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Q. 17

Expert-verified

Calculus

Found in: Page 871

Book edition 1st

Author(s) Peter Kohn, Laura Taalman

Pages 1155 pages

ISBN 9781429241861

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Short Answer

$Find the derivative of the vector - valued function r (t) = <t^{3}, 5 t^{3}, - 2 t^{3}>, t > 0, and show that the derivative at any point t_{0} is a scalar multiple of the derivative at any other point . What does this property say about the graph of r ? Sketch the graph of r .$

$r' (t) = \frac{d <t^{3}, 5 t^{3}, - 2 t^{3}>}{d t} r' (t) = <3 t^{2}, 15 t^{2}, - 6 t^{2}> The graph of r (t) is a portion of straight line .$

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1. Given information is:

$r (t) = <t^{3}, 5 t^{3}, - 2 t^{3}>, t > 0$

Step 2. Calculating derivative

$r' (t) = \frac{d <t^{3}, 5 t^{3}, - 2 t^{3}>}{d t} r' (t) = <3 t^{2}, 15 t^{2}, - 6 t^{2}>$

Step 3. Graph of r

$Let t_{0} and t_{1} be any two positive real numbers, r' (t_{0}) = <3 t_{0}^{2}, 15 t_{0}^{2}, - 6 t_{0}^{2}> r' (t_{0}) = 3 t_{0}^{2} <1, 5, - 2> r' (t_{1}) = <3 t_{1}^{2}, 15 t_{1}^{2}, - 6 t_{1}^{2}> r' (t_{0}) = 3 t_{1}^{2} <1, 5, - 2> The vectors r (t_{0}) and r' (t_{1}) are multiples of the vector <1, 5, - 2>, hence, they are scalar multiples of each other . Thus, the graph of r (t) is a portion of straight line .$

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Domain

$t \in ℝ, r (2) = (1, 1), r (0) = (- 3, 3), r (- 2) = (- 5, - 5) \lim_{t \to - \infty} x (t) = \infty, \lim_{t \to - \infty} y (t) = - \infty, and \lim_{t \to \infty} r (t) = (0, 0) .$

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