## Calculus 8th Edition

Published by Cengage

# Chapter 16 - Vector Calculus - 16.2 Line Integrals - 16.2 Exercises - Page 1126: 47

#### Answer

a) It has been proved that a constant force field does zero work on a particle that moves once uniformly around the circle $x^2 +y^2=1$ b) Yes

#### Work Step by Step

(a) Let us consider that $F(x,y)=A(x,y) i+B(x,y) j$ Since, $W=\int_C F\cdot ds=\int_C A dx +\int_C B dy$ or, $=\int_0^{2 \pi} A [-\sin t dt]+\int_0^{2 \pi} B [\cos t dt]=p(0)+q(0)=0$ This shows that a constant force field does zero work on a particle that moves once uniformly around the circle $x^2 +y^2=1$ b)Let us consider that $F(x,y)=A(x,y) i+B(x,y) j$ Since, $W=\int_C F\cdot ds=\int_C A dx +\int_C B dy=\int_0^{2 \pi} [k\cos t dt][-\sin t] dt+\int_0^{2 \pi} k \sin t=k \int_0^{2 \pi} -\cos t \sin t +\sin t \cos t dt=0$ Yes, for a force field the work done is zero and $F(x)=kx$.

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