Answer to Question #87918 in Calculus for ajay kumar

Question

Calculate the work done by a force F i j kˆ 2 ˆ ( ) ˆ 3 2 = x + xz − y + z r in moving a particle
along the curve x = t y = t z = t − t 2 2 2 from ; ; 4 t = 0 to t = 1. Is the force conservative?

Answer 1

Force field

$$F=3x^2\overrightarrow{i}+(2xz-y)\overrightarrow{j}+z\overrightarrow{k}$$

Work done along the space curve $x=2t^2, y=t, z=4t^2-t, from \ t=0\ to\ t=1$

$$Work=\int \overrightarrow{F}\cdot d\overrightarrow{r}=\displaystyle\int_{0}^1\overrightarrow{F}\cdot {d\overrightarrow{r} \over dt} dt=$$

$$=\displaystyle\int_{0}^1[3(2t^2)^2\overrightarrow{i}+(2(2t^2)(4t^2-t)-t)\overrightarrow{j}+(4t^2-t)\overrightarrow{k}]\cdot [4t\overrightarrow{i}+1\overrightarrow{j}+(8t-1)\overrightarrow{k}]dt=$$

$$=\displaystyle\int_{0}^1[48t^5+16t^4-4t^3-t+32t^3-4t^2-8t^2+t]dr=$$

$$=[8t^6+{16 \over 5}t^5+7t^4-4t^3]\begin{matrix} 1 \\ 0 \end{matrix}=8+{16 \over 5}+7-4-0={71 \over 5}$$

Work done along the straight line $x=2t, y=t, z=3t, from \ t=0\ to\ t=1$

$$Work=\int \overrightarrow{F}\cdot d\overrightarrow{r}=\displaystyle\int_{0}^1\overrightarrow{F}\cdot {d\overrightarrow{r} \over dt} dt=$$

$$=\displaystyle\int_{0}^1[3(2t)^2\overrightarrow{i}+(2(2t)(3t)-t)\overrightarrow{j}+3t\overrightarrow{k}]\cdot [2\overrightarrow{i}+1\overrightarrow{j}+3\overrightarrow{k}]dt=$$

$$=\displaystyle\int_{0}^1[24t^2+12t^2-t+9t]dr=$$

$$=[12t^3+4t^2]\begin{matrix} 1 \\ 0 \end{matrix}=12+4-0=16=\not {71 \over 5}$$

A conservative force is a force with the property that the total work done in moving a particle between two points is independent of the taken path.

Therefore, the force is not conservative.