## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 14 - Calculus of Vector-Valued Functions - 14.2 Calculus of Vector-Valued Functions - Exercises - Page 721: 50

#### Answer

\begin{align} r(t)&= \left\langle -\frac{1}{3}\cos 3t +2, -\frac{1}{3}\cos 3t+4, \frac{1}{2} t^2+ \pi^2/8 \right\rangle . \end{align}

#### Work Step by Step

By integration, we have \begin{align} r(t)&= \int \left\langle\sin 3t, \sin 3t, t \right\rangle dt \\ &= \left\langle \int \sin 3t dt, \int \sin 3tdt, \int t dt\right\rangle \\ &= \left\langle -\frac{1}{3}\cos 3t +c_1, -\frac{1}{3}\cos 3t+c_2, \frac{1}{2} t^2+c_3 \right\rangle. \\ \end{align} By the condition $r(\pi/2)=\lt 2,4,\pi^2/4\gt$, we get $$2=c_1, \quad 4=c_2, \quad \pi^2/8=c_3$$ Hence, we have \begin{align} r(t)&= \left\langle -\frac{1}{3}\cos 3t +2, -\frac{1}{3}\cos 3t+4, \frac{1}{2} t^2+ \pi^2/8 \right\rangle . \end{align}

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