Tangent Planes and Linear Approximations — Question 2

Question 2 – Solution

Strategy. Differentiate the height function, then evaluate its first-order change from the base point.

Step 1: Partial derivatives For $f(x,y)=(25-x^2-y^2)^{1/2}$, $$f_x=-\frac{x}{\sqrt{25-x^2-y^2}},\qquad f_y=-\frac{y}{\sqrt{25-x^2-y^2}}.$$ At $(3,0)$, these become $f_x=-3/4$ and $f_y=0$.

Step 2: Linearization $$\boxed{L(x,y)=4-\frac 34(x-3)}.$$ At the requested point, $$f(3.04,0.06)\approx 4-\frac 34(0.04)=\boxed{3.9700}.$$

Step 3: Geometry The sphere is symmetric under $y\mapsto-y$. At $y=0$, moving a small distance in either $y$-direction changes the height only to second order, so the tangent plane has no $y$-tilt.

Verification The exact height is $$\sqrt{25-3.04^2-0.06^2}=\sqrt{15.7548}\approx 3.9692317.$$ The estimate is high by about $0.0007683$.