Question

Calculate the work done by a particle under the influence of a force `y^2 hat(i) - x^2hat(j)` along the curve `y=4x^2` from `(0,0)` to `(1,4)`?

Answer 1

`1.2`

Explanation:

`C=(x,4x^2)` and `F=((4x^2)^2,-x^2)` but `dC=(1,8x)` so
`dW=<< F, dC >>dx = 16x^4-8x^3` then

`W=int_0^1 << F, dC >>dx = 16/5-8/4=1.2`

Answer 2

`int_C vec(F) * d vec(r) = 1.2`

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)),(d vec(r),=dx hat(i) +dy hat(j)) :}`

The integral is known as a line integral.

So we have:

`vec(F) = y^2hat(i) -x^2hat(j)`

and `C` is the arc of `y=4x^2` from `(0,0)` to `(1,4)`

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

On `C`, the variable `x` varies from `x=0` to `x=1`. Differentiating the equation for `C` wrt `x` gives:

`dy/dx = 8x`

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

`vec(F) = y^2hat(i) -x^2hat(j)`
`\ \ \ \ = (4x^2)^2hat(i) -x^2hat(j) \ \ \ \` (from the equation of `C`)
`\ \ \ \ = 16x^4 hat(i) -x^2 hat(j)`

And:

`d vec(r)=dx hat(i) + dy hat(j)`
`\ \ \ \ \ =dx hat(i) + 8x \ dx hat(j) \ \ \ \` (from derivative of eqn for `C`)

Hence,

`int_C \ vec(F) * d vec(r) = int_C \ ( 16x^4 hat(i) -x^2hat(j) ) * (dx hat(i) + 8x \ dx hat(j))`
`" "= int_0^1 \ 16x^4 \ dx -x^2(8x)dx`
`" "= int_0^1 \ 16x^4 -8x^3 \ dx`
`" "= [ 16/5 x^5 -2x^4 ]_0^1`
`" "= (16/5 -2) - 0`
`" "= 6/5`
`" "= 1.2`