Question 8

A particle moves in the plane from $A = [1, -2]$ to $B = [4, 4]$ along the straight line segment $AB$.
The particle is acted on by two forces, $\underline{G}$ and $\underline{H}$ at any point $[x, y]$ of the segment:

$$\underline{G} = [3, 5] ,$$

$$\underline{H} = [\sqrt{x}, 3y^2].$$

Calculate the work done by each of the forces as the particle moves from $A$ to $B$.

• For the constant force $\underline{G}$, the work $W$ Joules done in a displacement $\underline{D}$ is: $W = \underline{G}.\underline{D}$.
  Here $\underline{D} = \underline{AB} = \underline{B} - \underline{A} = [4, 4] - [1, -2] = [3, 6]$.
  So we get:
  $$W = \underline{G}.\underline{D} = \underline{G}.\underline{AB} = [3,5].[3, 6] = 3(3) + 5(6) = 9 + 30 = 39.$$

• For the variable force $\underline{H}$, the infinitesimal work $dW$ Joules, done in an infinitesimal displacement $[dx, dy]$ at $[x, y]$ is:
  $$dW = \underline{H}.[dx, dy] = [x^{\frac{1}{2}}, 3y^2].[dx, dy] = x^{\frac{1}{2}} dx + 3y^2 dy.$$

From here there are (at least) three routes to finish off the problem:

• First we recognize that $dW$ is an exact differential:
  $$dW = x^{\frac{1}{2}} dx + 3y^2 dy = d(\frac{2}{3} x^{\frac{3}{2}} + y^3).$$
  Then the total work $W$ Joules, done in going from $A$ to $B$, is, by FTC:
  $$W = \int_A^B dW = \int_A^B d(\frac{2}{3} x^{\frac{3}{2}} + y^3) =... ...{3}{2}} + y^3]_{A}^{B} = [\frac{2}{3} x^{\frac{3}{2}} + y^3]_{[1, -2]}^{[4,4]}$$
  $$= \frac{2}{3} 4^{\frac{3}{2}} + 4^3 - (\frac{2}{3} (1)^{\frac{3}{... ...64 - (\frac{2}{3} - 8) = \frac{14}{3} + 72 = \frac{230}{3} = 76.66\overline{6}.$$

• Secondly we parametrize the points $[x, y]$ of the line $AB$ as follows:
  $$[x, y] = \underline{A} + t\underline{AB} = [1, -2] + t[3, 6] = [1 + 3t, -2 + 6t],$$
  $$[dx, dy] = \frac{d}{dt}([x, y]) dt = [\frac{d}{dt} (1 + 3t), \frac{d}{dt} (-2 + 6t)] dt = [3, 6] dt.$$
  $$x = 1 + 3t, \hspace{10pt} y = -2 + 6t, \hspace{10pt} dx = 3dt, \hspace{10pt} dy = 6dt.$$
  We have $t = 0$ for the point $A$ and $t = 1$ for the point $B$.
  Substituting these relations into $dW$, we get:
  $$dW = x^{\frac{1}{2}} dx + 3y^2 dy = (1 + 3t)^{\frac{1}{2}}(3) dt + 3(-2 + 6t)^2 (6) dt$$
  $$= (3(1 + 3t)^{\frac{1}{2}} + 72(1 - 3t)^2) dt.$$
  Integrating we get the total work done $W$ in going from $A$ to $B$ along $AB$ as:
  $$W = \int_A^B dW = \int_0^1 (3(1 + 3t)^{\frac{1}{2}} + 72(1 - 3t)^2) dt$$
  $$= [\frac{2}{3}(1 + 3t)^{\frac{3}{2}} - 8(1 - 3t)^3)]_0^1 = (\frac{2}{3}(4)^{\frac{3}{2}} - 8(- 2)^3) - (\frac{2}{3}(1)^{\frac{3}{2}} - 8(1)^3))$$
  $$= \frac{16}{3} + 64 - (\frac{2}{3} - 8)= \frac{230}{3} = 76.66\overline{6}.$$

• Alternatively we can parametrize $y$ in terms of $x$.
  
  The line $AB$ has slope $\frac{6}{3} = 2$ and goes through $[1, -2]$, so is:
  $$y - (-2) = 2( x - 1), \hspace{10pt} y + 2 = 2x - 2, \hspace{10pt} y = 2x - 4,$$
  $$dy = \frac{dy}{dx} dx = 2dx.$$
  Substituting these relations into $dW$, we get:
  $$dW = x^{\frac{1}{2}} dx + 3y^2 dy = x^{\frac{1}{2}} dx + 3(2x - 4)^2 (2 dx) = (x^{\frac{1}{2}} + 24(x - 2)^2) dx.$$
  Integrating, the total work done $W$ in going from $A$ to $B$ along $AB$ is:
  $$W = \int_A^B dW = \int_1^4 (x^{\frac{1}{2}} + 24(x - 2)^2) dx$$
  $$= [\frac{2}{3} x^{\frac{3}{2}} + 8(x - 2)^3]_1^4 = [\frac{2}{3} 4^{\frac{3}{2}} + 8(2)^3 - (\frac{2}{3} 1^{\frac{3}{2}} + 8(1 - 2)^3)$$
  $$= \frac{16}{3} + 64 - (\frac{2}{3} - 8)= \frac{230}{3} = 76.66\overline{6}.$$