bf v}) = 
s({\bf u} \times {\bf v})\cdot {\bf u} + t({\bf u}\times {\bf v})\cdot {\bf v}.$$
Since ${\bf u}\times {\bf v}$ is orthogonal to both ${\bf u}$ and ${\bf v}$, 
both dot products here are zero.  Hence $({\bf u} \times {\bf v}) \cdot {\bf w} = 0$.
\bigskip
\noindent
III.  All parts of this question refer to 
$$\alpha(t) = (\cos(3t)\cos(t),\cos(3t)\sin(t))$$
called a {\it 3-leaved rose curve}.
\item{A)}  (15) Starting from $t = 0$, what is the first $t$ with
$\alpha(t) = (0,0)$?  At how many different $t$ is $\alpha(t) = (0,0)$
in the range $ 0 \le t < \pi$?
\medskip
\item{} {\it Solution:}  Starting from $t = 0$, the first $t$ with $\alpha(t) = (0,0)$
is $t = \pi/6$.  There are three $t$ in the range $0 \le t < \pi$ where $\alpha(t) = (0,0)$:
$t = \pi/6, \pi/2, 5\pi/6$.  These all come from zeroes of the function $\cos(3t)$.  
\medskip
\item{B)}  (15) Find a parametrization of the tangent line to the rose curve 
at $t = \pi/3$.
\medskip 
\item{} {\it Solution:}  We have 
$$\alpha(\pi/3) = (\cos(\pi)\cos(\pi/3),\cos(\pi)\sin(\pi/3)) = \left(-{1\over 2},-{\sqrt{3}\over 2}\right).$$
Then by the product rule in each component,
$$\alpha'(t) = (-3\sin(3t)\cos(t) - \cos(3t)\sin(t), -3\sin(3t)\sin(t) + \cos(3t)\cos(t)).$$
Hence 
$$\alpha'(\pi/3) = \left(0 + {\sqrt{3}\over 2}, 0 - {1\over 2}\right) = \left({\sqrt{3}\over 2},-{1\over 2}\right).$$
The tangent line is 
$$ \left(-{1\over 2},-{\sqrt{3}\over 2}\right) + s\left({\sqrt{3}\over 2},-{1\over 2}\right), 
s \in {\bf R}.$$
 
\item{C)}  (10) Give a parametrization of a circle surrounding
the ``leaf'' of $\alpha(t)$ in quadrants I and IV, {\it but not completely enclosing
either of the other leaves}.  (Any circle that does that is OK.)
\medskip
\item{} {\it Solution:}  If we place the center at $(1,0)$, then to enclose the 
``leaf'' in quadrants I and IV, but not the other two ``leaves,'' we can take a radius greater 
than or equal to $1$, but strictly less than the distance from $(1,0)$ to the point 
$\alpha(\pi/3)$, which is $\sqrt{3}$.  Note that $r = 1$ is in that range.  
Something like
$$\beta(t) = \left(1  + \cos(t), \sin(t)\right)$$
is a reasonable answer.
\bigskip
\noindent
{\bf Extra Credit}  (10) Let $\beta(t)$ be a parametric curve in 
${\bf R}^3$.  Suppose that $\beta(t)\cdot \beta(t) = 1$ (dot product)
for all $t$.  Show that $\beta'(t)$ is orthogonal
to $\beta(t)$ for all $t$.
\medskip
\item{}{\it Solution:}  By the product formula for derivatives, it follows that
$$(\beta(t) \cdot \gamma(t))' = \beta'(t)\gamma(t) + \beta(t)\cdot \gamma'(t)$$
for any vector valued functions $\beta$ and $\gamma$.  If $\gamma = \beta$ 
and $\beta(t) \cdot \beta(t)$ is {\it constant}, then this shows
$$2\beta(t)\cdot \beta'(t) = 0$$
for all $t$.  This shows that $\beta(t)$ and $\beta'(t)$ are orthogonal
for all $t$.   
\bye
