Q. 27 Find the directional derivative ... [FREE SOLUTION]

Chapter 12: Q. 27 (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.
$f (x, y) = \frac{x + y}{x^{2} + y^{2}}, P = (1, 2), v = 鉄� 3, 鈭� 2 鉄�$

Short Answer

Expert verified

$D_{u} f (1, 2) = \frac{11}{25 \sqrt{13}}$

Step by step solution

Step 1. Given Information

$The directional derivative of a function f (x, y) at a point P = (x_{0}, y_{0}) in the direction of a unit vector u = 鉄� a, b 鉄� is given by D_{u} f (x_{0}, y_{0}) = lim_{h 鈫� 0} 鈥� \frac{f (x_{0} + a h, y_{0} + b h) 鈭� f (x_{0}, y_{0})}{h} Here f (x, y) = \frac{x + y}{x^{2} + y^{2}}, P = (1, 2) .$

Step 2. Solution

$The unit vector in the direction of the given vector v = 鉄� 3, 鈭� 2 鉄� is given by \begin{array}{l} u = \frac{1}{鈭� v 鈭�} v \\ = \frac{1}{\sqrt{3^{2} + (鈭� 2)^{2}}} 鉄� 3, 鈭� 2 鉄� \\ = \frac{1}{\sqrt{13}} 鉄� 3, 鈭� 2 鉄� \\ Hence the directional derivative is given by \\ \begin{array}{l} D_{u} f (1, 2) = lim_{h 鈫� 0} 鈥� \frac{f (1 + \frac{3 h}{\sqrt{13}}, 2 鈭� \frac{2 h}{\sqrt{13}}) 鈭� f (1, 2)}{h} \\ = lim_{h 鈫� 0} 鈥� \frac{\frac{1 + \frac{3 h}{\sqrt{13}} + 2 鈭� \frac{2 h}{\sqrt{13}}}{{(1 + \frac{3 h}{\sqrt{13}})}^{2} + {(2 鈭� \frac{2 h}{\sqrt{13}})}^{2}} 鈭� \frac{1 + 2}{1^{2} + 2^{2}}}{h} \end{array} \\ = lim_{h 鈫� 0} 鈥� \frac{\frac{3 + \frac{h}{\sqrt{13}}}{5 + h^{2} 鈭� \frac{2 h}{\sqrt{13}}} 鈭� \frac{3}{5}}{h} \\ S i m p l i f y i n g N u m e r a t o r g i v e s \\ \begin{array}{l} D_{u} f (1, 2) = lim_{h 鈫� 0} 鈥� \frac{\frac{11 h}{\sqrt{13}} 鈭� 3 h^{2}}{5 h (5 + h^{2} 鈭� \frac{2 h}{\sqrt{13}})} \\ = lim_{h 鈫� 0} 鈥� \frac{\frac{11}{\sqrt{13}} 鈭� 3 h}{5 (5 + h^{2} 鈭� \frac{2 h}{\sqrt{13}})} \\ = \frac{11}{25 \sqrt{13}} \\ H e n c e D_{u} f (1, 2) = \frac{11}{25 \sqrt{13}} \end{array} \end{array}$

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

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)=x+yx2+y2,P=(1,2),v=鉄�3,鈭�2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

Logic and Functions

Discrete Mathematics

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Statistics

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