Science:Math Exam Resources/Courses/MATH200/December 2013/Question 01 (b) i/Solution 2

Note: There is another, perhaps easier, way of solving for the partial derivatives in the previous solution. If we move all the variables onto one side of the equation and leave the other as zero, we get:

$\displaystyle x^{2}z^{3}+y\sin(\pi x)+y^{2}=0$

Let's define a new function of three variables F(x,y,z),

$\displaystyle F(x,y,z)=x^{2}z^{3}+y\sin(\pi x)+y^{2},$

where the level surface is given by F(x,y,z)=0. Then, the partial derivatives we found earlier in this question can be expressed simply as:

${\begin{aligned}{\frac {\partial z}{\partial x}}&=-{\frac {F_{x}}{F_{z}}}\\{\frac {\partial z}{\partial y}}&=-{\frac {F_{y}}{F_{z}}}\end{aligned}}$

Let's test this out. If we find the partial derivatives of F with respect to x and to z, we get:

${\begin{aligned}{\frac {\partial F}{\partial x}}&=2xz^{3}+\pi y\cos(\pi x)\\{\frac {\partial F}{\partial z}}&=3x^{2}z^{2}\end{aligned}}$

Therefore, the partial derivative of z with respect to x is:

${\frac {\partial z}{\partial x}}=-{\frac {F_{x}}{F_{z}}}=-{\frac {2xz^{3}+\pi y\cos(\pi x)}{3x^{2}z^{2}}}$

This is the same equation as the one we found earlier using implicit differentiation. Try out the other partial derivative on your own to make sure both work.

Using the result obtained for ${\frac {\partial z}{\partial y}}$ and using the formula for the tangent plane and plugging the values of x,y, and z at the point P gives

${\color {blue}z+1={\frac {2+\pi }{3}}(x-1)-{\frac {2}{3}}(y-1)}.$