Science:Math Exam Resources/Courses/MATH200/December 2013/Question 01 (b) i/Solution 1

The equation of a tangent plane at $P(x_{0},y_{0},z_{0})$ is:

$\displaystyle z-z_{0}=f_{x}(x_{0},y_{0})(x-x_{0})+f_{y}(x_{0},y_{0})(y-y_{0})$

We need to find $\displaystyle f_{x}(x_{0},y_{0})$ and $\displaystyle f_{y}(x_{0},y_{0})$, but because z is defined implicitly, we need to be careful finding these partial derivatives.

First, let's keep y constant and differentiate implicitly with respect to x to find $f_{x}$ (or ${\frac {\partial z}{\partial x}}$):

$2xz^{3}+3x^{2}z^{2}{\frac {\partial z}{\partial x}}+\pi y\cos(\pi x)=0$

(Be careful differentiating the first term of the given equation - you must use the product rule!)

Rearranging to solve for the partial derivative, you get:

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

Therefore, at P(1,1,-1), you get:

${\frac {\partial z}{\partial x}}=-{\frac {2(1)(-1)^{3}+\pi (1)\cos(\pi (1))}{3(1)^{2}(-1)^{2}}}={\frac {2+\pi }{3}}$

Similarly, for $f_{y}$ (or ${\frac {\partial z}{\partial y}}$), we get:

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

Rearranging:

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

At P(1,1,-1):

$\displaystyle {\frac {\partial z}{\partial y}}=-{\frac {2(1)+\sin(\pi (1))}{3(1)^{2}(-1)^{2}}}=-{\frac {2}{3}}$

Now, plugging these values into the tangent plane formula, we get:

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

Making things nice and beautiful, the final answer is:

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