Quick Solve - Literally Stuck at the Last Step

  • Thread starter Thread starter IntegrateMe
  • Start date Start date
  • Tags Tags
    Stuck
IntegrateMe
Messages
214
Reaction score
1
The position of a particle moving in the xy-plane is given by the parametric equations x = t3 - 3t2 and y = 2t3 - 3t2 - 12t. For what values of t is the particle at rest?

So we're trying to find when the slope (dy/dx) is equal to 0.

dy = 6t2 - 6t - 12
dx = 3t2 - 3t

dy/dx = (6t2 - 6t - 12) / (3t2 - 3t)
dy/dx = [6(t2 - t - 2)] / [3t(t - 2)]
dy/dx = [2(t + 1)(t - 2)] / [t(t -2)]
dy/dx = 0 @ t = -1 or 2

So, i would assume the answer is "-1 or 2" but the answer is really just "2 only". Can anyone explain this to me?
 
Physics news on Phys.org
Assuming t is the time, if you find multiple values in the result you will discard the ones that are not useful, like the idea of something taking a negative amount of time.
 
Wow, that was simple. I'm sorry but i should've seen how obvious that was. Thanks a bunch!
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top