# Section 6: Directional Derivatives and the Gradient Vector

### Instructions

• First, you should watch the concepts videos below explaining the topics in the section.
• Second, you should attempt to solve the exercises and then watch the videos explaining the exercises.
• Last, you should attempt to answer the self-assessment questions to determine how well you learned the material.
• When you have finished the material below, you can start on the next section or return to the main several variable calculus page.

### Concepts

• The directional derivative

If you would like to see more videos on this topic, click the following link and see the related videos. Note the related videos at the link are not required viewing.

If you would like to see more videos on this topic, click the following link and see the related videos. Note the related videos at the link are not required viewing.

### Exercises

Directions: The following questions are an assessment of your understanding of the material above. If you are not sure of the answers, you may need to rewatch the videos.
1. Find $$D_{{\bf{u}}}f(x,y)$$ at the point $$(1,2)$$ in the direction of $$\left<1,-3\right>$$ to the surface $$f(x,y)=x^3+2x^2y^2.$$

$$D_{{\bf {u}}}f(1,2)=-\dfrac{5}{\sqrt{10}}$$

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2. Suppose $$f(x,y)=x^3-2xy+y^2$$. Find $$D_{{\bf{u}}}f(x,y)$$ at the point $$(1,2)$$ where $${\bf{u}}$$ is the unit vector corresponding to $$\dfrac{\pi}{3}.$$

$$D_{{\bf {u}}}f(1,2)=-\dfrac{1}{2}+\sqrt{3}$$

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3. If $$f(x,y,z)=z^3-x^2y$$, find $$D_{{\bf{u}}}f(1,6,2)$$ if $${\bf{u}}=\left<\dfrac{3}{13},\dfrac{4}{13},\dfrac{12}{13}\right>.$$

$$D_{{\bf {u}}}f(1,6,2)=\dfrac{104}{13}$$

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4. Let $$f(x,y)=xe^y$$
1. ​Find the rate of change of $$f$$ at the point $$(2,0)$$ in the direction of the point $$P(2,0)$$ to the point $$Q\left(\dfrac{1}{2},2\right).$$
2. At the point $$(2,0)$$, in what direction does $$f$$ have the maximum rate of change? What is the maximum rate of change?

1. $$D_{{\bf{u}}}f(2,0)=1$$
2. The direction of the maximum rate of change at the point $$(2,0)$$ is $$\nabla f(2,0)= \langle 1, 2\rangle$$, and the maximum rate of change is $$\sqrt{5}.$$

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5. ​Find the maximum rate of change of $$f(x,y)=\tan(3x+2y)$$ at the point $$\left(\displaystyle{{\pi}\over{6}}, -\displaystyle{{\pi}\over{8}}\right)$$ and the direction in which it occurs.

The direction of the maximum rate of change at $$\left(\dfrac{\pi}{6},-\dfrac{\pi}{8}\right)$$ is $$\nabla f\left(\dfrac{\pi}{6},-\dfrac{\pi}{8}\right)=\langle6,4\rangle$$, and the maximum rate of change is $$|\nabla f|=\sqrt{52}.$$

If you would like to see more videos on this topic, click the following link and see the related videos. Note the related videos at the link are not required viewing.

### Self-Assessment Questions

Directions: The following self-assessment question are a measure of how well you understood the material in this section.
1. How is the geometric interpretation of the directional derivative of $$z=f(x,y)$$ at a point $$P(x_0, y_0)$$ in the direction of the unit vector $${\bf{u}}$$ different from $$f_x(x_0, y_0)$$ and $$f_y(x_0, y_0)$$?
2. How can we use the directional derivative to find the steepest ascent or descent of a surface $$z=f(x,y)$$?
3. What is the difference between the direction of the maximum rate of change of $$z=f(x,y)$$ at a point $$P$$ and the maximum rate of the change of $$z=f(x,y)$$ at a point $$P$$?