Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 13 - Vector Functions - 13.4 Motion in Space: Velocity and Acceleration - 13.4 Exercises: 9

Answer

$$\mathbf{v}(t)=\langle{2t+1, 2t-1, 3t^2}\rangle\\ \mathbf{a}(t)=\langle{2, 2, 6t}\rangle\\ |\mathbf{v}(t)|=\sqrt{(2t+1)^2+(2t-1)^2+(3t^2)^2}$$

Work Step by Step

1) Differentiate $\mathbf{r}(t)$ to find $\mathbf{r}'(t)=\mathbf{v}(t)$ using power rule. $$\mathbf{v}(t)=\mathbf{r}'(t)=\frac{d(\langle{t^2+t, t^2-t, t^3}\rangle)}{dt}\\ \mathbf{v}(t)=\langle{2t+1, 2t-1, 3t^2}\rangle$$ 2) Differentiate $\mathbf{v}(t)$ to find $\mathbf{v}'(t)=\mathbf{a}(t)$ using power rule. $$\mathbf{a}(t)=\mathbf{v}'(t)=\frac{d(\langle{2t+1, 2t-1, 3t^2}\rangle)}{dt}\\ \mathbf{a}(t)=\langle{2, 2, 6t}\rangle$$ 3) Find magnitude $|\mathbf{v}(t)|$ of $\mathbf{v}(t)$ to find speed. $$speed=|\mathbf{v}(t)|=\sqrt{(2t+1)^2+(2t-1)^2+(3t^2)}$$
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