Chapter 12: Q. 35 (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 $v$ .
$f (x, y) = x^{2} - y^{2} at P = (3, 3), v = ⟨ - 1, 5 ⟩$

Short Answer

Expert verified

The function of directional derivative is $\nabla f (P) . u = \frac{- 18}{13} \sqrt{26}$ .

Step by step solution

Directional derivative of function.

For a given function $P = (x_{0}, y_{0}) = (3, 3)$ and $v = (- 1, 5)$ , we must find the directional derivative $f (x, y) = x^{2} - y^{2}$ .

$∥ v ∥ = \sqrt{- 1^{2} + 5^{2}} = \sqrt{26}$

$∴ u = (α, β) = (\frac{- \sqrt{26}}{26}, \frac{5 \sqrt{26}}{26})$

Directional unit vector.

The directional derivative of a function at point $P$ with direction unit vector $u$ is computed as follows:

localid="1650641425421" $\nabla f (P) \cdot u = \nabla f (3, 3) \times u = ({\frac{d f}{d x}|}_{(3, 3)} i + {\frac{d f}{d y}|}_{(3, 3)} j) \times (\frac{- \sqrt{26}}{26} i + \frac{5 \sqrt{26}}{26} j)$

localid="1650641448850" $= [(2 x)_{(3, 3)^{3}} i + (- 2 y)_{(3, 3)} j] \times (\frac{- \sqrt{26}}{26} i + \frac{5 \sqrt{26}}{26} j)$

localid="1650641466180" $= (6 i - 6 j) \times (\frac{- \sqrt{26}}{26} i + \frac{5 \sqrt{26}}{26} j)$

$= (\frac{- 6 \sqrt{26}}{26} - \frac{30 \sqrt{26}}{26})$

$= \frac{- 36 \sqrt{26}}{26}$

$\nabla f (P) . u = \frac{- 18}{13} \sqrt{26}$

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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 $v$ .
$f (x, y) = x^{2} - y^{2} at P = (3, 3), v = ⟨ - 1, 5 ⟩$

Short Answer

Step by step solution

Directional derivative of function.

Directional unit vector.

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In Exercises 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)=x2−y2atP=(3,3),v=⟨−1,5⟩

Short Answer

Step by step solution

Directional derivative of function.

Directional unit vector.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Pure Maths

Theoretical and Mathematical Physics

Decision Maths

Probability and Statistics

Statistics

Geometry

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 $v$ .
$f (x, y) = x^{2} - y^{2} at P = (3, 3), v = ⟨ - 1, 5 ⟩$