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\centerline
{
\bf Solutions to the ``QUIZ'' for Sept. 24, 2009
}


{\bf 1.} Compute the partial derivatives with respect to $x$ and $y$.
$$
z= \ln (x^2+y^3) \quad.
$$
{\bf Sol.} 
${{\partial z} \over {\partial x}}$, alias $z_x$ is obtained by treating
$x$ as the variable, and $y$ as {\bf constant}. By the chain-rule:
$$
z_x={{1} \over {x^2+y^3}} \cdot (x^2+y^3)'={{1} \over {x^2+y^3}} \cdot 2x =
{{2x} \over {x^2+y^3}} \quad .
$$
Analogously,
${{\partial z} \over {\partial y}}$, alias $z_y$ is obtained by treating
$y$ as the variable, and $x$ as {\bf constant}. By the chain-rule:
$$
z_x={{1} \over {x^2+y^3}} \cdot (x^2+y^3)'={{1} \over {x^2+y^3}} \cdot 3y^2 =
{{3y^2} \over {x^2+y^3}} \quad .
$$


{\bf Comments}: About \%90 of the people got it right. Some people need to review the chain-rule for
functions of a single variable.

{\bf 2.}
Find an equation of the tangent
plane to the given surface at the specified point.
$$
z=x^2+y^2+2 \quad , \quad (1,1,4) \quad .
$$

{\bf Sol.}: The formula is:
$$
(z-z_0)=f_x(x_0,y_0)(x-x_0)+ f_y(x_0,y_0)(y-y_0) \quad .
$$
Here $x_0=1, y_0=1, z_0=4$. To make sure that $z_0$ is correct we do
$f(1,1)=1^2+1^2+2=4$, that agrees.
Now
$$
f_x=2x+0+0 =2x
$$
$$
f_y=0+2y+0 =2y
$$
Plugging-in $x_0=1,y_0=1$, we get
$$
f_x(1,1)=2 \cdot 1=2 \quad ,
$$
$$
f_y(1,1)=2 \cdot 1=2 \quad .
$$
Putting it all together, we have
$$
(z-4)=2(x-1)+2(y-1) \quad .
$$
This is already a {\bf correct} answer, but you are supposed to simplify to get:
$$
z-4=2x-2+2y-2 \quad .
$$
yielding
$$
z=2x+2y \quad .
$$
{\bf Ans.}: An equation for the tanglent plane to $z=x^2+y^2+2$ at the point $(1,1,4)$ is $z=2x+2y$.


{\bf Comment}:About \%95 people got it right. Many people left it as
$(z-4)=2(x-1)+2(y-1)$. A few people messed-up the addition.

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