##### Home»Math Guides»Work done by vector field onto particle

##### Work Done By Force Field on Particle (Vector Fields, Two Points) Example 2

Find the work, W, done by the force field F in moving an object from point P to Q, where P is (1,1) and Q is (2,4).

We’re given \(F(x,y) = 2{y^{\frac{3}{2}}}\mathop i\limits^ \to + 3x\sqrt y \mathop j\limits^ \to \).

First, even if you aren’t told this, you should check and ask “Is F a conservative vector field?”

If it isn’t, the problem can’t be solved.

Let’s check to see if F is a conservative vector field.

To be conservative for two variables, the partial derivative for the first component of the vector field with respect to x must equal the partial derivative for the second component of the vector field with respect to y.

This is a mouthful, but basically derive what’s before i with x, and derive what’s before j with y and check if equal.

Let Q represent the i component and P represent the j component.

\[F(x,y) = 2{y^{\frac{3}{2}}}\mathop i\limits^ \to + 3x\sqrt y \mathop j\limits^ \to \]

\[\frac{{\partial Q}}{{\partial x}} = 3\sqrt y \]

\[\frac{{\partial P}}{{\partial y}} = 3\sqrt y \]

\[\frac{{\partial Q}}{{\partial x}} = \frac{{\partial P}}{{\partial y}}\]

So yes, the vector field given is indeed a conservative vector field.

We can find the work from the following integral.

\[W = \int {F \cdot dr} \]

We need f, which is a function that, when you find the gradient of it, equals F.

You can sometimes do this by “eyeballing it”, but a more systematic way is to integrate with respect to x, then derive that with respect to y.

\[f = \int {2{y^{\frac{3}{2}}}} dx\]

\[f = 2x{y^{\frac{3}{2}}} + g(y)\]

Where g(y) is an unknown function of y.

Now derive this f with respect to y.

\[\frac{{\partial f}}{{\partial y}} = 3x\sqrt y + g'(y)\]

This needs to be equal to what is in the j component of the vector field given.

It’s clear that g'(y) = 0, and integrating that, it’s clear that g(y) = K, where K is just a number constant.

So,

\[f(x,y) = 2x{y^{\frac{3}{2}}} + K\]

Did you get f correct?

To check, just find the gradient of f and you need to be able to get back the vector field given in the question.

\[\nabla f = \left\langle {\frac{{\partial f}}{{\partial x}},\frac{{\partial f}}{{\partial y}}} \right\rangle \]

\[\nabla f = \left\langle {2{y^{\frac{3}{2}}},3x\sqrt y } \right\rangle \]

So our f is good!

Remember, you can write out the components with i, j, k, or you can just put it in a vector with brackets as shown above.

For finding work, just substitute the points into f and find the difference between the endpoint and the first point.

\[W = \int {F \cdot dr} \]

\[W = f(2,4) – f(1,1)\]

\[W = \left( 2 \right)\left( 2 \right){\left( 4 \right)^{\frac{3}{2}}} + K – \left( 2 \right)\left( 1 \right){\left( 1 \right)^{\frac{3}{2}}} – K\]

\(W = 30\) \(N \cdot m\)

So the work done is 30 Nm.

We weren’t given units in the question, but in general work has units of Joules (J), which is just Newton-meters \((N \cdot m)\).

Return to the previous example, work done by vector field onto particle