Chapter 12: Q. 25 (page 964)

Use Theorem 12.32 to find the indicated derivatives in Exercises 21–26. Express your answers as functions of a single variable.
$\frac{d r}{d t} w h e n r = \sqrt{x^{2} + y^{2}}, x = \sqrt{t} a n d y = t^{2}$

Short Answer

Expert verified

The value is $\frac{d r}{d t} = \frac{1 + 4 t^{3}}{2 \sqrt{t + t^{4}}}$

Step by step solution

Step 1. Given Information:

Given:

$r = \sqrt{x^{2} + y^{2}}, x = \sqrt{t} a n d y = t^{2}$

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

Step 2. Solution:

$U \sin g x = \sqrt{t} a n d y = t^{2} i n r = \sqrt{x^{2} + y^{2}} w e g e t r = \sqrt{{(\sqrt{t})}^{2} + {(t^{2})}^{2}} r = \sqrt{t + t^{4}} D i f f . w . r . t . t w e g e t \frac{d r}{d t} = \frac{1}{2 \sqrt{t + t^{4}}} \cdot (1 + 4 t^{3}) \frac{d r}{d t} = \frac{1 + 4 t^{3}}{2 \sqrt{t + t^{4}}}$

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Use Theorem 12.32 to find the indicated derivatives in Exercises 21–26. Express your answers as functions of a single variable.
$\frac{d r}{d t} w h e n r = \sqrt{x^{2} + y^{2}}, x = \sqrt{t} a n d y = t^{2}$

Short Answer

Step by step solution

Step 1. Given Information:

Step 2. Solution:

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Use Theorem 12.32 to find the indicated derivatives in Exercises 21–26. Express your answers as functions of a single variable.drdtwhenr=x2+y2,x=tandy=t2

Short Answer

Step by step solution

Step 1. Given Information:

Step 2. Solution:

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Geometry

Theoretical and Mathematical Physics

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Study anywhere. Anytime. Across all devices.

Use Theorem 12.32 to find the indicated derivatives in Exercises 21–26. Express your answers as functions of a single variable.
$\frac{d r}{d t} w h e n r = \sqrt{x^{2} + y^{2}}, x = \sqrt{t} a n d y = t^{2}$