Chapter 12: Q. 30 (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.
$f (x, y, z) = \sqrt{\frac{x y}{z}}, P = (2, 3, 1), v = ⟨ 2, 1, - 2 ⟩$

Short Answer

Expert verified

$D_{u} f (2, 3, 1) = \frac{10}{\sqrt{6}}$

Step by step solution

Step 1. Given Information

$The directional derivative of a function f (x, y, z) at a point P = (x_{0}, y_{0}, z_{0}) in the direction of a unit vector u = ⟨ a, b, c ⟩ is given by D_{u} f (x_{0}, y_{0}, z_{0}) = lim_{h \to 0} \frac{f (x_{0} + a h, y_{0} + b h, z_{0} + c h) - f (x_{0}, y_{0}, z_{0})}{h} Here f (x, y, z) = \sqrt{\frac{x y}{z}}, and P = (2, 3, 1) . A l s o t h e v e c t o r i s v = (2, 1, - 2)$

Step 2. Solution

$The unit vector " u " in the direction of " v " is given by \begin{array}{l} u = \frac{1}{∥ v ∥} v \\ = \frac{1}{\sqrt{2^{2} + 1^{2} + (- 2)^{2}}} ⟨ 2, 1, - 2 ⟩ \\ = \frac{1}{3} ⟨ 2, 1, - 2 ⟩ \\ N o w b y d e f i n i t i o n : \\ \begin{array}{l} D_{u} f (2, 3, 1) = lim_{h \to 0} \frac{f (2 + \frac{2 h}{3}, 3 + \frac{h}{3}, 1 - \frac{2 h}{3}) - f (2, 3, 1)}{h} \\ = lim_{h \to 0} \frac{\sqrt{(2 + \frac{2 h}{3}) (3 + \frac{h}{3}) \div (1 - \frac{2 h}{3})} - \sqrt{\frac{6}{1}}}{h} \\ = lim_{h \to 0} \frac{\sqrt{6 + \frac{8 h}{3} + \frac{2 h}{9} - \sqrt{6 - 4 h}}}{h \sqrt{1 - \frac{2 h}{3}}} \\ R a t i o n a l i z e t h e n u m e r a t o r g i v e s : \\ \begin{array}{l} D_{u} f (2, 3, 1) = lim_{h \to 0} \frac{\sqrt{6 + \frac{8 h}{3} + \frac{2 h^{2}}{9}} - \sqrt{6 - 4 h}}{h \sqrt{1 - \frac{2 h}{3}}} \times \frac{\sqrt{6 + \frac{8 h}{3} + \frac{2 h^{2}}{9}} + \sqrt{6 - 4 h}}{\sqrt{6 + \frac{8 h}{3} + \frac{2 h^{2}}{9}} + \sqrt{6 - 4 h}} \\ = lim_{h \to 0} \frac{6 + \frac{8 h}{3} - \frac{2 h^{2}}{9} - 6 + 4 h}{h \sqrt{1 - \frac{2 h}{3}} (\sqrt{6 + \frac{8 h}{3} + \frac{2 h^{2}}{9}} + \sqrt{6 - 4 h})} \\ = lim_{h \to 0} \frac{\sqrt{\frac{20}{3} - \frac{2}{9} h}}{\sqrt{1 - \frac{2 h}{3}} (\sqrt{6 + \frac{8 h}{3} + \frac{2 h^{2}}{9}} + \sqrt{6 - 4 h})} \\ T a k i n g l i m i t a s h \to 0 g i v e s : D_{u} f (2, 3, 1) = \frac{10}{\sqrt{6}} \end{array} \end{array} \end{array}$

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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.
$f (x, y, z) = \sqrt{\frac{x y}{z}}, P = (2, 3, 1), v = ⟨ 2, 1, - 2 ⟩$

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Solution

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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)=xyz,P=(2,3,1),v=⟨2,1,−2⟩

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Solution

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Calculus

Probability and Statistics

Discrete Mathematics

Pure Maths

Decision Maths

Theoretical and Mathematical Physics

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.
$f (x, y, z) = \sqrt{\frac{x y}{z}}, P = (2, 3, 1), v = ⟨ 2, 1, - 2 ⟩$