Question

Calculate the work done by a particle under the influence of a force `x^2 hat(i) + y^2hat(j) + z^2hat(k)` along the curve `x=cost`, `y=sint` and `z=e^(2t)` from `(1,0,1)` to `(1,0,4)`?

Answer 1

`1/3 (e^(6 pi)-3)`

Explanation:

`F(r(t))=(cos^2t,sin^2t,e^(4t))`

`dr = (-sint,cost,2e^(2t))dt`

`F(r(t))cdotdr=(-sint cos^2t+cost sin^2t+2e^(6t))dt`

`int_(t=0)^(t=pi)(-sint cos^2t+cost sin^2t+2e^(6t))dt = 1/3 (e^(6 pi)-3)`

Answer 2

`int_C \ vec(F) * d vec(r) = 1/3e^(6pi)-1 ~~ 18829655`

Explanation:

The work done in moving a particle from the endpoints `A` to `B` along a curve `C` is.

`int_C \ vec(F) * d vec(r) \ \` where `\ \ {: (vec(F),=F_1 hat(i) + F_2 hat(j)+F_3 hat(k)),(d vec(r),=dx hat(i) +dy hat(j)+dz hat(k)) :}`

The integral is known as a line integral.

So we have:

`vec(F) = x^2 hat(i) + y^2hat(j) + z^2hat(k)`

and `C` is the arc of the parametrised curve:

`x=cost\ \`, `y=sint\ \` and `\ \ z=e^(2t)`

from `(1,0,1)` to `(1,0,4)`

To evaluate the line integral we convert it to a standard integral by choosing an appropriate integration variable, In this case integrating wrt the parameter variable `t` would seem to make sense.

On `C`, the variable `z` varies from `e^(2t)=1 => t=0` to `e^(2t)=e^(2pi)=>t=pi`, and these values of `t` are also consistent with the `x` and `y` initial and end values. Differentiating the equations for `C` wrt `t` gives:

`dx/dx = -sint \ \`; `dy/dt=cost \ \` and `\ \ dz/dt = 2e^(2t)`

And so we can express our vector fields in terms of `x` alone:

`vec(F) = x^2 hat(i) + y^2hat(j) + z^2hat(k)`
`\ \ \ \ = (cost)^2 hat(i) + (sint)hat(j) + (e^(2t))^2hat(k) \ \ \ \` (from the equation of `C`)
`\ \ \ \ = cos^2t hat(i) + sin^2t hat(j) + e^(4t) hat(k)`

And:

`d vec(r)=dx hat(i) + dy hat(j) + dz hat(j)`
`\ \ \ \ \ =(-sint \ dt) hat(i) + (cost \ dt) hat(j) + (2e^(2t) \ dt) hat(k)\ \ \ \` (from derivatives)
`\ \ \ \ \ =-sint \ dt hat(i) + cost \ dt hat(j) + 2e^(2t) \ dt hat(k)`

Hence,

`int_C \ vec(F) * d vec(r) = int_C \ (cos^2t hat(i) + sin^2t hat(j) + e^(4t) hat(k)) * (-sint \ dt hat(i) + cost \ dt hat(j) + 2e^(2t) \ dt hat(k))`
`" "= int_0^pi \ (cos^2t)(-sint)dt+(sin^2tcost)dt+(e^(4t)2e^(2t))dt`
`" "= int_0^pi \ -sintcos^2t+sin^2tcost+2e^(6t) \ dt`
`" "= [1/3cos^3t+1/3sin^3t+1/3e^(6t)]_0^pi`
`" "= 1/3[cos^3t+sin^3t+e^(6t)]_0^pi`
`" "= 1/3{(cos^3pi+sin^3pi+e^(6pi)) - (cos^3 0+sin^3 0+e^0)}`
`" "= 1/3{(-1+0+e^(6pi)) - (1+0+1)}`
`" "= 1/3(e^(6pi)-3)`
`" "= 1/3e^(6pi)-1`
`" "= 18829655.02 ...`