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\centerline
{
\bf Solutions to the ``QUIZ'' for Nov. 2, 2009
}
\bigskip

{\bf 1.} Sketch the planar vector field 
$$
{\bf F}= \langle x, y^2 \rangle \quad .
$$

{\bf Solution}: The way that you are supposed to do it is
to pick a few sample points.

At the point $(0,0)$, the vector is 
$\langle 0, 0^2 \rangle=\langle 0, 0 \rangle$, that is
the zero vector, so you only draw a dot. 

At the point $(1,0)$ the
vector is $\langle 1, 0^2 \rangle=\langle 1, 0 \rangle$, so you
draw an arrow starting at the point $(1,0)$ and ending at the point
$(2,0)$. 

At the point $(1,1)$ the vector is
$\langle 1, 1^2 \rangle=\langle 1, 1 \rangle$, so you draw
an arrow from the point $(1,1)$ to  the point $(1+1,1+1)=(2,2)$.

At the point $(2,3)$ the vector is
$\langle 2, 3^2 \rangle=\langle 2, 9 \rangle$, so you draw
an arrow from the point $(2,3)$ to the point $(2+2,3+9)=(4,12)$.
Etc.

{\bf Comments}: Only about \%50 of the students knew what to do,
but that was because I covered it too fast.  I will review this
again on Thursday.

{\bf 2.}  Find a potential function for the vector field ${\bf F}$
$$
{\bf F}= \langle \, y \cos(xy) \, , \, x \cos(xy) \, \rangle \quad .
$$

{\bf Solution}. Next time we will learn how to to do it
{\bf systematically}. Stricly speaking, we had to first
check that ${\bf F}$ is a {\bf gradient vector field}, by
checking that 
$$
{{\partial F_1} \over {\partial y}}
=
{{\partial F_2} \over {\partial x}} \quad .
$$
You are welcome to do it, but {\bf this} question didn't ask for that step.

Next time we will learn a {\bf method} for finding the potential function
$\phi(x,y)$ (whenever it exists), but in this section you are supposed
to do it ``by inspection'', and than verify that your guess is correct,
by checking that
$$
{\bf F} = \g \phi \quad \quad .
$$

The condition is that
$$
{{\partial \phi} \over {\partial x}}=F_1 \quad , \quad
{{\partial \phi} \over {\partial y}}=F_2 \quad .
$$
In other words
$$
{{\partial \phi} \over {\partial x}}=y \cos(xy) \quad , \quad
{{\partial \phi} \over {\partial y}} = x \cos(xy) \quad .
$$
Since the {\bf antiderivative} of {\bf cosine} is  {\bf sine},
a reasonable guess is that
$$
\phi(x,y)=\sin(xy) \quad .
$$
Let's confirm our guess.
$$
\phi_x= \cos(xy) \cdot y = y \cos(xy) \quad ,
$$
$$
\phi_y= \cos(xy) \cdot x = x \cos(xy) \quad .
$$
So it agrees.

{\bf Ans.} The potential function is $\phi(x,y)=\sin(xy)$.

{\bf Comments}. Only about \%30 of the students got it correctly.
It is a tricky question, and next time we will learn how to do it
systematically.



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