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{
\bf Solutions to the ``QUIZ'' for Oct. 1, 2009
}
\bigskip
{\bf 1.} 
Find ${{\partial f} \over {\partial r}}$
and ${{\partial f} \over {\partial s}}$ as 
{\bf functions of $r$ and $s$ }, if
$$
f(x,y)= x^2 +2xy^2+ 2y^3 \quad,
$$
and the variables are related by $x=r+2s$ and $y=3r+2s$.
You do not need to simplify!

{\bf Solution:}
$$
f_r=(f_x)(x_r)+(f_y)(y_r)=(2x+2y^2)(1)+(4xy+6y^2)(3)= 2x+2y^2+12xy+18y^2
$$
$$
f_s=(f_x)(x_s)+(f_y)(y_s)=(2x+2y^2)(2)+(4xy+6y^2)(2)= 4x+4y^2+8xy+12y^2
$$
So far this is correct, but you were asked to express everthing in terms of $r$ and $s$.
Plugging-in $r+2s$ for $x$ and $3r+2s$ for $y$, we get
$$
f_r=2(r+2s)+2(3r+2s)^2+12(r+2s)(3r+2s)+18(3r+2s)^2 \quad .
$$
$$
f_s=4(r+2s)+4(3r+2s)^2+8(r+2s)(3r+2s)+12(3r+2s)^2 \quad.
$$
That's the final answers.


{\bf Comment}: About half of the people got it perfectly. About \%40 almost got it,
but they didn't do the last part.




{\bf 2.} 
Find ${{\partial z} \over {\partial x}}$
and ${{\partial z} \over {\partial y}}$ if 
$$
x^2+y^2+z^2=5xyz+1 \quad .
$$


{\bf Solution}: First move everything to the left:
$$
x^2+y^2+z^2-5xyz-1=0 \quad .
$$
Call the left side $F(x,y,z)$. So $F(x,y,z)=x^2+y^2+z^2-5xyz-1$.
By the short-cut formulas for implicit differentiation
$$
{{\partial z} \over {\partial x}}= -{{F_x} \over {F_z}} \quad ,
$$
$$
{{\partial z} \over {\partial y}}= -{{F_y} \over {F_z}} \quad .
$$
We have
$$
F_x=2x-5yz \quad , \quad F_y=2y-5xz \quad , \quad F_z =2z-5xy \quad .
$$
So, we get
$$
{{\partial z} \over {\partial x}}= -{{2x-5yz} \over {2z-5xy}} \quad ,
$$
$$
{{\partial z} \over {\partial y}}= -{{2y-5xz} \over {2z-5xy}} \quad .
$$
That's the {\bf answer}.

{\bf Comment}: About \% 85 of the people got it perfectly.



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