The Gradient

The gradient is the direction of steepest increase in a function. To illustrate, take the function $f(x,y) = 9-x^2-y^2$, plotted below.

The plot shows the function, a downward-opening paraboloid. Below it is its contour plot, a projection of the level curves on the $xy$ plane.

We can define the vector on the left side of (1) as $\nabla f$, commonly pronounced "del f", the gradient of function $f(x,y)$.

$$\nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right)$$

And the vector on the right is just

$$\frac{d \vec r}{dt} = \left( \frac{dx}{dt}, \frac{dy}{dt} \right)$$

So we can rewrite expression (1) as

$$\nabla f \cdot \frac{d \vec r}{dt} = 0$$

If $\vec r(t)$ is one of our level curves, then its derivative, $\frac{d \vec r}{dt}$, is tangent to that curve at any given value of $t$. Because the dot product of $\nabla f$ and $r'(t)$ is zero, $\nabla f$ must be normal to the level curve.

Here is a plot of just the contours of our function. The derivative along any level curve is tangent to the curve, and by definition, the gradient vector is always normal to that. In fact, the gradient always points in the direction of steepest increase of the slope of the function.

Gradient: direction of steepest ascent

The gradient of a vector function is a vector at a point which gives the direction and the rate of greatest increase of the function.

The vectors $(3,3)$ and $(4,4)$ are gradient vectors. They show the direction of steepest increase of this 3D function at those points.

Example 1

Calculate the gradient of the function $f(x,y)=x^2+y^2$.

The function looks like this:

The gradient is

$$ \begin{align} \nabla f(x,y) &= \bigg\langle \frac{\partial f(x,y)}{\partial x}, \; \frac{\partial f(x,y)}{\partial y} \bigg\rangle \\[5pt] &= \langle 2x, \; 2y\rangle \end{align}$$

Example 2

Calculate the gradient of the function $f(x,y)=-\frac{2}{3}xy^2$.

Here's a plot of the function:

The gradient is

$$ \begin{align} \nabla f(x,y) &= \bigg\langle \frac{\partial f(x,y)}{\partial x}, \; \frac{\partial f(x,y)}{\partial y} \bigg\rangle \\[5pt] &= \bigg\langle -\frac{2}{3}y^2, \; -\frac{4}{3}xy\bigg\rangle \end{align}$$

Example 3

Find and characterize any critical points of the function $f(x,y) = x \, e^{-2x^2-2y^2}$

First calculate the first partial derivatives — this is a tricky function requiring the product and chain rules; be careful. I've skipped some steps for brevity in calculating these partial derivatives.

$$ \begin{align} f_x &= e^{-2x^2-2y^2}[1-4x^2] = 0 \\[5pt] 1 - 4x^2 &= 0 \tag{1} \\[5pt] x &= \pm \frac{1}{2} \\[8pt] f_y &= -4xy \, e^{-2x^2-2y^2} = 0 \\[5pt] y &= 0 \end{align}$$

Notice that in (1), we used the fact that $e^{-2x^2-2y^2}$ is never equal to zero, which makes finding the zeros of the first partials. So our critical points are $\left( \frac{1}{2}, 0 \right), \; \left( -\frac{1}{2}, 0 \right)$.

Now our second derivatives are

$$ \begin{align} f_{xx} &= e^{-2x^2-2y^2}[16x^3 - 12x] \\[5pt] f_{yy} &= e^{-2x^2-2y^2}[-4x+16y^2] \\[5pt] f_{xy} &= -4y \, e^{-2x^2-2y^2}[1-4x^2] \end{align}$$

As a shorcut, we can calculate $e^{-2x^2-2y^2}$ for $x = \pm \frac{1}{2}, \; y = 0$. It's value (regardless of the sign of $x$) is $e^{1/2}$. What will matter most to us is that it's a positive number.

Now we can calculate

$$ \begin{align} f_{xx}\left( \frac{1}{2}, 0 \right) &= \frac{-4}{\sqrt{e}} \\[5pt] f_{xx}\left( -\frac{1}{2}, 0 \right) &= \frac{4}{\sqrt{e}} \\[5pt] f_{yy}\left(\pm \frac{1}{2}, 0 \right) &= \frac{-2}{\sqrt{e}} \\[5pt] f_{xy} \left(\pm \frac{1}{2}, 0 \right) &= 0 \end{align}$$

Practice problems

Find the gradient of the function and evaluate it to a specific vector at the point given.