Solutions

• Find the differentials of the following functions:
  • $f = t - \sin(t)\cos(t))$.
    $$df = f' dt = (1 - \cos(t)\cos(t) - \sin(t)(-\sin(t))) dt$$
    $$= (1 - \cos^2(t) + \sin^2(t)) dt = 2\sin^2(t) dt.$$

  • $g = \sin(e^t)$.
    $$dg = g'(t) dt = \cos(e^t)e^t dt.$$

  • $h = \frac{t-1}{t+1}$.
    $$dh = h'(t) dt = \frac{1(t+1) - (1)(t - 1)}{(t+1)^2} dt = \frac{2dt}{(t+1)^2}.$$

  • $k = \frac{e^t - 1}{e^t +1}$.
    $$dk = d(h(e^t)) = h'(e^t))e^t dt = \frac{2e^t dt}{(e^t + 1)^2}.$$

  • $j = \int_x^{x^2} e^{t^2} dt$
    $$dj = j'(x) dx = \frac{d}{dx} (\int_0^{x^2} e^{t^2} dt - \int_0^x e^{t^2} dt) dx$$
    $$= ([e^{u^2} u']_{u = x^2} - e^{x^2}) dx = (2xe^{x^4} - e^{x^2})dx .$$

• A particle moves in a force field $\underline{F} = [y^2, -x^2]$.
  Its motion is given by $\underline{X} = [\cos(t), \sin(t)]$.
  The associated work differential is $dW = \underline{F}.d\underline{X} = \underline{F}.(\frac{d}{dt}\underline{X}) dt = \underline{F}.\underline{V} dt$.
  Find $dW$ for this motion.
  
  We have $x = \cos(t)$ and $y = \sin(t)$, so $\underline{F} = [\sin^2(t), -\cos^2(t)]$.
  Also $\underline{V} = \frac{d}{dt}\underline{X} = [-\sin(t), \cos(t)]$.
  So we get:
  $$dW = \underline{F}.\underline{V} dt = [\sin^2(t), - \cos^2(t)].[-\sin(t), \cos(t)] dt = - (\sin^3(t) + \cos^3(t) )dt .$$