Question 8 of 8: Tangent Plane to a Logarithmic Surface and Linear Approximation

Question 8: Tangent Plane to a Logarithmic Surface 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 $(x_0,y_0)=(2,3)$ (note $x_0y_0-5=1$, so $\ln(1)=0$ simplifies $f$ there); target point $(2.1,2.95)$.

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

Approach. Compute $f,f_x,f_y$ at $(2,3)$ (the log term vanishes there, simplifying both the value and the partials), then use the standard tangent-plane formula $z=f(x_0,y_0)+f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)$ and evaluate it at the target point.

1. Value at the base point. $x_0y_0-5=6-5=1$, so $f(2,3)=1+2\ln(1)=1+0=1$.
2. Partial derivatives. $$f_x=\ln(xy-5)+\frac{xy}{xy-5},\qquad f_y=\frac{x^2}{xy-5}.$$ At $(2,3)$, with $xy-5=1$: $f_x(2,3)=\ln1+\dfrac{6}{1}=0+6=6$, and $f_y(2,3)=\dfrac{4}{1}=4$.
3. Tangent plane. $$z=f(2,3)+f_x(2,3)(x-2)+f_y(2,3)(y-3)=1+6(x-2)+4(y-3).$$ $$\boxed{z=1+6(x-2)+4(y-3)}$$
4. Linear-approximation estimate. At $(x,y)=(2.1,2.95)$: $x-2=0.1$, $y-3=-0.05$. $$f(2.1,2.95)\approx1+6(0.1)+4(-0.05)=1+0.6-0.2=1.4.$$

$$f(2.1,2.95)\approx\boxed{1.4}$$