Question

Find the work done by the force field \({\rm{F}}\left( {{\rm{x,y,z}}} \right){\rm{ = }}\left\langle {{\rm{x - }}{{\rm{y}}^{\rm{2}}}{\rm{,y - }}{{\rm{z}}^{\rm{2}}}{\rm{,z - }}{{\rm{x}}^{\rm{2}}}} \right\rangle \)on a particle that moves along the line segment from\(\left( {{\rm{0,0,1}}} \right)\)to\(\left( {{\rm{2,1,0}}} \right)\).

Short answer

The work done is \(\frac{{\rm{7}}}{{\rm{3}}}\).

Step-by-step solution

Step 1:Find the parametric representation of a line segment

Remember that,

The parametric representation of a line segment joining the points\({\rm{(a, b, c)}}\)and \({\rm{(l, m, n)}}\)is \({\rm{r(t) = (1 - t)}} \cdot \left\langle {{\rm{a,b,c}}} \right\rangle {\rm{ + t}} \cdot \left\langle {{\rm{l,m,n}}} \right\rangle \;\;\;{\rm{Where t}} \in {\rm{(0,1)}}\).

Therefore, the parametric representation of the line segment joining the points\({\rm{(0,0,1) }}\)and\(\left( {{\rm{2,1,0}}} \right)\) is.

\(\begin{array}{l}{\rm{r(t) = (1 - t)}} \cdot \langle {\rm{0,0,1}}\rangle {\rm{ + t}} \cdot \langle {\rm{2,1,0}}\rangle \;\;{\rm{Where t}} \in \left( {{\rm{0,1}}} \right)\\{\rm{r(t) = }}\langle {\rm{0,0,(1 - t)}}\rangle + \langle 2t,t,0\rangle \;\;\;{\rm{ Where t}} \in {\rm{(0,1)}}\\{\rm{r(t)}} = \langle {\rm{2t,t,(1 - t)}}\rangle \;\;\;{\rm{ Where }}t \in {\rm{(0,1)}}\end{array}\)

Given that

\({\rm{F(x,y,z) = x - }}{{\rm{y}}^{\rm{2}}}{\rm{,y - }}{{\rm{z}}^{\rm{2}}}{\rm{,z - }}{{\rm{x}}^{\rm{2}}}\)

Therefore,

\(\begin{array}{l}{\rm{F(r(t)) = (2t) - }}{{\rm{t}}^{\rm{2}}}{\rm{,t - (1 - t}}{{\rm{)}}^{\rm{2}}}{\rm{,(1 - t) - (2t}}{{\rm{)}}^{\rm{2}}}\\{\rm{F(r(t)) = 2t - }}{{\rm{t}}^{\rm{2}}}{\rm{,t - }}\left( {{{\rm{t}}^{\rm{2}}}{\rm{ - 2t + 1}}} \right){\rm{,(1 - t) - 4}}{{\rm{t}}^{\rm{2}}}\\{\rm{F(r(t)) = 2t - }}{{\rm{t}}^{\rm{2}}}{\rm{,}}\;{\rm{ - }}{{\rm{t}}^{\rm{2}}}{\rm{ + 3t - 1,1 - t - 4}}{{\rm{t}}^{\rm{2}}}\end{array}\)

Evaluate the work

Note that

\({\rm{dr = }}\left\langle {{\rm{2,1, - 1}}} \right\rangle {\rm{dt}}\)

Work Done\( = {\bf{F}} \cdot d{\bf{r}}\)

Note that Integrate from\({\rm{0 to 1}}\), because\({\rm{t}} \in {\rm{(0,1)}}\)

\(\begin{array}{c}{\bf{F}} \cdot d{\bf{r}} = \int_{\rm{0}}^{\rm{1}} {{\rm{2t - }}{{\rm{t}}^{\rm{2}}}} {\rm{, - }}{{\rm{t}}^{\rm{2}}}{\rm{ + 3t - 1,1 - t - 4}}{{\rm{t}}^{\rm{2}}} \cdot \left\langle {2,1, - 1} \right\rangle {\rm{dt}}\\ = \int_{\rm{0}}^{\rm{1}} {{\rm{2 \times 2t - }}{{\rm{t}}^{\rm{2}}}{\rm{ + - }}{{\rm{t}}^{\rm{2}}}{\rm{ + 3t - 1 - 1 - t - 4}}{{\rm{t}}^{\rm{2}}}{\rm{dt}}} \\ = \int_{\rm{0}}^{\rm{1}} {{\rm{4t - 2}}{{\rm{t}}^{\rm{2}}}{\rm{ + - }}{{\rm{t}}^{\rm{2}}}{\rm{ + 3t - 1 + - 1 + t + 4}}{{\rm{t}}^{\rm{2}}}{\rm{dt}}} \\ = \int_{\rm{0}}^{\rm{1}} {{\rm{4t - 2}}{{\rm{t}}^{\rm{2}}}{\rm{ - }}{{\rm{t}}^{\rm{2}}}{\rm{ + 3t - 1 - 1 + t + 4}}{{\rm{t}}^{\rm{2}}}{\rm{dt}}} \end{array}\)

\({\bf{F}} \cdot d{\bf{r}} = \int_{\rm{0}}^{\rm{1}} {{{\rm{t}}^{\rm{2}}}{\rm{ + 8t - 2dt}}} \)

\(\begin{array}{c}{\bf{F}} \cdot d{\bf{r}}{\rm{ = }}\left( {\frac{{{{\rm{t}}^{\rm{3}}}}}{{\rm{3}}}{\rm{ + 4}}{{\rm{t}}^{\rm{2}}}{\rm{ - 2t}}} \right)_0^1\\{\rm{ = }}\frac{{\rm{1}}}{{\rm{3}}}{\rm{ + 4 - 2}}\\{\rm{ = }}\frac{{\rm{1}}}{{\rm{3}}}{\rm{ + 2}}\\{\rm{ = }}\frac{{\rm{7}}}{{\rm{3}}}\end{array}\)

Therefore, the work done is\(\frac{{\rm{7}}}{{\rm{3}}}\).