Question

Find the work done by the force field f in moving an object from a to

b. f(x, y) = 6y3/2 i + 9x y j a(1, 1), b(4, 9)

Answer 1

Presumably the path from $a$ to $b$ is a line segment, in which case we can parameterize it by

$\mathbf r(t)=(1-t)(1,1)+t(4,9)=(1+3t,1+8t)$

with $0\le t\le1$. Then the work done along this path - denoted by $\mathcal C$ - is

$\displaystyle\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r=\int_{t=0}^{t=1}(6(1+8t)^{3/2},9(1+3t)(1+8t))\cdot(3,8)\,\mathrm dt$
$=\displaystyle\int_0^1(72(1+3t)(1+8t)+18(1+8t)^{3/2})\,\mathrm dt$
$=\dfrac{6309}5$

Answer 2

By direct calculation, we will see that the work is equal to 2786.4 J.

Remember that work is equal to force times displacement, then we have that:

$W = \int\limits^b_a {F*dl}$

Obviously, the path that we take from a to b matters here, so we assume that we take the simplest path, which is a line segment:

$l(t) = (1 - t)*a + t*b = (1 + 3*t, 1 + 8*t)$

Now we just perform the integration, instead of x and y we use:

x = 1 + 3*t

y = 1 + 8*t

And the integral is performed between 0 and 1.

We perform the dot product between the field with the vector b - a = (3, 8), this also makes the end result a scalar.

$W = \int\limits^1_0 {(6*(1 + 8*t)^{3/2}, 9*(1 + 3*t)*(1 + 8*t))}.(3, 8) \, dt\\\\W = \int\limits^1_0 {(3*6*(1 + 8*t)^{3/2} + 9*8*(1 + 3*t)*(1 + 8*t))}\, dt\\\\W = \int\limits^1_0 {(18*(1 + 8*t)^{3/2} + 72*(1 + 3*t)*(1 + 8*t))}\, dt\\\\W = 18*(\frac{2}{5}*(9^{5/2} - 1)) + 72*(1 + \frac{11}{2} + 8)\\\\W = 2786.4$

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