Chapter 12: Q. 26 (page 964)

Use Theorem 12.32 to find the indicated derivatives in Exercises 21–26. Express your answers as functions of a single variable.
$\frac{d θ}{d t} w h e n θ = \tan^{- 1} \frac{y}{x}, x = e^{t}, a n d y = e^{2 t}$

Short Answer

Expert verified

The value is $\frac{d θ}{d t} = \frac{e^{t}}{1 + e^{2 t}}$

Step by step solution

Step 1. Given Information:

Given:

$θ = t a n^{- 1} \frac{y}{x}, x = e^{t} a n d y = e^{2 t}$

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

Step 2. Solution:

$U \sin g x = e^{t}, a n d y = e^{2 t} i n θ = \tan^{- 1} \frac{y}{x} w e g e t θ = \tan^{- 1} \frac{e^{2 t}}{e^{t}} θ = \tan^{- 1} e^{t} D i f f . w . r . t . t w e g e t \frac{d θ}{d t} = \frac{1}{1 + e^{2 t}} \cdot e^{t} \frac{d θ}{d t} = \frac{e^{t}}{1 + e^{2 t}}$

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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.
$\frac{d θ}{d t} w h e n θ = \tan^{- 1} \frac{y}{x}, x = e^{t}, a n d y = e^{2 t}$

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.dθdtwhenθ=tan−1yx,x=et,andy=e2t

Short Answer

Step by step solution

Step 1. Given Information:

Step 2. Solution:

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Most popular questions from this chapter

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Geometry

Applied Mathematics

Calculus

Discrete Mathematics

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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.
$\frac{d θ}{d t} w h e n θ = \tan^{- 1} \frac{y}{x}, x = e^{t}, a n d y = e^{2 t}$