Q. 38 In exercise 35-38, find the dire... [FREE SOLUTION]

Chapter 12: Q. 38 (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 $v$ .
$f (x, y, z) = x^{2} + y^{2} 鈭� z^{3} a t P = (0, - 2, 5), v = - i + 3 j + 5 k$

Short Answer

Expert verified

The directional derivative of the given function $D_{w} f (0, - 2, 5) = - \frac{137}{\sqrt{35}}$

Step by step solution

Directional of derivative.

For given $P = (x_{0}, y_{0}, z_{0}) = (0, - 2, 5)$ and $v = - i + 3 j = 5 k$ , we must compute the directional derivative of function $f (x, y, z) = x^{2} + y^{2} - z^{2}$ .

Think this as directional derivative

$D_{u} f (x_{0}, y_{0}, z_{0}) = {Lim}_{h 鈫� 0} 鈦� \frac{f (x_{0} + 伪 h, y_{0} + 尾 h, z_{0} + 纬 h) 鈭� f (x_{0}, y_{0}, z_{0})}{h} . . . . (1)$

$||v|| = \sqrt{{(- 1)}^{2} + 3^{2} + 5^{2}} = \sqrt{1 + 9 + 25}$

$||v|| = \sqrt{35}$

$u = (伪, 尾, 纬) = (\frac{- 1}{\sqrt{35}}, \frac{3}{\sqrt{35}}, \frac{5}{\sqrt{35}})$

Equation.

Therefore,

$f (x_{0} + 伪 h, y_{0} + 尾 h, z_{0} + 纬 h) = f (0 鈭� \frac{1}{\sqrt{35}} h, 鈭� 2 + \frac{3}{\sqrt{35}} h, 5 + \frac{5}{\sqrt{35}} h)$

$= {(0 鈭� \frac{1}{\sqrt{35}} h)}^{2} + {(鈭� 2 + \frac{3}{\sqrt{35}} h)}^{2} 鈭� {(5 + \frac{5}{\sqrt{35}} h)}^{3}$

$= \frac{1}{35} h^{2} + (4 - \frac{12}{\sqrt{35}} h + \frac{9}{35} h^{2}) - (125 + \frac{375}{\sqrt{35}} h + \frac{375}{35} h^{2} + \frac{125}{{(\sqrt{35})}^{3}} h^{3})$

$= \frac{1}{35} h^{2} + 4 - \frac{12}{\sqrt{35}} h + \frac{9}{35} h^{2} - 125 - \frac{375}{\sqrt{35}} h - \frac{375}{35} h^{2} - \frac{125}{{(\sqrt{35})}^{3}} h^{3}$

$= - 121 - \frac{365}{35} h^{2} - \frac{137}{\sqrt{35}} h - \frac{125}{{(\sqrt{35})}^{3}} h^{3} . . . . . (2)$

$f (x_{0}, y_{0}, z_{0}) = 0^{2} + {(- 2)}^{2} - 5^{3} = - 121 . . . . . . (3)$

Directional derivatives

When equations $2$ and $3$ are substituted for equation $1$ , we get

$D_{w} f (0, 鈭� 2, 5) = {Lim}_{h 鈫� 0} 鈦� \frac{鈭� 121 鈭� \frac{137}{\sqrt{35}} h 鈭� \frac{365}{35} h^{2} 鈭� \frac{125}{(\sqrt{35})^{3}} h^{3} 鈭� 121}{h}$

$= {Lim}_{h 鈫� 0} 鈭� \frac{137}{\sqrt{35}} 鈭� \frac{365}{35} h 鈭� \frac{125}{(\sqrt{35})^{3}} h^{2}$

$D_{w} f (0, - 2, 5) = - \frac{137}{\sqrt{35}}$

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91影视

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 $v$ .
$f (x, y, z) = x^{2} + y^{2} 鈭� z^{3} a t P = (0, - 2, 5), v = - i + 3 j + 5 k$

Short Answer

Step by step solution

Directional of derivative.

Equation.

Directional derivatives

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91影视

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,z)=x2+y2鈭�z3atP=(0,-2,5),v=-i+3j+5k

Short Answer

Step by step solution

Directional of derivative.

Equation.

Directional derivatives

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Statistics

Theoretical and Mathematical Physics

Probability and Statistics

Decision Maths

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Calculus

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 $v$ .
$f (x, y, z) = x^{2} + y^{2} 鈭� z^{3} a t P = (0, - 2, 5), v = - i + 3 j + 5 k$