## Problem on level curves and directional derivatives Let $f(x,y) = x^2 + y^2$.
1. Describe the shape of the $f(x,y)=2$ level curve.

2. Without calculation, find the directional derivative at $(1,1)$ in the direction $-\bfi+\bfj$.
Hint: consider the level curve at $(1,1).$

3. By computation, find the directional derivative at $(1,1)$ in the direction of $-\bfi + \bfj$.

• ## Solution #### Part a Recall that The $f=2$ level curve consists of the points such that $x^2 + y^2 = 2$. This is a circle of radius $\sqrt{2}$ centered at the origin. #### Part b To use the hint, we need to relate directional derivatives to level sets. Recall that Because a function has constant value along a level curve, the directional derivative is zero in the direction tangent to the level curve. We draw the level curve with the tangent vector at $(1,1)$.   The level curve is tangent to $-\bfi + \bfj$ at the point $(1,1)$. Hence $f$ does not change in the direction of $-\bfi +\bfj$, and the directional derivative $$D_{(-\bfi + \bfj)} f = 0.$$ #### Part c Recall that We can now compute the directional derivative at $(1,1)$ as $$D_{(-\bfi + \bfj)}f = (2 \ \bfi + 2 \ \bfj) \cdot \frac{ -\bfi + \bfj}{\sqrt{2}}$$