Q. 29 Find the directional derivative ... [FREE SOLUTION]

Chapter 12: Q. 29 (page 989)

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.
$f (x, y, z) = x y^{2} z^{3}, P = (0, 0, 0), v = 鉄� 1, 鈭� 2, 鈭� 1 鉄�$

Short Answer

Expert verified

$D_{u} f (0, 0, 0) = 0$

Step by step solution

Step 1. Given Information

$The directional derivative of a function f (x, y, z) at a point P = (x_{0}, y_{0}, z_{0}) in the direction of a unit vector u = 鉄� a, b, c 鉄� is given by D_{u} f (x_{0}, y_{0}, z_{0}) = lim_{h 鈫� 0} 鈥� \frac{f (x_{0} + a h, y_{0} + b h, z_{0} + c h) 鈭� f (x_{0}, y_{0}, z_{0})}{h} Here f (x, y, z) = x y^{2} z^{3}, and P = (0, 0, 0) .$

Step 2. Solution

$Also the vector is v = 鉄� 1, 鈭� 2, 鈭� 1 鉄� The unit vector " u " in the direction of " v " is given by \begin{array}{l} u = \frac{1}{鈭� v 鈭�} v \\ = \frac{1}{\sqrt{1^{2} + (鈭� 2)^{2} + (鈭� 1)^{2}}} 鉄� 1, 鈭� 2, 鈭� 1 鉄� \\ = \frac{1}{\sqrt{6}} 鉄� 1, 鈭� 2, 鈭� 1 鉄� \\ Hence the directional derivative is: \\ \begin{array}{l} D_{u} f (0, 0, 0) = lim_{h 鈫� 0} 鈥� \frac{f (\frac{h}{\sqrt{6}}, 鈭� \frac{2 h}{\sqrt{6}}, 鈭� \frac{h}{\sqrt{6}}) 鈭� f (0, 0, 0)}{h} \\ = lim_{h 鈫� 0} 鈥� \frac{\frac{h}{\sqrt{6}} {(\frac{鈭� 2 h}{\sqrt{6}})}^{2} {(\frac{鈭� h}{\sqrt{6}})}^{3} 鈭� 0}{h} \\ = lim_{h 鈫� 0} 鈥� \frac{鈭� 4 h^{5}}{6^{3}} \\ = 0 \\ H e n c e D_{u} f (0, 0, 0) = 0 \end{array} \end{array}$

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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.
$f (x, y, z) = x y^{2} z^{3}, P = (0, 0, 0), v = 鉄� 1, 鈭� 2, 鈭� 1 鉄�$

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Solution

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91影视

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,z)=xy2z3,P=(0,0,0),v=鉄�1,鈭�2,鈭�1鉄�

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Solution

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Most popular questions from this chapter

Recommended explanations on Math Textbooks

Applied Mathematics

Pure Maths

Decision Maths

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Discrete Mathematics

Study anywhere. Anytime. Across all devices.

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.
$f (x, y, z) = x y^{2} z^{3}, P = (0, 0, 0), v = 鉄� 1, 鈭� 2, 鈭� 1 鉄�$