Tangent Planes and Linear Approximations — Question 5

Question 5 – Solution

Strategy. Existing partial derivatives only suggest a plane; differentiability requires the remainder to be negligible compared with the distance to the base point.

Step 1: Axis partial derivatives Along either coordinate axis, $f$ is zero. Hence $$f_x(0,0)=\lim_{h\to 0}\frac{f(h,0)-0}{h}=0, \quad f_y(0,0)=\lim_{k\to 0}\frac{f(0,k)-0}{k}=0.$$

Step 2: Candidate plane The value and both proposed slopes are zero, so the formal candidate is $$\boxed{z=0}.$$

Step 3: Remainder test Along $(x,y)=(t,t^2)$ with $t\ne 0$, $$f(t,t^2)=\frac{t^4}{t^4+t^4}=\frac 12.$$ For the candidate linearization $L=0$, the required quotient is $$\frac{|f(t,t^2)-L(t,t^2)|}{\sqrt{t^2+t^4}} =\frac{1/2}{|t|\sqrt{1+t^2}}\longrightarrow\infty.$$ Therefore the error is not $o(\sqrt{x^2+y^2})$.

Conclusion The function is not even continuous at the origin along this curve, so it is not differentiable there. Thus the graph has .