Q. 38 In Exercises 35-39 a vector func... [FREE SOLUTION]

Chapter 11: Q. 38 (page 872)

In Exercises 35-39 a vector function $r (t)$ and scalar function $t = f (蟿)$ are given. Find $\frac{d r}{d 蟿}$ .
$38. r (t) = 鉄� \sin 鈦� t, \cos 鈦� t, 2 \sin 鈦� 2 t 鉄�, t = \sqrt{蟿^{2} + 1}$

Short Answer

Expert verified

$\frac{d r}{d 蟿} = 鉄� \frac{蟿 \cos 鈦� \sqrt{蟿^{2} + 1}}{\sqrt{蟿^{2} + 1}}, \frac{鈭� 蟿 \sin 鈦� \sqrt{蟿^{2} + 1}}{\sqrt{蟿^{2} + 1}}, \frac{4 \cos 鈦� 2 \sqrt{蟿^{2} + 1}}{\sqrt{蟿^{2} + 1}} 鉄� 鈭�$

Step by step solution

Step 1. Given data

The given vector function is $r (t) = 鉄� \sin 鈦� t, \cos 鈦� t, 2 \sin 鈦� 2 t 鉄�, t = \sqrt{蟿^{2} + 1}$

We have to find $\frac{d r}{d 蟿}$

Step 2. Use chain rule

If $r (t)$ is a vector function and $t = f (蟿)$ , a scalar function. Then the chain rule states that $\frac{d r}{d 蟿} = \frac{d r}{d t} 鈰� \frac{d t}{d 蟿}$
We have, $\begin{aligned} r (t) = 鉄� \sin 鈦� t, \cos 鈦� t, 2 \sin 鈦� 2 t 鉄� \\ \frac{d r}{d t} = r^{鈥�} (t) = 鉄� \cos 鈦� t, 鈭� \sin 鈦� t, 4 \cos 鈦� 2 t 鉄� \end{aligned}$

It is given $t = \sqrt{蟿^{2} + 1}$

$\begin{aligned} \frac{d t}{d 蟿} = \frac{1}{2 \sqrt{蟿^{2} + 1}} (2 蟿) \\ = \frac{蟿}{\sqrt{蟿^{2} + 1}} \end{aligned}$

By the chain rule,

$\begin{aligned} \frac{d r}{d 蟿} = \frac{d r}{d t}, \frac{d t}{d 蟿} \\ = 鉄� \cos 鈦� t, 鈭� \sin 鈦� t, 4 \cos 鈦� 2 t 鉄� \frac{蟿}{\sqrt{蟿^{2} + 1}} \\ = 鉄� \cos 鈦� \sqrt{蟿^{2} + 1}, 鈭� \sin 鈦� \sqrt{蟿^{2} + 1}, 4 \cos 鈦� 2 \sqrt{蟿^{2} + 1} 鉄� \frac{蟿}{\sqrt{蟿^{2} + 1}} \\ = 鉄� \frac{蟿 \cos 鈦� \sqrt{蟿^{2} + 1}}{\sqrt{蟿^{2} + 1}}, \frac{鈭� \sin 鈦� \sqrt{蟿^{2} + 1}}{\sqrt{蟿^{2} + 1}}, \frac{4 \cos 鈦� 2 \sqrt{蟿^{2} + 1}}{\sqrt{蟿^{2} + 1}} 鉄� \end{aligned}$

Thus $\frac{d r}{d 蟿} = 鉄� \frac{蟿 c o s 鈦� \sqrt{蟿^{2} + 1}}{\sqrt{蟿^{2} + 1}}, \frac{鈭� 蟿 s i n 鈦� \sqrt{蟿^{2} + 1}}{\sqrt{蟿^{2} + 1}}, \frac{4 c o s 鈦� 2 \sqrt{蟿^{2} + 1}}{\sqrt{蟿^{2} + 1}} 鉄�$

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In Exercises 35-39 a vector function $r (t)$ and scalar function $t = f (蟿)$ are given. Find $\frac{d r}{d 蟿}$ .
$38. r (t) = 鉄� \sin 鈦� t, \cos 鈦� t, 2 \sin 鈦� 2 t 鉄�, t = \sqrt{蟿^{2} + 1}$

Short Answer

Step by step solution

Step 1. Given data

Step 2. Use chain rule

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91影视

In Exercises 35-39 a vector function r(t)and scalar function t=f(蟿)are given. Find drd蟿.38.r(t)=鉄�sin鈦�t,cos鈦�t,2sin鈦�2t鉄�,t=蟿2+1

Short Answer

Step by step solution

Step 1. Given data

Step 2. Use chain rule

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Statistics

Theoretical and Mathematical Physics

Decision Maths

Pure Maths

Discrete Mathematics

Calculus

Study anywhere. Anytime. Across all devices.

In Exercises 35-39 a vector function $r (t)$ and scalar function $t = f (蟿)$ are given. Find $\frac{d r}{d 蟿}$ .
$38. r (t) = 鉄� \sin 鈦� t, \cos 鈦� t, 2 \sin 鈦� 2 t 鉄�, t = \sqrt{蟿^{2} + 1}$