# Find the work done by the force F = xyi + (y - x)j over the straight line from (0, 0) to (1, -3)....

## Question:

Find the work done by the force F = xyi + (y - x)j over the straight line from (0, 0) to (1, -3).

The amount of work done is.

(Type an integer or a simplified fraction.)

## Work done by a Force :

According to definition, Work done by a force on an object in moving it from one position to another is given by {eq}Work=Force\times Distance {/eq}. In Calculus, work done on a particle across a curve is given by integration of force on that path {eq}W=\int F\cdot dr {/eq}. We solve this integral by following formula {eq}\int F\cdot dr=\int_{a}^{b} F(r(t))\cdot r'(t)dt {/eq}

## Answer and Explanation:

Parametrization of the line joining the points {eq}(0,0), (1,-3) {/eq} is given by

{eq}r(t)=\langle 0+(1-0)t, 0+(-3-0)t, \rangle {/eq}

{eq}r(t)=\langle t, -3t \rangle {/eq}

Now {eq}F(x,y,z)=\langle xy, (y-x) \rangle {/eq}

Writing the function in parametric form

{eq}F(r(t))= \langle (t)(-3t) , -3t-t \rangle {/eq}

{eq}F(r(t))= \langle -3t^2 , -4t \rangle {/eq}

Take the derivative

{eq}r'(t)=\langle 1, -3 \rangle {/eq}

As we know work done by force is given by {eq}W=\int F\cdot dr {/eq}

Also {eq}\int F\cdot dr=\int_{a}^{b} F(r(t))\cdot r'(t)dt {/eq}

{eq}W=\int_{a}^{b} F(r(t))\cdot r'(t)dt {/eq}

{eq}W=\int_{0}^{1} \langle -3t^2 , -4t \rangle \cdot \langle 1,-3 \rangle dt {/eq}

{eq}W=\int_{0}^{1}[ -3t^2+12t ]dt {/eq}

Solving we get

{eq}W= (-t^3+6t^2)_{0}^{1} {/eq}

{eq}W= -1+6 {/eq}

{eq}W=5 {/eq}