Chapter 11: Q. 65 (page 873)

Let $r_{1} (f)$ and $r_{2} (t)$ be differentiable vector functions with three components each. Prove that
$\frac{d}{d t} (r_{1} (t) - r_{2} (t)) = r_{1}^{'} (t) - r_{2} (t) + r_{1} (t) - r_{2}^{'} (t) ∣$ (This is Theorem 11.11 (c).)

Short Answer

Expert verified

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

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 :

Then

r_{1} (t) r_{2} (t) = <x_{1} (t), y_{1} (t), z_{1} (t)> \cdot <x_{2} (t), y_{2} (t), z_{2} (t)> = x_{1} (t) x_{2} (t) + y_{1} (t) y_{2} (t) + z_{1} (t) z_{2} (t) \frac{d}{d t} (r_{1} (t) \cdot r_{2} (t)) = \frac{d}{d t} (x_{1} (t) x_{2} (t) + y_{1} (t) y_{2} (t) + z_{1} (t) z_{2} (t)) = \frac{d}{d t} (x_{1} (t) x_{2} (t)) + \frac{d}{d t} (y_{1} (t) y_{2} (t)) + \frac{d}{d t} (z_{1} (t) z_{2} (t)) = x_{1} {}^{'}(t) x_{2} (t) + x_{1} (t) x_{2}^{'} (t) + y_{1} {}^{'}(t) y_{2} (t) + y_{1} (t) y_{2} {}^{'}(t) + z_{1}^{'} (t) z_{2} (t) + z_{1} (t) z_{2}^{'} (t) = x_{1} {}^{'}(t) x_{2} (t) + y_{1}^{'} (t) y_{2} (t) + z_{1}^{'} (t) z_{2} (t) + x_{1} (t) x_{2}^{'} (t) + y_{1} (t) y_{2}^{'} (t) + z_{1} (t) z_{2}^{'} (t) = <x_{1} {}^{'}(t), y_{1} {}^{'}(t), z^{'} (t)> <x_{2} (t), y_{2} (t), z_{2} (t)> + <x_{1} (t), y_{1} (t), z_{1} (t)> <x_{2}^{'} (t), y_{2}^{'} (t), z_{2}^{'} (t)> = r_{1} {}^{'}(t) \cdot r_{2} (t) + r_{1} (t) r_{2}^{'}

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

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Let $r_{1} (f)$ and $r_{2} (t)$ be differentiable vector functions with three components each. Prove that
$\frac{d}{d t} (r_{1} (t) - r_{2} (t)) = r_{1}^{'} (t) - r_{2} (t) + r_{1} (t) - r_{2}^{'} (t) ∣$ (This is Theorem 11.11 (c).)

Short Answer

Step by step solution

Step 1. Given information:

Step 2. Proving :

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Let r1(f)and r2(t) be differentiable vector functions with three components each. Prove thatddtr1(t)-r2(t)=r1'(t)-r2(t)+r1(t)-r2'(t)∣(This is Theorem 11.11 (c).)

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

Decision Maths

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Study anywhere. Anytime. Across all devices.

Let $r_{1} (f)$ and $r_{2} (t)$ be differentiable vector functions with three components each. Prove that
$\frac{d}{d t} (r_{1} (t) - r_{2} (t)) = r_{1}^{'} (t) - r_{2} (t) + r_{1} (t) - r_{2}^{'} (t) ∣$ (This is Theorem 11.11 (c).)