%qANov5.tex: 
%begin macros
\baselineskip=14pt
\parskip=10pt
\magnification=\magstephalf
\parindent=0pt
\overfullrule=0in
\def\frac#1#2{{#1 \over #2}}
\nopagenumbers
%end macros
 

\centerline
{
\bf Solutions to the ``QUIZ'' for Nov. 5, 2009
}
\bigskip

{\bf 1.} 
Let $C$ be the line segment from $(0,1)$ to $(2,3)$, find
$\int_C xy \, ds \, $.

{\bf Sol.}: We first find a {\it paramteric representation} for the line segment joining
$(0,1)$ and $(2,3)$:
$$
{\bf r}(t)= (1-t)\langle 0, 1 \rangle + t \langle 2 , 3 \rangle=
\langle 0, 1-t \rangle +  \langle 2t , 3t \rangle= \langle 2t, 1+2t \rangle
$$
Of course {\bf always} (for line segments): $0 \leq t \leq 1$.
So $x=2t, y=1+2t$.

Next we find ${\bf r}'(t):\langle 2, 2 \rangle$, so $|{\bf r}'(t)|=\sqrt{2^2+2^2}=\sqrt{8}$,
and so $ds=\sqrt{8} dt$.
Not we plug in $x=2t$, $y=1+2t$ into the integrand, to get:
$$
\int_{0}^{1} (2t)(1+2t) \sqrt{8} \, dt=
\sqrt{8} \int_{0}^{1} (2t)(1+2t)  \, dt= 
$$
$$
\sqrt{8} \int_{0}^{1} 2t+4t^2  \, dt=  \sqrt{8} ( t^2 +4{{t^3} \over {3}} ) \Bigl \vert_{0}^{1}=
\sqrt{8} ( 1^2 +4{{1^3} \over {3}} )- 0=
\sqrt{8} {{7} \over {3}}=\sqrt{2} {{14} \over {3}} \quad .
$$

{\bf Ans.}: $\sqrt{2} {{14} \over {3}}$.

{\bf Comments}: About \%30 got it completely. Another \%40 followed the right way, but messed up the
calculations. Some people had trouble with the first part of finding the paratmetric representation
of the line segment.

{\bf 2.} 
Evaluate
$$
\int_C \, xy^2 \, dx + x^2 y \, dy \quad,
$$
where $C$ is $x:t^2 \, , \, y=t^3$, $ 0 \leq t \leq 1$.

{\bf Sol.} $dx= (t^2)' \, dt=2t\, dt$, $dy= (t^3)' \, dt= 3t^2 \, dt$.
So we have
$$
\int_{0}^{1} (t^2)(t^3)^2 (2t) \, dt + (t^2)^2 (t^3) (3t^2) \, dt=
\int_{0}^{1} 2t^9 \, dt + 3t^9 \, dt=
\int_{0}^{1}  5t^9 \, dt= {{t^{10}} \over {2}} \Bigl \vert_{0}^{1}= {{1} \over {2}} \quad .
$$

{\bf Ans.}: ${{1} \over {2}}$.

{\bf Comment:} About \%85 of the students got it completely.


\end








