Tangent Planes and Linear Approximations — Question 10

Question 10 – Solution

Strategy. Express the value and slopes at $(1,1)$ in terms of $a,b$, then turn the two tangent-plane conditions into a linear system.

Step 1: General linearization At $(1,1)$, $$f=a+b+1,\qquad f_x=2a+b,\qquad f_y=b+2.$$ Therefore $$L(x,y)=a+b+1+(2a+b)(x-1)+(b+2)(y-1).$$ Passing through $(2,0,5)$ requires $$3a+b-1=5,\qquad\text{so}\qquad 3a+b=6.$$ The prediction at $(1.02,0.99)$ requires $$1.04a+1.01b+0.98=3.06.$$ Solving the two equations gives $$\boxed{a=2,\qquad b=0}.$$

Step 2: Tangent plane Now $f(1,1)=3$, $f_x(1,1)=4$, and $f_y(1,1)=2$. Hence $$\boxed{z=3+4(x-1)+2(y-1)}.$$

Step 3: New estimate and verification At $(0.98,1.03)$, $$f(0.98,1.03)\approx 3+4(-0.02)+2(0.03)=\boxed{2.98}.$$ Also $L(2,0)=3+4-2=5$, and $L(1.02,0.99)=3+0.08-0.02=3.06$, verifying both supplied constraints.