Science:Math Exam Resources/Courses/MATH200/December 2011/Question 01 (c)/Solution 1

From part (b), we found that the tangent plane to the surface $z=f(x,y)$ at the point $(2,1)$ was

$-4x+8y-z=-1.$

Since the point $(1.99,1.01)$ is very close to $(2,1)$, the tangent plane approximation gives a good estimate for the true value of $f(1.99,1.01)$. Plugging in $(x,y)=(1.99,1.01)$ into the tangent plane equation we get

$\begin{array}{rl}-4(1.99)+8(1.01)-z& =-1\\ -7.96+8.08-z& =-1\\ 0.12-z& =-1\to z=1.12\end{array}$

Therefore,

$f(1.99,1.01)\approx 1.12.$