Science:Math Exam Resources/Courses/MATH200/December 2012/Question 03 (b)/Solution 1

What to do

The linear approximation of a function $\displaystyle z=f(x,y)$ in a point $\displaystyle (p_{x},p_{y})$ is

$\displaystyle z=f(x,y)\approx f(p_{x},p_{y})+(x-p_{x}){\frac {\partial z}{\partial x}}(p_{x},p_{y})+(y-p_{y}){\frac {\partial z}{\partial y}}(p_{x},p_{y})$

Find $\displaystyle f(-1,1)$

We have that $\displaystyle (p_{x},p_{y})=(-1,1)$. To find $\displaystyle f(-1,1)$, we must solve the equation $\displaystyle xyz+x+y^{2}+z^{3}=0$, where $\displaystyle (x,y)=(-1,1)$ for the variable $\displaystyle z$.

$\displaystyle {\begin{aligned}(-1)1z+(-1)+1^{2}+z^{3}&=0\\-z-1+1+z^{3}&=0\\z(z^{2}-1)&=0\\z(z-1)(z+1)&=0\end{aligned}}$

Hence, $\displaystyle z_{1}=-1,\ z_{2}=0,\ z_{3}=1$ are solution to this equations. With the knowledge,that $\displaystyle f(-1,1)<0$, the value must be $\displaystyle z=f(-1,1)=-1$

Find $\displaystyle {\frac {\partial z}{\partial x}}(-1,1)$

We already know from part a), that $\displaystyle {\frac {\partial z}{\partial x}}=-{\frac {yz+1}{xy+3z^{2}}}$. Pluging in the point $\displaystyle (x,y,z)=(-1,1,-1)$ gives $\displaystyle {\frac {\partial z}{\partial x}}(-1,1)=0$

Find $\displaystyle {\frac {\partial z}{\partial y}}(-1,1)$

We know that $\displaystyle {\frac {\partial z}{\partial y}}=-{\frac {F_{y}}{F_{z}}}=-{\frac {xz+2y}{xy+3z^{2}}}$ Pluging in the point $\displaystyle (x,y,z)=(-1,1,-1)$ gives $\displaystyle {\frac {\partial z}{\partial y}}(-1,1)=-{\frac {3}{2}}$

The linear approximation

Hence the linear approximation is $\displaystyle f(x,y)\approx -1-{\frac {3}{2}}(y-1)$