Chapter 11: Q. 69 (page 874)

Let $r (t)$ be a differentiable vector function such that $r (t) \cdot r^{'} (t) = 0$ for every value of $t$ . Prove that $‖ r (t) ‖$ is a constant.

Short Answer

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Ans:

$\frac{d}{d t} (‖ r ‖^{2}) = \frac{d}{d t} (r \cdot r) = r \frac{d}{d t} (r) + r \frac{d}{d t} (r) = 0$

Step by step solution

Step 1. GIven information:

Consider a differentiable vector function $r (t)$ such that $r (t) \cdot r^{'} (t) = 0$ For every value of $t$ .

Step 2. Proving:

Consider

$\frac{d}{d t} (‖ r ‖^{2}) = \frac{d}{d t} (r \cdot r) s i c n e ‖ r ‖^{2} = r \cdot r = r \frac{d}{d t} (r) + r \frac{d}{d t} (r) Derivativeforproductof 2 functions$

$\begin{array}{l} = r \cdot r^{'} + r \cdot r^{'} & \frac{d r}{d r} = r^{'} \\ = 2 r \cdot r^{'} & add \\ = 2 (0) & since r \cdot r^{'} = 0 \\ = 0 & Multiply \end{array} \frac{d}{d t} (‖ r ‖^{2}) = 0 . That means ‖ r ‖ is a constant, since the derivative of a constant is zero.$

Thus $‖ r (t) ‖$ is a constant if $r (t) \cdot r^{'} (t) = 0$

Note: The converse is always true. That is, if $r (t)$ is a differentiable vector function such that $‖ r (t) ‖$ is a constant then $r (t) \cdot r^{'} (t) = 0$ .

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Let $r (t)$ be a differentiable vector function such that $r (t) \cdot r^{'} (t) = 0$ for every value of $t$ . Prove that $‖ r (t) ‖$ is a constant.

Short Answer

Step by step solution

Step 1. GIven information:

Step 2. Proving:

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Let r(t) be a differentiable vector function such that r(t)·r'(t)=0 for every value of t. Prove that ‖r(t)‖ is a constant.

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

Calculus

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Study anywhere. Anytime. Across all devices.

Let $r (t)$ be a differentiable vector function such that $r (t) \cdot r^{'} (t) = 0$ for every value of $t$ . Prove that $‖ r (t) ‖$ is a constant.