Tangent Plane and Linear Approximation: 04-BS-1, December 2019, Q5

Question 5: 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 $(2,3)$.

Find. The tangent-plane equation at $(2,3)$, then $f(2.1,2.95)$ approximated by that plane.

Approach. Evaluate $f,f_x,f_y$ at $(2,3)$ (noting $xy-5=1$ there, which zeroes the $\ln$ term), form the tangent plane $z=f(2,3)+f_x(2,3)(x-2)+f_y(2,3)(y-3)$, then substitute the nearby point.

Evaluate $f$ at $(2,3)$. $xy-5=6-5=1$, so $\ln(xy-5)=\ln1=0$ and $f(2,3)=1+2(0)=1$ — the base point is chosen so this log term vanishes.
Partial derivatives. $$f_x=\ln(xy-5)+\frac{xy}{xy-5},\qquad f_y=\frac{x^2}{xy-5}.$$ At $(2,3)$: $f_x=0+\dfrac{6}{1}=6$, $\quad f_y=\dfrac{4}{1}=4$.
Tangent plane. $$z=1+6(x-2)+4(y-3).$$
Approximate $f(2.1,2.95)$. $\Delta x=0.1,\ \Delta y=-0.05$: $$f(2.1,2.95)\approx1+6(0.1)+4(-0.05)=1+0.6-0.2=1.4.$$

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

Quantity	Result
$f(2,3)$	$1$
$f_x(2,3),\,f_y(2,3)$	$6,\ 4$
Tangent plane	$z=1+6(x-2)+4(y-3)$
$f(2.1,2.95)\approx$	$1.4$

Concept & Theory

Topic. Tangent-plane linear approximation of a function of two variables.

Key relations. The tangent plane to $z=f(x,y)$ at $(x_0,y_0)$ is $z=f(x_0,y_0)+f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)$; for $(x,y)$ near $(x_0,y_0)$, $f(x,y)\approx$ this plane.

Why this works. The base point $(2,3)$ is engineered so that $xy-5=1$, zeroing the logarithm and making $f,f_x$ both evaluate cleanly — recognize this simplification before differentiating, since it eliminates the need to handle $\ln$ of a non-trivial argument.

Common pitfall. Forgetting the product-rule term when differentiating $x\ln(xy-5)$ with respect to $x$ (both the $\ln$ term itself and the chain-rule term from the argument contribute to $f_x$); mixing up $\Delta x,\Delta y$ signs when the target point has a coordinate below the base point (here $y$ decreases from $3$ to $2.95$).

Source. Stewart, Calculus: Early Transcendentals, Ch. 14 (§14.4 Tangent Planes and Linear Approximations).

Question 5 of 8: Tangent Plane and Linear Approximation

Question 5: Tangent Plane and Linear Approximation (20 marks)