Question

Find the work done by the force field \({\rm{F}}\left( {{\rm{x,y}}} \right){\rm{ = }}{{\rm{x}}^{\rm{2}}}{\rm{i + y}}{{\rm{e}}^{\rm{x}}}{\rm{j}}\) on a particle that moves along the parabola \({\rm{x = }}{{\rm{y}}^{\rm{2}}}{\rm{ + 1}}\)from \(\left( {{\rm{1,0}}} \right)\)to \(\left( {{\rm{2,1}}} \right)\)

Short answer

The work done by the force field is \({\rm{ = }}\frac{{\rm{7}}}{{\rm{3}}}{\rm{ + }}\frac{{{{\rm{e}}^{\rm{2}}}{\rm{ - e}}}}{{\rm{2}}}\).

Step-by-step solution

Find the work along the path.

Work done along a path is the line Integral along that path.

That is

Work done\({\rm{ = }}\mathop {{\rm{F}} \cdot {\rm{dr}}}\limits_{\rm{C}} \)

Note that

\({\rm{dr = dxi + dyj}}\)

Given that

\({\rm{F(x,y) = }}{{\rm{x}}^{\rm{2}}}{\rm{i + y}}{{\rm{e}}^{\rm{x}}}{\rm{j}}\)

Therefore,

Work done

\(\begin{array}{l}{{\rm{ = }}_{\rm{C}}}{{\rm{x}}^{\rm{2}}}{\rm{i + y}}{{\rm{e}}^{\rm{x}}}{\rm{j}} \cdot {\rm{(dxi + dyj)}}\\{{\rm{ = }}_{\rm{C}}}{{\rm{x}}^{\rm{2}}}{\rm{dx + y}}{{\rm{e}}^{\rm{x}}}{\rm{dy}}\end{array}\)

Given that \({\rm{C:x = }}{{\rm{y}}^{\rm{2}}}{\rm{ + 1}}\)

Therefore,

\({\rm{y = }}\sqrt {{\rm{x - 1}}} \;\;\;{\rm{ and }}\;\;\;{\rm{dy = }}\frac{{{\rm{dx}}}}{{{\rm{2}}\sqrt {{\rm{x - 1}}} }}\)

Replace the value of \({\rm{y}}\)and\({\rm{dy}}\),To get an integral completely in terms of \({\rm{x}}\)

Work done

\(\begin{array}{l}{\rm{ = }}\;\;\;{{\rm{x}}^{\rm{2}}}{\rm{dx + }}\sqrt {{\rm{x - 1}}} {{\rm{e}}^{\rm{x}}}\frac{{{\rm{dx}}}}{{{\rm{2}}\sqrt {{\rm{x - 1}}} }}{\rm{ }}\\{{\rm{ = }}_{\rm{C}}}{{\rm{x}}^{\rm{2}}}{\rm{ + }}\frac{{\rm{1}}}{{\rm{2}}}{{\rm{e}}^{\rm{x}}}{\rm{dx}}\end{array}\)

Evaluate the work is done

Integrate the works done with respect to \({\rm{x}}\) and move from \({\rm{(1,0) to (2,1)}}\), \({\rm{x}}\) increases from the works done in step1.

Work done

\(\begin{array}{l}{\rm{ = }}_{\rm{1}}^{\rm{2}}{{\rm{x}}^{\rm{2}}}{\rm{ + }}\frac{{\rm{1}}}{{\rm{2}}}{{\rm{e}}^{\rm{x}}}{\rm{dx}}\\{\rm{ = }}\frac{{{{\rm{x}}^{\rm{3}}}}}{{\rm{3}}}{\rm{ + }}\frac{{\rm{1}}}{{\rm{2}}}{{\rm{e}}^{\rm{x}}}^{\rm{2}}\\{\rm{ = }}\frac{{{{\rm{2}}^{\rm{3}}}}}{{\rm{3}}}{\rm{ + }}\frac{{\rm{1}}}{{\rm{2}}}{{\rm{e}}^{\rm{2}}}{\rm{ - }}\frac{{{{\rm{1}}^{\rm{3}}}}}{{\rm{3}}}{\rm{ + }}\frac{{\rm{1}}}{{\rm{2}}}{{\rm{e}}^{\rm{1}}}\\{\rm{ = }}\frac{{\rm{8}}}{{\rm{3}}}{\rm{ + }}\frac{{\rm{1}}}{{\rm{2}}}{{\rm{e}}^{\rm{2}}}{\rm{ - }}\frac{{\rm{1}}}{{\rm{3}}}{\rm{ - }}\frac{{\rm{e}}}{{\rm{2}}}\\{\rm{ = }}\frac{{\rm{7}}}{{\rm{3}}}{\rm{ + }}\frac{{{{\rm{e}}^{\rm{2}}}{\rm{ - e}}}}{{\rm{2}}}\end{array}\)

Therefore, the work done is\({\rm{ = }}\frac{{\rm{7}}}{{\rm{3}}}{\rm{ + }}\frac{{{{\rm{e}}^{\rm{2}}}{\rm{ - e}}}}{{\rm{2}}}\).