Question

What is the work done by moving in the force field f~ (x, y) = h2x 3 + 1, 2y 4 i along the parabola y = x 2 from (−1, 1) to (1, 1)?

a.compute directly

b.use the theorem?

Answer 1

The force field seems to be


`\mathbf f(x,y)=\langle2x^3+1,2y^4\rangle`

(though it's not entirely clear whether that's it...)

The parabolic path can be parameterized by


`\mathbf r(t)=\langle t,t^2\rangle`

where
`-1\le t\le1`. Then the work done by the field along the path is


`\displaystyle\int_C\mathbf f(x,y)\cdot\mathrm d\mathbf r=\int_(t=-1)^(t=1)\langle2t^3+1,2t^8\rangle\cdot\langle1,2t\rangle\,\mathrm dt`

`=\displaystyle\int_(-1)^1(2t^3+1+4t^9)\,\mathrm dt`

`=2`

Not sure what theorem is being referred to in part (b). Perhaps you mean the gradient theorem? In which case you would need to show that there's a scalar function
`f(x,y)` such that
`\\abla f(x,y)=\mathbf f(x,y)`. We have


`\\abla f(x,y)=\left\langle(\partial f)/(\partial x),(\partial f)/(\partial y)\right\rangle=\left\langle2x^3+1,2y^4\right\rangle`

from which we have


`(\partial f)/(\partial x)=2x^3+1\implies f(x,y)=\displaystyle\int(2x^3+1)\,\mathrm dx`

`\implies f(x,y)=\frac12x^4+x+g(y)`


`\implies(\partial f)/(\partial y)=g'(y)=2y^4`

`\implies g(y)=\frac25y^5+C`


`\implies f(x,y)=\frac12x^4+x+\frac25y^5+C`

We have a candidate for a potential function, so we apply the gradient theorem, which asserts that the value of the line integral is path independent and can be determined by evaluating the potential function at the endpoints of the path. We then have


`\displaystyle\int_C\mathbf f(x,y)\cdot\mathrm d\mathbf r=f(1,1)-f(-1,1)`

`=(19)/(10)-\left(-\frac1{10}\right)`

`=2`

as expected.