Multivariable calculus and vector analysis

Directional derivative and gradient examples

Example 1

Let $f\left(x,y\right)=x^{2}y.$ (a) Find $\nabla f\left(3,2\right)$. (b) Find the derivative of $f$ in the direction of (1,2) at the point (3,2).

Solution: (a) The gradient is just the vector of partial derivatives. The partial derivatives of $f$ at the point $\left(x,y\right)=\left(3,2\right)$ are:

$$\begin{array}{llllllll}\hfill \frac{\partial f}{\partial x}\left(x,y\right)& =2xy& \hfill \frac{\partial f}{\partial y}\left(x,y\right)& =x^2& \hfill & & \hfill & \\ \hfill \frac{\partial f}{\partial x}\left(3,2\right)& =12& \hfill \frac{\partial f}{\partial y}\left(3,2\right)& =9& \hfill & & \hfill & \end{array}$$

Therefore, the gradient is

$$\begin{array}{lll}\hfill \nabla f\left(3,2\right)=12\mathbf{i}+9\mathbf{j}=\left(12,9\right).& & \hfill \end{array}$$

(b) Let $\mathbf{u}=u_1\mathbf{i}+u_2\mathbf{j}$ be a unit vector. The directional derivative at (3,2) in the direction of $\mathbf{u}$ is

$$\begin{array}{llll}\hfill D_{\mathbf{u}}f\left(3,2\right)& =\nabla f\left(3,2\right)\cdot \mathbf{u}& \hfill & \\ \hfill & =\left(12\mathbf{i}+9\mathbf{j}\right)\cdot \left(u_1\mathbf{i}+u_2\mathbf{j}\right)& \hfill & \\ \hfill & =12u_1+9u_2.& \hfill \text{ (1)}\end{array}$$

To find the directional derivative in the direction of the vector (1,2), we need to find a unit vector in the direction of the vector (1,2). We simply divide by the magnitude of $\left(1,2\right)$.

$$\begin{array}{lll}\hfill \mathbf{u}=\frac{\left(1,2\right)}{\parallel \left(1,2\right)\parallel }=\frac{\left(1,2\right)}{\sqrt{1^2+2^2}}=\frac{\left(1,2\right)}{\sqrt{5}}=\left(1∕\sqrt{5},2∕\sqrt{5}\right).& & \hfill \end{array}$$

Plugging this expression for $\mathbf{u}=\left(u_1,u_2\right)$ into equation (1) for the directional derivative, and we find that the directional derivative at the point $\left(3,2\right)$ in the direction of $\left(1,2\right)$ is

$$\begin{array}{llll}\hfill D_{\mathbf{u}}f\left(3,2\right)& =12u_1+9u_2& \hfill & \\ \hfill & =\frac{12}{\sqrt{5}}+\frac{18}{\sqrt{5}}=\frac{30}{\sqrt{5}}.& \hfill & \end{array}$$

Example 2

For the $f$ of Example 1, find the directional derivative of $f$ at the point (3,2) in the direciton of $\left(2,1\right)$.

Solution: The unit vector in the direction of $\left(2,1\right)$ is

$$\begin{array}{lll}\hfill \mathbf{u}=\frac{\left(2,1\right)}{\sqrt{5}}=\left(2∕\sqrt{5},1∕\sqrt{5}\right).& & \hfill \end{array}$$

Since we are still at the point (3,2), equation (1) is still valid. We plug in our new $\mathbf{u}$ to obtain

$$\begin{array}{llll}\hfill D_{\mathbf{u}}f\left(3,2\right)& =12u_1+9u_2& \hfill & \\ \hfill & =\frac{24}{\sqrt{5}}+\frac{9}{\sqrt{5}}=\frac{33}{\sqrt{5}}& \hfill & \end{array}$$

Example 3

For the $f$ of Example 1 at the point (3,2), (a) in which direction is the directional derivative maximal, (b) what is the directional derivative in that direction?

Solution: (a) The gradient points in the direction of the maximal directional derivative. The directional derivative is maximal in the direction of (12,9). (A unit vector in that direction is $\mathbf{u}=\left(12,9\right)∕\sqrt{{12}^2+9^2}=\left(4∕5,3∕5\right)$.)

(b) The magnitude of the gradient is this maximal directional derivative, which is $\left|\right|\left(12,9\right)\left|\right|=\sqrt{{12}^2+9^2}=15$. Hence the directional derivative at the point (3,2) in the direction of (12,9) is 15.

We could double-check by calculating the result using equation (1) and the unit vector $\mathbf{u}=\left(4∕5,3∕5\right)$. Then we find that

$$\begin{array}{llll}\hfill D_{\mathbf{u}}f\left(3,2\right)& =12u_1+9u_2& \hfill & \\ \hfill & =\frac{48}{5}+\frac{27}{15}=\frac{75}{5}=15,& \hfill & \end{array}$$

which agrees with our result.

Example 4

For the $f$ of Example 1 at the point (3,2), (a) what is the directional derivative in the direction (-3,4) (which is perpendicular to $\nabla f\left(3,2\right)$), and (b) what is the directional derivative in the direction (-4,-3) (which is opposite of the direction of $\nabla f\left(3,2\right)$)?

Solution: (a) The directional derivative must be zero. (b) The directional derivative must be $-\left|\right|\nabla f\left(3,2\right)\left|\right|$, which is $-15$. (You can verify these by calculating the results directly using equation (1).)

Contents by topic

1. Geometry in higher dimensions

2. Differential Calculus

2a. The partial derivative

2b. The derivative and differentiability

2c. The chain rule, directional derivative and gradient

• The chain rule
• Chain rule examples
• The directional derivative and the gradient
• Directional derivative and gradient examples

2d. Applications of differential calculus

3. Integral calculus

4. Vector calculus

5. The fundamental theorems of vector calculus

6. Review material