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Chapter 12: Q 64. (page 965)

Use at least two methods to prove that $\frac{d r}{d t} = 0$ when $r = \sqrt{x^{2} + y^{2}}$ $x = 伪 \cos t, and y = 伪 \sin t$ if $伪$ is constant.

Short Answer

Expert verified

Above relation is proved using chain rule.

$\frac{d r}{d t} = \frac{鈭� r}{鈭� x} \frac{d x}{d t} + \frac{鈭� r}{鈭� y} \frac{d y}{d t}$

Step by step solution

01

Given Information

It is given that

$r = \sqrt{x^{2} + y^{2}}, x = 伪 \cos t and y = 伪 \sin t$

02

Applying Chain Rule

Using chain rule

$\frac{d r}{d t} = \frac{鈭� r}{鈭� x} \frac{d x}{d t} + \frac{鈭� r}{鈭� y} \frac{d y}{d t}$

Solving for $\frac{鈭� r}{鈭� x}$

$\frac{鈭� r}{鈭� x} = \frac{鈭�}{鈭� x} (\sqrt{x^{2} + y^{2}})$

$= \frac{1}{2} {(x^{2} + y^{2})}^{- \frac{1}{2}} \frac{鈭�}{鈭� x} (x^{2} + y^{2})$

$= \frac{1}{2 \sqrt{x^{2} + y^{2}}} (2 x + 0)$

$= \frac{x}{\sqrt{x^{2} + y^{2}}}$

Finding $\frac{鈭� r}{鈭� y}$

$\frac{鈭� r}{鈭� y} = \frac{鈭�}{鈭� y} (\sqrt{x^{2} + y^{2}})$

$= \frac{1}{2} {(x^{2} + y^{2})}^{- \frac{1}{2}} \frac{鈭�}{鈭� y} (x^{2} + y^{2})$

$= \frac{1}{2 \sqrt{x^{2} + y^{2}}} (0 + 2 y)$

$= \frac{y}{\sqrt{x^{2} + y^{2}}}$

03

Differentiating w.r.t t

$\frac{d x}{d t} = \frac{d}{d t} 伪 \cos t = - 伪 \sin t$

Also

$\frac{d y}{d t} = \frac{d}{d t} 伪 \sin t = 伪 \cos t$

Solving for $\frac{d z}{d t} = \frac{鈭� z}{鈭� x} \frac{d x}{d t} + \frac{鈭� z}{鈭� y} \frac{d y}{d t}$

$= (\frac{x}{\sqrt{x^{2} + y^{2}}}) (- 伪 \sin t) + (\frac{y}{\sqrt{x^{2} + y^{2}}}) (伪 \cos t)$

$= \frac{伪}{\sqrt{x^{2} + y^{2}}} (- x \sin t + y \cos t)$

$= \frac{伪}{\sqrt{(伪 \cos t)^{2} + (伪 \sin t)^{2}}} (- 伪 \cos t \sin t + 伪 \sin t \cos t)$

$= \frac{伪}{\sqrt{伪^{2} (\cos^{2} t + \sin^{2} t)}} 路 0 = 0$

04

Solve using direct derivative

Use $x = 伪 \cos t and y = 伪 \sin t in r = \sqrt{x^{2} + y^{2}}$

Hence,

$r = \sqrt{x^{2} + y^{2}}$

$= \sqrt{(伪 \cos t)^{2} + (伪 \sin t)^{2}}$

$= \sqrt{伪^{2}}$

Differentiating $r = 伪$ wrt $t$

$\frac{d r}{d t} = \frac{d}{d t} 伪 = 0$

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Most popular questions from this chapter

Let $f (x, y)$ be a differentiable function such that $鈭� f (x, y) 鈮� 0$ for every point in the domain of f, and let $R$ be a closed, bounded subset of role="math" localid="1649887954022" ${鈩�}^{2} .$ Explain why the maximum and minimum of f restricted to $R$ occur on the boundary ofrole="math" localid="1649888770915" $R .$

Find the gradient of the given function, and find the direction in which the function increases most rapidly at the specified point P.

$f (x, y) = \frac{x}{y^{鈥�}} P = (4, 3)$

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

Use Theorem 12.32 to find the indicated derivatives in Exercises 21鈥�26. Express your answers as functions of a single variable.

$\frac{d y}{d t} w h e n y = r \sin 胃, r = t^{3}, a n d 胃 = \sqrt{t}$

In Exercises, find the maximum and minimum of the function f subject to the given constraint. In each case explain why the maximum and minimum must both exist.

$f (x, y) = x + y$ when $x^{2} + y^{2} = 16$

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