Answer to Question #125195 in Calculus for Sunil

Question

Determine the work done by a force
F=(xy+3z)i(cap) + (2y^2-x^2)j(cap) + (z-2y) k(cap) in taking a particle from x=0 to x=1 along a curve defined by equation

x^2=2y ; 2x^3=3z

Answer 1

$\vec F=(xy+3z)\vec i +(2y^2-x^2)\vec j+(z-2y)\vec k$

Curve $C$: $x^2=2y, 2x^3=3z$ from $x=0$ to $x=1$

This curve is given by vector function $r(t)=t\vec i +\frac{ t^2}2 \vec j +\frac{2t^3}{3}\vec k, \ \ 0\leq t\leq 1$

$r^{\prime}(t)=\vec i +t\vec{j}+2t^2\vec k$

$\vec F(r(t))=\frac{5t^3}{2}\vec i +\left( \frac{t^4}{2}-t^2 \right)\vec j+\left (\frac{2t^3}{3}-t^2\right)\vec k$

Work done by the force $\vec F$ along curve $C$ is

$W=\int _C\vec F d\vec r=\int_0^1 \vec F(r(t))r^{\prime}(t) dt=\int_0^1 \left(\frac{5t^3}{2}+\frac{t^5}{2}-t^3+\frac{4t^5}{3}-2t^4 \right)dt=\int _0^1 \left( \frac{11t^5}{6}+\frac{3t^3}{2}-2t^4\right)dt=\frac{11t^6}{36}+\frac{3t^4}{8}-\frac{2t^5}{5}\bigg|_0^1= \frac{11}{36}+\frac{3}{8}-\frac{2}{5}=\frac{101}{360}.$

Answer: $W=\frac{101}{360}.$