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 $$F(x,y) = 2{y^{\frac{3}{2}}}\mathop i\limits^ \to + 3x\sqrt y \mathop j\limits^ \to$$. Is F conservative?

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.$\begin{array}{l}\frac{{\partial Q}}{{\partial x}} = 3\sqrt y \\\frac{{\partial P}}{{\partial y}} = 3\sqrt y \\\frac{{\partial Q}}{{\partial x}} = \frac{{\partial P}}{{\partial y}}\end{array}$

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. $\begin{array}{l}f = \int {2{y^{\frac{3}{2}}}} dx\\f = 2x{y^{\frac{3}{2}}} + g(y)\end{array}$

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. $\begin{array}{l}W = \int {F \cdot dr} \\W = f(2,4) - f(1,1)\\W = \left( 2 \right)\left( 2 \right){\left( 4 \right)^{\frac{3}{2}}} + K - \left( 2 \right)\left( 1 \right){\left( 1 \right)^{\frac{3}{2}}} - K\\W = 30\end{array}$

So the work done is 30. We weren't given units in the question, but in general work has units of Joules (J), which is just Newton-meters ($$N \cdot m$$).

We have examples on determining on whether a vector field is conservative or not, finding the potential of a vector field, and finding the work involving particles and vector fields below:

Work Done By Force Field on Particle (Vector Fields) Ex1

Finding Potential of Conservative Vector Field and Work Ex1

Finding Potential of Conservative Vector Field Ex2

Finding Potential of Conservative Vector Field Ex3