Question 7 of 8: Tangent Plane and Linear Approximation

Question 7: Tangent Plane and Linear Approximation (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. $f(x,y)=1+x\ln(xy-5)$, base point $(x_0,y_0)=(2,3)$.

Find. The tangent-plane equation at $(2,3)$, and the approximation of $f(2.1,2.95)$ it gives.

Approach. Note that $x_0y_0-5=1$, which zeroes the logarithm and simplifies every derivative evaluation. Compute $f,f_x,f_y$ at $(2,3)$, assemble the tangent plane $z=f(x_0,y_0)+f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)$, then substitute the nearby point.

1. Function value at the base point. $xy-5=(2)(3)-5=1$, so $\ln(xy-5)=\ln1=0$ and $$f(2,3)=1+2\ln1=1.$$
2. Partial derivatives. $$f_x=\ln(xy-5)+\frac{xy}{xy-5},\qquad f_y=\frac{x^2}{xy-5}.$$ At $(2,3)$, with $xy-5=1$: $$f_x(2,3)=\ln1+\frac{6}{1}=6,\qquad f_y(2,3)=\frac{4}{1}=4.$$
3. Tangent plane. $$z=f(2,3)+f_x(2,3)(x-2)+f_y(2,3)(y-3)=1+6(x-2)+4(y-3).$$
4. Approximate $f(2.1,2.95)$. With $\Delta x=0.1,\ \Delta y=-0.05$, $$f(2.1,2.95)\approx1+6(0.1)+4(-0.05)=1+0.6-0.2=1.4.$$ (For reference, the exact value is $f(2.1,2.95)=1+2.1\ln(1.195)\approx1.374$ — the linear approximation is close since the step is small.)

$$z=1+6(x-2)+4(y-3),\qquad f(2.1,2.95)\approx\boxed{1.4}$$