Question #108052
Determine the work done by the force F(xy+3z)i^+(2y^2 -x^2)j+(z-2y)k in taking a particle from  x = 0 to  x =1 along a curve defined by the equations: x^2= 2y ; 2x^3= 3z
1
Expert's answer
2020-04-06T08:56:14-0400
dydx=x,dzdx=2x2\frac{dy}{dx}=x,\frac{dz}{dx}=2x^2

W=01((xy+3z)+(2y2x2)x+(z2y)2x2)dxW=\int_0^1((xy+3z)+(2y^2 -x^2)x+(z-2y)2x^2)dx

W=01IdxW=\int_0^1Idx

(xy+3z)=xx22+2x3=5x32(xy+3z)=x\frac{x^2}{2}+2x^3=\frac{5x^3}{2}

(2y2x2)x=(2(x22)2x2)x=(x52x3)(2y^2 -x^2)x=\left(2\left(\frac{x^2}{2}\right)^2 -x^2\right)x=\left(\frac{x^5}{2} -x^3\right)

(z2y)2x2=(2x33x2)2x2=(4x532x4)(z-2y)2x^2=\left(\frac{2x^3}{3}-x^2\right)2x^2=\left(\frac{4x^5}{3}-2x^4\right)

I=5x32+(x52x3)+(4x532x4)=(11x56+3x322x4)I=\frac{5x^3}{2}+\left(\frac{x^5}{2} -x^3\right)+\left(\frac{4x^5}{3}-2x^4\right)=\left(\frac{11x^5}{6}+\frac{3x^3}{2}-2x^4\right)

W=01(11x56+3x322x4)dxW=\int_0^1\left(\frac{11x^5}{6}+\frac{3x^3}{2}-2x^4\right)dx

W=(116(6)+32(4)215)=1013600.28W=\left(\frac{11}{6(6)}+\frac{3}{2(4)}-2\frac{1}{5}\right)=\frac{101}{360}\approx 0.28


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