2020-04-04T10:25:55-04:00
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
2020-04-06T08:56:14-0400
d y d x = x , d z d x = 2 x 2 \frac{dy}{dx}=x,\frac{dz}{dx}=2x^2 d x d y = x , d x d z = 2 x 2
W = ∫ 0 1 ( ( x y + 3 z ) + ( 2 y 2 − x 2 ) x + ( z − 2 y ) 2 x 2 ) d x W=\int_0^1((xy+3z)+(2y^2 -x^2)x+(z-2y)2x^2)dx W = ∫ 0 1 (( x y + 3 z ) + ( 2 y 2 − x 2 ) x + ( z − 2 y ) 2 x 2 ) d x
W = ∫ 0 1 I d x W=\int_0^1Idx W = ∫ 0 1 I d x
( x y + 3 z ) = x x 2 2 + 2 x 3 = 5 x 3 2 (xy+3z)=x\frac{x^2}{2}+2x^3=\frac{5x^3}{2} ( x y + 3 z ) = x 2 x 2 + 2 x 3 = 2 5 x 3
( 2 y 2 − x 2 ) x = ( 2 ( x 2 2 ) 2 − x 2 ) x = ( x 5 2 − x 3 ) (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) ( 2 y 2 − x 2 ) x = ( 2 ( 2 x 2 ) 2 − x 2 ) x = ( 2 x 5 − x 3 )
( z − 2 y ) 2 x 2 = ( 2 x 3 3 − x 2 ) 2 x 2 = ( 4 x 5 3 − 2 x 4 ) (z-2y)2x^2=\left(\frac{2x^3}{3}-x^2\right)2x^2=\left(\frac{4x^5}{3}-2x^4\right) ( z − 2 y ) 2 x 2 = ( 3 2 x 3 − x 2 ) 2 x 2 = ( 3 4 x 5 − 2 x 4 )
I = 5 x 3 2 + ( x 5 2 − x 3 ) + ( 4 x 5 3 − 2 x 4 ) = ( 11 x 5 6 + 3 x 3 2 − 2 x 4 ) 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) I = 2 5 x 3 + ( 2 x 5 − x 3 ) + ( 3 4 x 5 − 2 x 4 ) = ( 6 11 x 5 + 2 3 x 3 − 2 x 4 )
W = ∫ 0 1 ( 11 x 5 6 + 3 x 3 2 − 2 x 4 ) d x W=\int_0^1\left(\frac{11x^5}{6}+\frac{3x^3}{2}-2x^4\right)dx W = ∫ 0 1 ( 6 11 x 5 + 2 3 x 3 − 2 x 4 ) d x
W = ( 11 6 ( 6 ) + 3 2 ( 4 ) − 2 1 5 ) = 101 360 ≈ 0.28 W=\left(\frac{11}{6(6)}+\frac{3}{2(4)}-2\frac{1}{5}\right)=\frac{101}{360}\approx 0.28 W = ( 6 ( 6 ) 11 + 2 ( 4 ) 3 − 2 5 1 ) = 360 101 ≈ 0.28
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