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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
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  3. Nov 26, 2006 #2

    HallsofIvy

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    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: Nov 27, 2006
  4. Nov 26, 2006 #3

    benorin

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    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]
    [tex]=-\frac{6}{27}\frac{t^2+4}{(t^2-4)^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.


    Absolutely.

    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.

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

    HallsofIvy

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    [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: 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

    HallsofIvy

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    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.
     
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