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\centerline
{
\bf Solutions to the `QUIZ'' for Lecture 21
}
\bigskip
{\bf 1.} Compute the line-integral of the scalar function
$f(x,y,z)=z^2$ over the curve ${\bf c}(t)= (2t,3t,4t)$
for $0 \leq t \leq 2$.

{\bf Sol.} 
$$
{\bf c}'(t)= \langle 2, 3, 4 \rangle \quad
$$
So
$$
|{\bf c}'(t)|=| \langle 2, 3, 4 \rangle|=
\sqrt{2^2+3^2+4^2}= \sqrt{4+9+16}=\sqrt{29}
$$
So
$$
\int_C z^2 \, ds= \int_{0}^{2} (4t)^2 \sqrt{29} \, dt=
\int_{0}^{2} 16t^2 \sqrt{29} \, dt=
$$
$$
16 \sqrt{29}  \frac{t^3}{3} \Bigl \vert_{0}^{2}=
16 \sqrt{29}  \frac{2^3}{3}= \frac{128 \sqrt{29}}{3} \quad .
$$

{\bf Ans.}: $\frac{128 \sqrt{29}}{3}$.

{\bf Comment}:  About \%90 of the people got it right.

{\bf 2.} Compute the Jacobian of 
$$
\Phi(u,v)= (3u+4v,u-2v) \quad .
$$

{\bf Sol.} Here $x=3u+4v,y=u-2v$.
$$
J=(x_u)(y_v)-(x_v)(y_u)= (3)(-2)-(4)(1)=-6-4 = -10 \quad .
$$

{\bf Ans.} $-10$  .
(Type: function of $(u,v)$ that happens to be
a constant function. )

{\bf Comment}:  About \%95 of the people got it right.


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