1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Parametric Equations homework

  1. Nov 26, 2006 #1
    1 If [tex] x = t^{3} - 12t [/tex], [tex] y = t^{2} - 1 [/tex]

    find [tex] \frac{dy}{dx} [/tex] and [tex] \frac{d^{2}y}{dx^{2}} [/tex]. For what values of [itex] t [/itex] is the curve concave upward.

    So [tex] \frac{dy}{dx} = \frac{2t}{3t^{2}-12} [/tex] and

    [tex] \frac{d^{2}y}{dx^{2}} = \frac{2}{3t^{2}-12} [/tex]

    So [tex] 3t^{2}-12 > 0 [/tex] and [tex] t > 2 [/tex] for the curve to be concave upward?

    Is this correct?

    2 If [tex] x = 2\cos \theta [/tex] and [tex] y = \sin 2\theta [/tex] find the points on the curve where the tangent is horizontal or vertical.

    So [tex] \frac{dy}{dx} = -\frac{\cos 2\theta}{\sin \theta} [/tex]. The tangent is horizontal when [tex] -\cos 2\theta = 0 [/tex] and vertical when [tex] \sin \theta = 0 [/tex].

    So [tex] \theta = \frac{\pi}{4}+ \pi n [/tex] when the tangent is horizontal and [tex] \theta = \pi n [/tex] when the tangent is vertical? Is this correct?

    3 At what point does the curve [tex] x = 1-2\cos^{2} t [/tex], [tex] y = (\tan t )(1-2\cos^{2}t) [/tex] cross itself? Find the equations of both tangent at that point. So I set [tex] \tan t = 0 [/tex] and [tex] 1-2\cos^{2}t = 0 [/tex]
    Last edited: Nov 26, 2006
  2. jcsd
  3. Nov 26, 2006 #2


    User Avatar
    Science Advisor

    Yes, [tex]\frac{dy}{dx}= \frac{\frac{dy}{dt}}{\frac{dx}{dt}}[/tex]

    How did you get this? [tex]\frac{d^2y}{dx^2}= \frac{\frac{d(\frac{dy}{dx})}{dt}}{\frac{dx}{dt}}[/tex]
    I get [tex]\frac{d^2y}{dx^2}= -\frac{2}{3}\frac{1}{x^2- 4}[/tex]

    Okay, that looks good.

    Why? Are you assuming that y= 0 where the curve crosses itself? "Crossing itself" only means that x and y are the same for two or more different values of t. Solve the pair of equations [itex]1- 2cos^2 t= 1- 2cos^2 s[/itex] and [itex](tan t)(1- 2cos^2 t)= (tan s)(1- 2cos^2 s)[/itex] for t and s.
    Last edited by a moderator: Nov 27, 2006
  4. Nov 26, 2006 #3


    User Avatar
    Homework Helper

    For parametric equations, we have [tex] \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{2t}{3t^{2}-12}[/tex]

    so [tex] \frac{dy}{dx} = \frac{2t}{3t^{2}-12} [/tex] is correct, but

    for the second derivative, we consider y as a function of x and apply the chain rule to the first derivative to get

    [tex] \frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{d}{dt}\left(\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\right) [/tex]

    [tex] \Rightarrow\frac{d^2y}{dx^2}\cdot\frac{dx}{dt} = \frac{\frac{d^2y}{dt^2}\frac{dx}{dt}- \frac{dy}{dt}\frac{d^2x}{dt^2}}{\left(\frac{dx}{dt}\right)^2 }[/tex]

    which gives

    [tex]\frac{d^2y}{dx^2}= \frac{\frac{d^2y}{dt^2}\frac{dx}{dt}- \frac{dy}{dt}\frac{d^2x}{dt^2}}{\left(\frac{dx}{dt}\right)^3} =
    \frac{2(3t^2-12)- 2t(6t)}{(3t^2-12)^3}[/tex]

    So [tex] t^{2}-4 < 0 [/tex] which gives [tex] t<-2\mbox{ or }t > 2 [/tex] for the curve to be concave upward.


    Notice that [tex] y = x\tan t [/tex]. Now for two points to be the same for different values of t, say [tex]t_1[/tex] and [tex]t_2[/tex], and the y and x values are the same for the values of t so we have

    [tex] y-y = x\left(\tan (t_1)-\tan (t_2)\right) \Rightarrow \tan (t_1)-\tan (t_2)=0[/tex]​

    so we solve [tex]\tan (t_1)=\tan (t_2),\, t_1<>t_2[/tex] which has the solution [tex]t_1 =t_2+k\pi,\, \, k=\pm 1, \pm2,\ldots[/tex], you can get it from here...
  5. Nov 26, 2006 #4
    So [tex] y = x\tan t_{1} [/tex] and [tex] y = x\tan t_{2} \Rightarrow \tan t_{1} = \tan t_{2} [/tex].

    What does [tex] t_1<>t_2 [/tex] mean? Also how did you get the solution [tex]t_1 =t_2+k\pi,\, \, k=\pm 1, \pm2,\ldots[/tex]. Do I just substitute this back in for the equations for [tex] x [/tex] and [tex] y [/tex]? How would you find the equations of both tangents? I have only one point.

    Last edited: Nov 26, 2006
  6. Nov 27, 2006 #5


    User Avatar
    Science Advisor

    [itex]t_1<>t_2[/itex] means "t1 is NOT equal to t2. It's a computer programming notation.

    He got [itex]t_1 =t_2+k\pi,\, \, k=\pm 1, \pm2,\ldots[/itex] by remembering that tan(x) is periodic with period [itex]\pi[/itex]!

    Well, you can't substitute [itex]t_1 =t_2+k\pi,\, \, k=\pm 1, \pm2,\ldots[/itex] because you don't know what t2 is! Substitute those back into your other equation to solve for both t1 and t2. Use either to determine x and y.
    Last edited by a moderator: Nov 27, 2006
  7. Nov 27, 2006 #6
    Sorry for my stupidity, but do you mean to do this:

    [tex] \tan t_{1} = \tan (t_{1}-k\pi) [/tex]?

    [tex] t_{1} = \arctan (t_{1}-k\pi) [/tex]
  8. Nov 27, 2006 #7


    User Avatar
    Science Advisor

    Yes, that's true because tangent has period [itex]\pi[/itex]

    No, that's not true. Even
    [tex]arctan(tan t_1)= \arctan(tan(t_1- k\pi))[/tex]
    so that [itex]t_1= t_1- k\pi[/itex] isn't correct. Since tangent is periodic it is not one-to-one and arctan is not a true inverse.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook