Q. 23 In Exercises 21鈥�23 you are giv... [FREE SOLUTION]

Chapter 11: Q. 23 (page 872)

In Exercises 21鈥�23 you are given a vector function $r$ and a scalar function $t = f (蟿)$ . Compute $\frac{d r}{d 蟿}$ in the following two ways:
(a) By using the chain rule $\frac{d r}{d 蟿} = \frac{d r}{d t} \frac{d t}{d 蟿}$ .
(b) By substituting $t = f (蟿)$ into the formula for $r$ . Ensure that your two answers are consistent.

Short Answer

Expert verified

Part (a) The answer is $\frac{d r}{d T} = <T \cos T + \sin T, \frac{1}{2 \sqrt{T}}, - T \sin T + \cos T>$

Part (b) The value is $\frac{d r}{d T} = <T \cos T + \sin T, \frac{1}{2 \sqrt{T}}, - T \sin T + \cos T>$

The two answers are consistent, that is

\frac{d r}{d 蟿} = <蟿 \cos 鈦� 蟿 + \sin 鈦� 蟿, \frac{1}{2 \sqrt{蟿}}, 鈭� 蟿 \sin 鈦� 蟿 + \cos 鈦� 蟿>

Step by step solution

Part (a) Step 1. Given data

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

We have to find $\frac{d r}{d 蟿}$ in two ways,

Part (a) Step 2. Finddrdτ

By using the chain rule, $\frac{d r}{d 蟿} = \frac{d r}{d t} 鈰� \frac{d t}{d 蟿}$

$\begin{aligned} \frac{d r}{d 蟿} & = \frac{d}{d t} r (t) 鈰� \frac{d}{d 蟿} (t) \\ = \frac{d}{d t} 鉄� t^{2} \sin 鈦� t^{2}, t, t^{2} \cos 鈦� t^{2} 鉄� 鈰� \frac{d}{d 蟿} (\sqrt{蟿}) \\ = 鉄� t^{2} (2 t) \cos 鈦� t^{2} + 2 t \sin 鈦� t^{2}, 1, 鈭� t^{2} (2 t) \sin 鈦� t^{2} + 2 t \cos 鈦� t^{2} 鉄� \frac{1}{2 \sqrt{蟿}} \\ = 鉄� 2 t^{3} \cos 鈦� t^{2} + 2 t \sin 鈦� t^{2}, 1, 鈭� 2 t^{3} \sin 鈦� t^{3} + 2 t \cos 鈦� t^{2} 鉄� \frac{1}{2 \sqrt{蟿}} \end{aligned} \begin{aligned} = 鉄� 2 蟿^{\frac{3}{2}} \cos 鈦� 蟿 + 2 \sqrt{蟿} \sin 鈦� 蟿, 1, 鈭� 2 蟿^{\frac{3}{2}} \sin 鈦� 蟿 + 2 \sqrt{蟿} \cos 鈦� 蟿鉄� \frac{1}{2 \sqrt{蟿}} \\ = 鉄� 蟿 \cos 鈦� 蟿 + \sin 鈦� 蟿, \frac{1}{2 \sqrt{蟿}}, 鈭� 蟿 \sin 鈦� 蟿 + \cos 鈦� 蟿鉄� \end{aligned}$

Part (b) Step 1. Find drdτ by substituting t=f(τ)

By substituting $t = \sin 鈦� 蟿$ in $r (t)$

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

Here, results obtained in part (a) and part (b) are equal.

Therefore, the two answers are consistent.

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Short Answer

Step by step solution

Part (a) Step 1. Given data

Part (a) Step 2. Finddrdτ

Part (b) Step 1. Find drdτ by substituting t=f(τ)

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91影视

In Exercises 21鈥�23 you are given a vector function rand a scalar function t=f(蟿). Compute drd蟿in the following two ways:(a) By using the chain rule drd蟿=drdtdtd蟿.(b) By substituting t=f(蟿)into the formula for r. Ensure that your two answers are consistent.

Short Answer

Step by step solution

Part (a) Step 1. Given data

Part (a) Step 2. Finddrdτ

Part (b) Step 1. Find drdτ by substituting t=f(τ)

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Mechanics Maths

Logic and Functions

Probability and Statistics

Applied Mathematics

Decision Maths

Pure Maths

Study anywhere. Anytime. Across all devices.