Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 14 - Calculus of Vector-Valued Functions - 14.2 Calculus of Vector-Valued Functions - Exercises - Page 721: 52


\begin{align} r(t)&= \left\langle \frac{1}{4} e^{2t-2}+\frac{3}{2}t-\frac{7}{4}, \frac{1}{12}t^4-\frac{1}{2}t^2+ \frac{2}{3}t -\frac{1}{4}, \frac{1}{2}t^2 -t+\frac{3}{2}\right\rangle . \end{align}

Work Step by Step

By integration, we have \begin{align} r'(t)& = \left\langle \frac{1}{2} e^{2t-2}+c_1, \frac{1}{3}t^3-t+ c_2, t +c_3 \right\rangle . \end{align} By the condition $r'(1)=\lt 2,0,0\gt$, we get $$\frac{3}{2}=c_1, \quad \frac{2}{3}=c_2, \quad -1=c_3$$ Hence we have \begin{align} r'(t)& = \left\langle \frac{1}{2} e^{2t-2}+\frac{3}{2}, \frac{1}{3}t^3-t+ \frac{2}{3}, t -1 \right\rangle . \end{align} Again, by integration we have \begin{align} r(t)& = \left\langle \frac{1}{4} e^{2t-2}+\frac{3}{2}t+a, \frac{1}{12}t^4-\frac{1}{2}t^2+ \frac{2}{3}t+b, \frac{1}{2}t^2 -t+c \right\rangle . \end{align} By the condition $r(1)=\lt0,0,1\gt$, we get $$-\frac{7}{4}=a, \quad -\frac{1}{4}=b, \quad \frac{3}{2}=c$$ Hence we have \begin{align} r(t)&= \left\langle \frac{1}{4} e^{2t-2}+\frac{3}{2}t-\frac{7}{4}, \frac{1}{12}t^4-\frac{1}{2}t^2+ \frac{2}{3}t -\frac{1}{4}, \frac{1}{2}t^2 -t+\frac{3}{2}\right\rangle . \end{align}
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