Chapter 12: Q. 37 (page 953)

In Exercise $35 - 38$ , find the directional derivative of the given function at the specified point $P$ and in the direction of the given vector $v$ .
$f (x, y) = \sqrt{\frac{y}{x}} a t P = (1, 16), v = (2, 1)$

Short Answer

Expert verified

The directional derivative is $\nabla f (P) . u = \frac{- 33}{40} \sqrt{5}$ .

Step by step solution

Directional derivative.

For a given function $P = (x_{0}, y_{0}) = (1, 16)$ and $v = (2, - 1)$ , we must find the directional derivative $f (x, y) = \sqrt{\frac{y}{x}}$ .

$||v|| = \sqrt{2^{2} + 1^{2}} = \sqrt{5}$

Therefore;

$u = (α, β) = (\frac{2 \sqrt{5}}{5}, \frac{- \sqrt{5}}{5})$

Directional unit vector.

The function's directional derivative at point $P$ with directional unit vector $u$ is,

localid="1650642585491" $\nabla f (P) \cdot u = \nabla f (1, 16) \times u = ({\frac{d f}{d x}|}_{(1, 16)} i + {\frac{d f}{d y}|}_{(1, 16)} j) \times (\frac{2 \sqrt{5}}{5} i - \frac{\sqrt{5}}{5} j)$

localid="1650642610945" $= [{(\frac{1}{2 \sqrt{\frac{y}{x}}} \frac{d}{d x} (\frac{y}{x}))}_{(1, 16)} i + {(\frac{1}{2 \sqrt{\frac{y}{x}}} \frac{d}{d x} (\frac{y}{x}))}_{(1, 16)} j] \times (\frac{2 \sqrt{5}}{5} i - \frac{\sqrt{5}}{5} j)$

localid="1650642627783" $= {[\frac{\frac{- y}{x^{2}}}{2 \sqrt{\frac{y}{x}}})}_{(1, 16)} i + (\frac{\frac{1}{x}}{{2 \sqrt{\frac{y}{x}})}_{(1, 16)}} j] \times (\frac{2 \sqrt{5}}{5} i - \frac{\sqrt{5}}{5} j)$

localid="1650642643659" $= [(\frac{\frac{- 16}{1^{2}}}{2 \sqrt{\frac{16}{1}}}) i + (\frac{\frac{1}{1}}{2 \sqrt{\frac{16}{1}}}) j] \times (\frac{2 \sqrt{5}}{5} i - \frac{\sqrt{5}}{5} j)$

localid="1650642674743" $= [(\frac{- 16}{8}) i + (\frac{1}{8}) j] \times (\frac{2 \sqrt{5}}{5} i - \frac{\sqrt{5}}{5} j)$

localid="1650642705305" $= (- 2 i + \frac{1}{8} j) \times (\frac{2 \sqrt{5}}{5} i - \frac{\sqrt{5}}{5} j)$

$= (\frac{- 2 \cdot 2 \sqrt{5}}{5} - \frac{\sqrt{5}}{8 \cdot 5})$

$= \frac{\sqrt{5}}{5} (- 4 - \frac{1}{8})$

$= \frac{\sqrt{5}}{5} (- \frac{33}{8})$

$\nabla f (P) . u = \frac{- 33}{40} \sqrt{5}$

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In Exercise $35 - 38$ , find the directional derivative of the given function at the specified point $P$ and in the direction of the given vector $v$ .
$f (x, y) = \sqrt{\frac{y}{x}} a t P = (1, 16), v = (2, 1)$

Short Answer

Step by step solution

Directional derivative.

Directional unit vector.

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In Exercise 35-38, find the directional derivative of the given function at the specified point Pand in the direction of the given vector v.f(x,y)=yxatP=(1,16),v=(2,1)

Short Answer

Step by step solution

Directional derivative.

Directional unit vector.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Discrete Mathematics

Calculus

Theoretical and Mathematical Physics

Probability and Statistics

Applied Mathematics

Mechanics Maths

Study anywhere. Anytime. Across all devices.

In Exercise $35 - 38$ , find the directional derivative of the given function at the specified point $P$ and in the direction of the given vector $v$ .
$f (x, y) = \sqrt{\frac{y}{x}} a t P = (1, 16), v = (2, 1)$