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Chapter 12: Q. 23 (page 964)

Use Theorem 12.32 to find the indicated derivatives in Exercises 21鈥�26. Express your answers as functions of a single variable. $\frac{d x}{d t} w h e n x = r \cos 胃, r = t^{2} 鈭� 5, a n d 胃 = t^{3} + 1$

Short Answer

Expert verified

The value is $\frac{d x}{d t} = 2 t 路 \cos (t^{3} + 1) - 3 t^{2} (t^{2} 鈭� 5) \sin (t^{3} + 1)$

Step by step solution

01

Step 1. Given information:

Given:

$x = r \cos 胃, r = t^{2} 鈭� 5, a n d 胃 = t^{3} + 1$

We have to find the indicated derivatives and express your answers as functions of a single variable.

02

Step 2. Solution:

$U \sin g r = t^{2} 鈭� 5 a n d 胃 = t^{3} + 1 i n x = r \cos 胃 w e g e t x = (t^{2} 鈭� 5) 路 \cos (t^{3} + 1) D i f f . w . r . t . t w e g e t \frac{d x}{d t} = (t^{2} 鈭� 5) \frac{d}{d t} \cos (t^{3} + 1) + \cos (t^{3} + 1) \frac{d}{d t} (t^{2} 鈭� 5) \frac{d x}{d t} = (t^{2} 鈭� 5) (- \sin (t^{3} + 1) 路 3 t^{2}) + \cos (t^{3} + 1) 路 2 t \frac{d x}{d t} = 2 t 路 \cos (t^{3} + 1) - 3 t^{2} (t^{2} 鈭� 5) \sin (t^{3} + 1)$

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Most popular questions from this chapter

In Exercises 21鈥�26, find the discriminant of the given function.

$f (x, y) = e^{2 x} \cos y$ .

Use Theorem 12.33 to find the indicated derivatives in Exercises 27鈥�30. Express your answers as functions of two variables.

$\frac{鈭� z}{鈭� 胃} w h e n z = (x^{2} + x y) e^{y}, x = r \cos 胃 a n d y = r \sin 胃$

$Let p (x) be a polynomial of degree n, q (y) be any function of y, and f (x, y) = p (x) q (y) . Prove that \frac{{鈭�}^{n + 1} f}{鈭� x^{n + 1}} = 0$

In Exercises $21 - 28$ , find the directional derivative of the given function at the specified point $P$ and in the direction of the given unit vector $u$ .

$f (x, y) = \sqrt{\frac{y}{x}}$ at $P = (4, 9), u = (- \frac{\sqrt{17}}{17}, - \frac{4 \sqrt{17}}{17})$

Explain how you could use the method of Lagrange multipliers to find the extrema of a function of two variables, $f (x, y),$ subject to the constraint that $(x, y)$ is on the boundary of the rectangle $R$ defined by $a 鈮� x 鈮� b and c 鈮� y 鈮� d .$

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