Chapter 12: Q. 37 (page 953)

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

Short Answer

Expert verified

The directional derivative function is $\frac{- 33}{40} \sqrt{5}$

Step by step solution

Given data

$f (x, y) = \sqrt{\frac{y}{x}}$

$P = (x_{0}, y_{0}) = (1, 16) and v = (2, - 1)$

Solution

Therefore

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

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

Directional derivative of function is given by

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

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

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

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

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

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

Step 3

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

localid="1650377777679" $= \frac{\sqrt{5}}{5} (- 4 - \frac{1}{8})$

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

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

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

Short Answer

Step by step solution

Given data

Solution

Step 3

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In Exercises 35-38, find the directional derivative of the givenfunction at the specified point Pand in the direction of thegiven vector v.f(x,y)=yxatP=(1,16),v=⟨2,−1⟩

Short Answer

Step by step solution

Given data

Solution

Step 3

One App. One Place for Learning.

Most popular questions from this chapter

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Statistics

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Study anywhere. Anytime. Across all devices.

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