Tangent Plane and Linear Approximation: 04-BS-1, May 2013, Q4

Question 4: 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 $(x_0,y_0)=(2,3)$; target evaluation point $(2.1,2.95)$.

Find. (a) the tangent-plane equation at $(2,3,f(2,3))$; (b) the linear-approximation estimate of $f(2.1,2.95)$.

Approach. Evaluate $f$ and both first partials 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 target point.

Evaluate $f$ at the base point. $xy-5=(2)(3)-5=1$, so $\ln(1)=0$ and $$f(2,3)=1+2\ln(1)=\boxed{1}.$$
Partial derivative $f_x$. By the product rule, $f_x=\ln(xy-5)+x\cdot\dfrac{y}{xy-5}$. At $(2,3)$: $\ln(1)=0$ and $\dfrac{(2)(3)}{1}=6$, so $$f_x(2,3)=0+6=\boxed{6}.$$
Partial derivative $f_y$. $f_y=x\cdot\dfrac{x}{xy-5}=\dfrac{x^2}{xy-5}$. At $(2,3)$: $$f_y(2,3)=\frac{4}{1}=\boxed{4}.$$
Assemble the tangent plane and approximate. $$z=f(2,3)+f_x(2,3)(x-2)+f_y(2,3)(y-3)=1+6(x-2)+4(y-3).$$ Substituting $(x,y)=(2.1,2.95)$, i.e. $\Delta x=0.1,\ \Delta y=-0.05$: $$f(2.1,2.95)\approx 1+6(0.1)+4(-0.05)=1+0.6-0.2=\boxed{1.4}.$$

Quantity	Result
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 for functions of two variables.

Key relations. $z=f(x_0,y_0)+f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)$; product rule for $x\ln(xy-5)$: treat $\ln(xy-5)$ as a chained function of the inner argument $xy-5$.

Why this works. The tangent plane is the first-order Taylor expansion of $f$ about $(x_0,y_0)$; because $(2.1,2.95)$ is close to $(2,3)$, the plane's value closely tracks $f$'s actual value, with error $O(\|\Delta\|^2)$. This is the two-variable analogue of using $f(x_0)+f'(x_0)\Delta x$ to approximate a single-variable function near $x_0$.

Common pitfall. Forgetting the product rule on $x\ln(xy-5)$ (differentiating only the logarithm and dropping the leading $\ln(xy-5)$ term in $f_x$); also evaluating $\ln(xy-5)$ at the wrong point or missing that $xy-5=1$ makes $\ln(1)=0$, which conveniently simplifies both partials.

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

Question 4 of 8: Tangent Plane and Linear Approximation

Question 4: Tangent Plane and Linear Approximation (20 marks)