Tangent Planes and Linear Approximations — Question 6

Question 6 – Solution

Strategy. Write $t=x+2y$ so the two-variable approximation reduces to the first-order Taylor approximation of $e^t$ at zero.

Step 1: Linearization At the origin, $f=1$, $f_x=1$, and $f_y=2$. Therefore $$\boxed{L(x,y)=1+x+2y}.$$ At $(0.02,0.01)$, $t=0.04$, so $$f(0.02,0.01)\approx 1.04.$$

Step 2: Uniform bound Taylor’s theorem gives $$e^t=1+t+\frac{e^\xi}{2}t^2$$ for some $\xi$ between $0$ and $t$. In the given rectangle, $$|t|=|x+2y|\le 0.06,$$ so $e^\xi\le e^{0.06}$. Consequently, $$\boxed{|f-L|\le\frac{e^{0.06}}2(0.06)^2<0.001912}.$$

Step 3: Actual error $$|e^{0.04}-1.04|\approx\boxed{0.0008108},$$ which lies below the uniform bound. The bound is larger because it must cover the entire rectangle, not only the requested point.