Tangent Planes and Linear Approximations — Question 1

Question 1 – Solution

Strategy. Use the value and both first partial derivatives at the base point to build the unique affine approximation.

Step 1: Local data $$f(1,0)=1,\qquad f_x=2xe^y,\qquad f_y=x^2e^y.$$ Thus $f_x(1,0)=2$ and $f_y(1,0)=1$.

Step 2: Plane and linearization The tangent-plane formula gives $$z-1=2(x-1)+(y-0).$$ Consequently, $$\boxed{z=1+2(x-1)+y},\qquad \boxed{L(x,y)=1+2(x-1)+y}.$$

Step 3: Estimate Here $\Delta x=0.02$ and $\Delta y=-0.03$, so $$f(1.02,-0.03)\approx L(1.02,-0.03) =1+2(0.02)-0.03=\boxed{1.0100}.$$

Verification Direct evaluation gives $$1.0404e^{-0.03}\approx 1.0096515.$$ The linearization is high by about $0.0003485$, a small error consistent with the small displacement from $(1,0)$.