Calculus 8th Edition

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

Chapter 13 - Vector Functions - 13.2 Derivatives and Integrals of Vector Functions - 13.2 Exercises - Page 900: 17


$\mathbf{T}(2)=\langle\frac{2}{7}, \frac{3}{7}, \frac{6}{7}\rangle$

Work Step by Step

1) Differentiate $\mathbf{r}(t)$ by differentiating each component with power rule --> $\mathbf{r}'(t)=\frac{d(\langle{t^2-2t, 1+3t, \frac{1}{3}t^3+{1}{2}t^2}\rangle)}{dt}=\langle{2t-2, 3, t^2+t}\rangle$ 2) Evaluate $\mathbf{r}'(t)$ at given parameter value $t=2$ -->$\mathbf{r}'(2)=\langle2, 3, 6\rangle$ 3) Find magnitude of $\mathbf{r}'(2)$ --> $|\mathbf{r}'(2)|=\sqrt{2^2+3^2+6+2}=\sqrt{4+9+36}=\sqrt{49}=7$ 4) By definition, $\mathbf{T}(t)=\frac{\mathbf{r}'(t)}{|\mathbf{r}'(t)|}$ 5) Therefore, $\mathbf{T}(2)=\frac{\mathbf{r}'(2)}{|\mathbf{r}'(2)|}=\frac{\langle2, 3, 6\rangle}{7}=\langle\frac{2}{7}, \frac{3}{7}, \frac{6}{6}\rangle$
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