Problem on level curves and directional derivatives - Leading Lesson

Problem on level curves and directional derivatives

Let $f(x,y) = x^2 + y^2$.

Describe the shape of the $f(x,y)=2$ level curve.

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

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

Solution
Part a
Recall that
Level curves and surfaces

A level curve of $f(x,y)$ is a curve in the $xy$-plane along which $f$ has a constant value.
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
Directional derivative definition

The directional derivative of a function is the rate of change of that function if the argument is changed in a particular direction.
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
Directional derivative and gradient

The directional derivative of $f(x,y)$ in the direction of the vector $\mathbf{v}$ can be computed by
$$D_\mathbf{v} f = \nabla f(x,y) \cdot \frac{\mathbf{v}}{|\mathbf{v}|}$$
Length of a vector

We identify $\bfv = -\bfi + \bfj$ and compute that $|\bfv| = \sqrt{2}$.
Definition of the gradient

Evaluating the gradient of $f$ at $(1,1)$ we get, \begin{align} \nabla f(x,y) &= 2 x \ \mathbf{i} + 2 y \ \mathbf{j} \\ \nabla f(1,1) &= 2 \ \bfi + 2 \ \bfj \end{align}
We can now compute the directional derivative at $(1,1)$ as $$D_{(-\bfi + \bfj)}f = (2 \ \bfi + 2 \ \bfj) \cdot \frac{ -\bfi + \bfj}{\sqrt{2}}$$
Direct computation of dot product

Computing the dot product, we get that the directional derivative is zero.

Related topics

Multivariable calculus
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