## Calculus 8th Edition

Published by Cengage

# Chapter 13 - Vector Functions - 13.2 Derivatives and Integrals of Vector Functions - 13.2 Exercises - Page 901: 41

#### Answer

$r(t)=(2(\frac{1}{2})t^2+0)i+(3(\frac{1}{3})t^3+0)j+((\frac{1}{3/2})t^{3/2}-\dfrac{2}{3})k$ or, $r(t)=t^2i+t^3j+(\dfrac{2}{3}t^{3/2}-\dfrac{2}{3})k$

#### Work Step by Step

Suppose $I=\int (2ti+3t^2j+\sqrt t k)dt$ Rewrite this equation as: $I=Ai+ Bj+Ck$ ... (a) Here, $P=2t$ and $Q=3t^2, R=\sqrt t$ $P=2t=2(\frac{1}{2})t^2+C$ and $Q=3t^2=3(\frac{1}{3})t^3+C'$ and $R=\sqrt t =(\frac{1}{3/2})t^{3/2}+C''$ From equation (a): $I=(2(\frac{1}{2})t^2+C)i+(3(\frac{1}{3})t^3+C')j+((\frac{1}{3/2})t^{3/2}+C'')k$;(C,C',C'' are the constant of integration.) From the question, $r(1)=i+j$ gives $\lt 1,1,0 \gt$. After simplifications, we get: $C=0,C'=0,C''=-\dfrac{2}{3}$ The desired result is: $r(t)=(2(\frac{1}{2})t^2+0)i+(3(\frac{1}{3})t^3+0)j+((\frac{1}{3/2})t^{3/2}-\dfrac{2}{3})k$ or, $r(t)=t^2i+t^3j+(\dfrac{2}{3}t^{3/2}-\dfrac{2}{3})k$

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