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Chapter 12: Q. 49 (page 964)

In Exercises 49鈥�54, find the directional derivative of the given function at the specified point P and in the specified direction v. Note that some of the direction vectors are not unit vectors.
$z = e^{x} \tan y, P = (0, \frac{蟺}{4}), v = 3 i 鈭� j$

Short Answer

Expert verified

The directional derivative of the given function is $\frac{\sqrt{10}}{10}$ .

Step by step solution

01

Step 1. Given information.

The given function is

$z = e^{x} \tan y$ .

02

Step 2. Calculation.

The given vector is $v = 3 i 鈭� j$ .

First we find the magnitude of the given vector. The magnitude of the given vector is:

$\sqrt{3^{2} + {(- 1)}^{2}} = \sqrt{10}$

so, the unit vector is $\hat{n} = \frac{1}{\sqrt{10}} (3 i - j)$

Now we have to find the gradient of the function.

$鈭� z = \frac{鈭� (e^{x} \tan y)}{鈭� x} i + \frac{鈭� (e^{x} \tan y)}{鈭� y} j = e^{x} tany i + e^{x} \sec^{2} y j$

Therefore the required directional derivative is equal to:

$\hat{n} 路 (鈭� z) = \frac{1}{\sqrt{10}} (3 i - j) 路 e^{x} tany i + e^{x} \sec^{2} y j = \frac{3}{\sqrt{10}} e^{x} tany - \frac{1}{\sqrt{10}} e^{x} \sec^{2} y$

03

Step 3. Calculation.

Now we find directional derivative of the given function at the point $(0, \frac{蟺}{4})$ .

role="math" localid="1650433084002" $\hat{n} 路 {鈭� z}_{(0, \frac{蟺}{4})} = \frac{3}{\sqrt{10}} e^{0} \tan (\frac{蟺}{4}) - \frac{1}{\sqrt{10}} e^{0} s e c^{2} \frac{蟺}{4} = \frac{1}{\sqrt{10}} = \frac{\sqrt{10}}{10}$

04

Step4. Conclusion.

The directional derivative of the given function is $\frac{\sqrt{10}}{10}$ .

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Most popular questions from this chapter

Partial derivatives: Find all first- and second-order partial derivatives for the following functions:

$f (x, y, z) = \ln 鈦� (x + y + z)$

Find the directional derivative of the given function at the specified point P in the direction of the given vector. Note: The given vectors may not be unit vectors.

$f (x, y) = \frac{x + y}{x^{2} + y^{2}}, P = (1, 2), v = 鉄� 3, 鈭� 2 鉄�$

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