Question

5. Find the work done by the force f bar = (3x^2– 6yz)i +(2y + 3xz) j +(1- 4xyz^2)k
in moving particle from the point (0, 0, 0) to the point (1,1,1,)
along the curve C: x=t, y=t^2,z= t^3​

Answer 1

Answer:

Don't know the answer

Type properly

Answer 2

Answer:

The work done by the force $f$ bar is $2.$

Step-by-step explanation:

Given, $f$ bar $=(3x^{2} -6yz)i+(2y+3xz)j+(1-4xyz^{2} )k$

Formula is, $\int\limitsa_c {A} \, dr$$=\int\limitsa_c {(xi+yj+zk)} \, (dxi+dyj+dzk)$

Then,$\int\limitsa_c {A} \, dr=\int\limitsa_c {((3x^{2} -6yz)i+(2y+3xz)j+(1-4xyz^{2} )k)} \, (dxi+dyj+dzk)$

$\int\limitsa_c {A} \, dr=\int\limitsa_c {((3x^{2} -6yz)dx+(2y+3xz)dy+(1-4xyz^{2} )dz)} \,$

If $x=t, y=t^2, z=t^3,$ points $(0,0,0)$ and $(1,1,1)$ corresponding to $t=0$ and $t=1$ respectively.

$\int\limitsa_c {A} \, dr=\int\limits^1_0 {((3t^{2} -6t^{2} t^{3} )dt+(2t^{2} +3tt^{3} )d(t^{2} )+(1-4tt^{2} (t^{3})^{2} )d(t^{3}))$

$\int\limitsa_c {A} \, dr=\int\limits^1_0 {((3t^{2} -6t^{5} )dt+(4t^{3} +6t^{5} )dt)+(3t^{2}-12(t^{11})dt)=2.$