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\def\R{\bf R}
\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 
$P = (1,0,1)$, $Q = (1,2,-2)$, and $R = (1,-1,1)$ 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 $Q - P = (0,2,-3)$
and the vector from $P$ to $R$ is $R - P = (0,-1,0)$.  Hence 
$N = (Q-P) \times (R-P) = (-3,0,0)$ is a normal vector for the plane 
we want.  Using ${-1\over 3} N$ and the point $P$, equation is  
$$x = 1.$$
(Note that this could also be derived by observation without any calculations!)
\item{B)}  (15) Give a parametrization of the line segment from $R$ to $Q$
(in that direction), including the proper range of $t$-values.  Which $t$-value
gives the midpoint?
\medskip
\item{} {\it Solution:}  The direction vector we want is $Q - R = (0,3,-3)$.
The line segment is 
$$(1,-1,1) + t(0,3,-3) = (1,-1+3t,1-3t) \quad {\rm where}\ 0 \le t \le 1.$$
The midpoint is obtained with $t = 1/2$.  (The line segment is parametrized
with a constant speed curve, so over the interval $[0,1]$, the midpoint
is obtained for $t = 1/2$.)
\medskip  
\item{C)}  (10)  Compute the angle between the vectors $v = Q - P$ and 
$w = R - P$.
\medskip
\item{} {\it Solution:}  $Q - P = (0,2,-3)$ and $R - P = (0,-1,0)$ from part A.  
The angle satisfies 
$$\cos(\theta) = {(0,2,-3)\cdot (0,-1,0)\over \sqrt{13}\cdot 1} = {-2 \over \sqrt{13}}.$$
The angle is 
$$\theta = \cos^{-1}\left({-2\over \sqrt{13}}\right) \doteq 2.16\ {\rm radians}.  $$
\bigskip
\noindent
II. II. Let $\alpha(t) = (u(t),v(t))$ and $\beta(t) = (w(t),z(t))$ be parametric curves in $\R^2$.
\bigskip
\item{A)} (10)  Show that if $\alpha(t)$ and $\beta(t)$ are differentiable, then
$${d\over dt} (\alpha(t) \cdot \beta(t)) = \alpha'(t) \cdot \beta(t) + \alpha(t) \cdot \beta'(t),$$
(where $\cdot$ denotes the dot product). 
\item{} {\it Solution:}  We have 
$$\alpha(t) \cdot \beta(t) = u(t) w(t) + v(t) z(t).$$
So by the product and sum rules for derivatives:
$$
\eqalign{
{d\over dt} (\alpha(t) \cdot \beta(t)) &= u'(t)w(t) + u(t)w'(t) + v'(t) z(t) + v(t) z'(t)\cr
&= ( u'(t) w(t) + v'(t) z(t) ) + (u(t) w'(t) + v(t) z'(t) )\cr
&= \alpha'(t)\cdot \beta(t) + \alpha(t)\cdot \beta(t).\cr}$$
This is what we wanted to show.
\item{B)} (10)  Let $\beta(t)$ be a parametric curve in 
$\R^2$.  Suppose that $\beta(t)\cdot \beta(t) = 1$ (dot product)
for all $t$.  Show that the vector $\beta'(t)$ is orthogonal to the vector $\beta(t)$ for all $t$.
\item{} {\it Solution:}  Using part A and the commutativity of the dot product, we see 
$$0 = {d\over dt}(\beta(t) \cdot \beta(t)) = 2\beta(t) \cdot \beta'(t).$$
Therefore $\beta(t) \cdot \beta'(t) = 0$, which shows that they are orthogonal.  
\bigskip
\noindent
III.  All parts of this question refer to 
$$\alpha(t) = (\cos(2t)\cos(t),\cos(2t)\sin(t))$$
called a {\it 4-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 $t = \pi/4$.  There are two $t$ in the range $0 \le t < \pi$ where $\alpha(t) = (0,0)$:
$t = \pi/4, 3\pi/4$.  These all come from zeroes of the function $\cos(2t)$.  
\medskip
\item{B)}  (15) Find a parametrization of the tangent line to the rose curve 
at $t = \pi/4$.
\medskip 
\item{} {\it Solution:}  We have 
$$\alpha(\pi/4) = (\cos(\pi/2)\cos(\pi/4),\cos(\pi/2)\sin(\pi/4)) = (0,0).$$
Then by the product rule in each component,
$$\alpha'(t) = (-2\sin(2t)\cos(t) - \cos(2t)\sin(t), -2\sin(2t)\sin(t) + \cos(2t)\cos(t)).$$
Hence 
$$\alpha'(\pi/4) = (-\sqrt{2}, -\sqrt{2}).$$
The tangent line is 
$$ \{(-s\sqrt{2},-s\sqrt{2}) \mid 
s \in {\bf R}\}.$$
 
\item{C)}  (10)Give a parametrization of a circle that will fit completely inside
the ``leaf'' of $\alpha(t)$ in quadrants I and IV.  (Any circle that does that is OK.)
\medskip
\item{} {\it Solution:}  If we place the center at $(1/2,0)$, then to fit inside the 
``leaf'' in quadrants I and IV, the radius must be small, something like $1/10$.  
Something like
$$\beta(t) = \left(1/2  + \cos(t)/10, \sin(t)/10\right)$$
is a reasonable answer.
\bigskip
\noindent
{\bf Extra Credit} (10) Identify and sketch the slices of the surface
$$\{(x,y,z) \in {\R}^3 \mid x^2 - y^2 + z^2 = 1\}$$
in the planes $x = 1$ and $x = 2$.
\medskip
\item{}{\it Solution:}  The slice in the plane $x = 1$ is given by $-y^2 + z^2 = 0$. This is
the union of two lines: $x = 1, y = z$, and $x = 1, y = -z$.  The slice in the plane
$x = 2$ is given by $-y^2 + z^2 = -3$, or ${y^2\over 3} - {z^2\over 3} = 1$.  This is
a hyperbola opening in the direction of the positive and negative $y$-axes.   
\bye
