Question

In Exercises 27 and \(28,\) find the work done by the force field \(F\) in moving an object from \(P\) to \(Q\). $$ \mathbf{F}(x, y)=9 x^{2} y^{2} \mathbf{i}+\left(6 x^{3} y-1\right) \mathbf{j} ; P(0,0), Q(5,9) $$

Short answer

To obtain the total work done by the force field in moving the object from point P to Q, evaluate the line integral. The result of the line integral represents the work done by the force field.

Step-by-step solution

Define the Path Vector

We first need to define the path vector along which the object is moving from P to Q. Since the coordinates of P are (0,0) and Q are (5,9), we can describe a linear path from P to Q parametrically as \( r(t) = \langle t, 1.8t \rangle \) for \( 0 \leq t \leq 5 \). This path vector \( r(t) \) represents the x and y coordinates of the object at any time \( t \) while it is moving along the path.

Compute dr/dt

Next, compute the derivative of the path vector \( r'(t) = dr/dt \). Using the rule that the derivative of \( t \) is 1 and of \( 1.8t \) is \( 1.8 \), we find that \( r'(t) = \langle 1, 1.8 \rangle \).

Substitute in the Force Field

Now we substitute the path \( r(t) \) into the force field. The given force field is \( F(x, y) = 9x^2y^2i + (6x^3y - 1)j \). Substituting \( x = t \) and \( y = 1.8t \) into the force field, we get \( F(t) = 9t^2(1.8t)^2i + (6t^3(1.8t) - 1)j \). This simplifies to \( F(t) =25.92t^4i + (19.44t^4 - 1)j \).

Calculate the Line Integral

Now, we can calculate the work done by the force field along the path from P to Q by computing the line integral \( Work = \int_0^5 F(t) \cdot r'(t) dt = \int_0^5 (F_x(t),F_y(t)) \cdot (1,1.8) dt \). Substituting the components of \( F(t) \) from the previous step, we get \( Work = \int_0^5 (25.92t^4 \cdot 1 + (19.44t^4 - 1) \cdot 1.8) dt \). Compute this integral to obtain the total work done by the force field.

Compute the Integral

Finally, compute the definite integral to get the work done by the force field in moving the object from P to Q. The integration can be simplified to \( Work = \int_0^5 (25.92t^4 + 34.992t^4 - 1.8) dt \), and evaluating this expression gives the work done by the force field.