Compute the work done by the force field F(x, y) = (x^2, ye^x) on a particle that moves along the...

Question:

Compute the work done by the force field {eq}\vec{F}(x, y) = \left \langle x^2, \; ye^x \right \rangle {/eq} on a particle that moves along the parabola {eq}x = y^2 + 1 {/eq} from {eq}(1, 0) {/eq} to {eq}(2, 1) {/eq}.

Work:

Work is the amount of energy it takes to move something from point a to point b. We compute it using a line integral:

{eq}W = \int_C \vec F \cdot d\vec r {/eq}

We will first write the position vector for the particle on the path, then we will evaluate the inetgral.

To write the position vector, we can just assign {eq}x \to t {/eq}:

{eq}\begin{align*} \vec r &= \left< t,\ t^2+1 \right> \end{align*} {/eq}

We plug this directly into the line integral to get the work

{eq}\begin{align*} W &= \int_1^2 \left< x^2, ye^x \right>\cdot d\left< t, t^2+1 \right> \\ &= \int_1^2 \left< t^2, (t^2+1)e^t \right>\cdot \left< 1,2t \right>\ dt \\ &= \int_1^2 t^2 + 2t^3e^t + 2te^t\ dt \\ &= \left [ \frac13t^3 + 2e^t \left( t^3-3t^2+6t-6 \right) + 2e^t \left( t-1 \right) \right ]_1^2 \\ &= \frac13(2^3-1^3) + 2e^2 \left( 2^3-3(2)^2+6(2)-6 \right) - 2e^1 \left( 1^3-3(1)^2+6(1)-6 \right) + 2e^2 \left( 2-1 \right) - 2e^1 \left( 1-1 \right) \\ &= \frac73 +4e+6e^2 \\ &\approx 57.54 \end{align*} {/eq}