Q. 66 Let r1(t)聽and r2(t) both be dif... [FREE SOLUTION]

Chapter 11: Q. 66 (page 873)

Let $r_{1} (t)$ and $r_{2} (t)$ both be differentiable three-component vector functions. Prove that
$\frac{d}{d t} (r_{1} (t) 脳 r_{2} (t)) = r_{1}^{'} (t) 脳 r_{2} (t) + r_{1} (t) 脳 r_{2}^{'} (t)$ (This is Theorem 11.11 (d).)

Short Answer

Expert verified

Ans: $r_{1} (t) 脳 r_{2} (t) = <x_{1} (t), y_{1} (t), z_{1} (t)> 脳 <x_{2} (t), y_{2} (t), z_{2} (t)> = r_{1} {}^{'}(t) 脳 r_{2} (t) + r_{1} (t) 脳 r_{2} {}^{'}(t)$

Step by step solution

Step 1. Given information:

$r_{1} (t)$ and $r_{2} (t)$ are differentiable vector functions with three components each.

Step 2. Proving :

Let $r_{1} (t) = <x_{1} (t), y_{1} (t), z_{1} (t)>$ and $r_{2} (t) = <x_{2} (t), y_{2} (t), z_{2} (t)>$

Then $r_{1} {}^{'}(t) = <x_{1} {}^{'}(t), y_{1} {}^{'}(t), z_{1} {}^{'}(t)>$ and $r_{2} {}^{'}(t) = <x_{2} {}^{'}(t), y_{2} {}^{'}(t), z_{2} {}^{'}(t)>$

$r_{1} (t) 脳 r_{2} (t) = <x_{1} (t), y_{1} (t), z_{1} (t)> 脳 <x_{2} (t), y_{2} (t), z_{2} (t)>$

$= <y_{1} (t) z_{2} (t) - y_{2} (t) z_{1} (t), x_{2} (t) z_{1} (t) - x_{1} (t) z_{2} (t), x_{1} (t) y_{2} (t) - x_{2} (t) y_{1} (t)>$

$= \frac{d}{d t} (<y_{1} (t) z_{2} (t) - y_{2} (t) z_{1} (t), x_{2} (t) z_{1} (t) - x_{1} (t) z_{2} (t), x_{1} (t) y_{2} (t) - x_{2} (t) y_{1} (t)>)$

$= <\begin{array}{l} y_{1} {}^{'}(t) z_{2} (t) + y_{1} (t) z_{2} {}^{'}(t) - y_{2} {}^{'}(t) z_{1} (t) - y_{2} (t) z_{1} {}^{'}(t), \\ x_{2} {}^{'}(t) z_{1} (t) + x_{2} (t) z_{1} {}^{'}(t) - x_{1} {}^{'}(t) z_{2} (t) - x_{1} (t) z_{2} {}^{'}(t), \\ x_{1} {}^{'}(t) y_{2} (t) + x_{1} (t) y_{2} (t) - x_{2} {}^{'}(t) y_{1} (t) - x_{2} (t) y_{1} {}^{'}(t) \end{array}>$

$= <y_{1} {}^{'}(t) z_{2} (t) - y_{2} (t) z_{1} {}^{'}(t), x_{2} (t) z_{1} {}^{'}(t) - x_{1} {}^{'}(t) z_{2} (t), x_{1} {}^{'}(t) y_{2} (t) - x_{2} (t) y_{1} {}^{'}(t)>$

$+ <y_{1} (t) z_{2} {}^{'}(t) - y_{2} {}^{'}(t) z_{1} (t), x_{2} {}^{'}(t) z_{1} (t) - x_{1} (t) z_{2} {}^{'}(t), x_{1} (t) y_{2} {}^{'}(t) - x_{2} {}^{'}(t) y_{1} (t)>$

$= r_{1} {}^{'}(t) 脳 r_{2} (t) + r_{1} (t) 脳 r_{2} {}^{'}(t)$

Thus

$\frac{d}{d t} (r_{1} (t) 脳 r_{2} (t)) = r_{1} {}^{'}(t) 脳 r_{2} (t) + r_{1} (t) 脳 r_{2} {}^{'}(t)$

Unlock Step-by-Step Solutions & Ace Your Exams!

Full Textbook Solutions
Get detailed explanations and key concepts
Unlimited Al creation
Al flashcards, explanations, exams and more...
Ads-free access
To over 500 millions flashcards
Money-back guarantee
We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

91影视

Let $r_{1} (t)$ and $r_{2} (t)$ both be differentiable three-component vector functions. Prove that
$\frac{d}{d t} (r_{1} (t) 脳 r_{2} (t)) = r_{1}^{'} (t) 脳 r_{2} (t) + r_{1} (t) 脳 r_{2}^{'} (t)$ (This is Theorem 11.11 (d).)

Short Answer

Step by step solution

Step 1. Given information:

Step 2. Proving :

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Logic and Functions

Decision Maths

Statistics

Mechanics Maths

Pure Maths

Calculus

Study anywhere. Anytime. Across all devices.

91影视

Let r1(t)and r2(t) both be differentiable three-component vector functions. Prove thatddtr1(t)脳r2(t)=r1'(t)脳r2(t)+r1(t)脳r2'(t) (This is Theorem 11.11 (d).)

Short Answer

Step by step solution

Step 1. Given information:

Step 2. Proving :

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Logic and Functions

Decision Maths

Statistics

Mechanics Maths

Pure Maths

Calculus

Study anywhere. Anytime. Across all devices.

Let $r_{1} (t)$ and $r_{2} (t)$ both be differentiable three-component vector functions. Prove that
$\frac{d}{d t} (r_{1} (t) 脳 r_{2} (t)) = r_{1}^{'} (t) 脳 r_{2} (t) + r_{1} (t) 脳 r_{2}^{'} (t)$ (This is Theorem 11.11 (d).)