Find the total work done by the force field F

Question

Find the total work done in moving a particle in a force field given by:

$F=(y^2-x^2)i+(2-x+y)j$ along the curve $y=x^3$, from $(-1,-1)$ to $(1,1)$

Help is appreciated!

Answer 1

In the case of a constant force and a linear motion, work is given by finding the component of the force acting in the direction of motion and then multiplying that by the distance over which you travel. Things are more difficult when you're moving in the plane.

Let ${\bf x}(t)$ give the path along which you move, then $\dot{\bf x}(t) = {\bf v}(t)$ gives the direction of (instantaneous) motion and so the scaler product $\langle {\bf F},{\bf v}\rangle$, also called the dot product and written ${\bf F} \cdot {\bf v}$, gives the component of the force in the direction of (instantaneous) motion. The instantaneous version of multiplication is integration, and so $W = F\times d$ becomes $$W=\int_{t_1}^{t_2} {\bf F} \cdot {\bf v} \, \operatorname{d}\!t$$

Your paths has the equation $y=x^3$ with $-1 \le x \le 1$. We can parametrise this as $x(t)=t$ and $y(t)=t^3$ where $-1 \le t \le 1$. In vector form: ${\bf x}(t) = t{\bf i} + t^3{\bf j}$. The force can be re-written: \begin{eqnarray*} {\bf F}(x,y) &=& (y^2-x^2){\bf i} + (2-x+y){\bf j} \\ \\ {\bf F}(t) &=& (t^6-t^2){\bf i} + (2-t+t^3){\bf j} \end{eqnarray*} Finally, ${\bf v}(t) = \dot{\bf x}(t) = {\bf i} + 3t^2{\bf j}$. Hence: \begin{eqnarray*} W &=& \int_{t_1}^{t_2} {\bf F} \cdot {\bf v} \, \operatorname{d}\!t \\ \\ &=& \int_{-1}^1 \left((t^6-t^2){\bf i} + (2-t+t^3){\bf j}\right) \cdot \left({\bf i} + 3t^2{\bf j}\right) \, \operatorname{d}\!t \\ \\ &=& \int_{-1}^1 t^6-t^2+6t^2-3t^3+3t^5 \, \operatorname{d}\!t \\ \\ &=& \int_{-1}^1 t^6+5t^2-3t^3+3t^5 \, \operatorname{d}\!t \\ \\ &=& \left[ \tfrac{1}{7}t^7+\tfrac{5}{3}t^3-\tfrac{3}{4}t^4+\tfrac{1}{2}t^6 \right]_{-1}^1 \\ \\ &=& \tfrac{76}{21} \end{eqnarray*}

Answer 2

The work done by a force $f$ on a particle moving through a suitably smooth curve $\gamma$ on the interval $[t_0,t_1]$ is given by $W = \int_{t_0}^{t_1} \langle f(t), \dot{\gamma}(t) \rangle dt$.

Here we can parameterize the curve by $\gamma(t) = (t,t^3)$ for $t \in [-1,1]$.

We have $f(t) = F(\gamma(t)) = (t^6-t^2,t^3-t+2)$, and $\dot{\gamma}(t) = (1,3t^2)$.

Hence $ W = \int_{-1}^1 (t^6-t^2+3 t^2(t^3-t+2)) dt = \frac{76}{21}$.