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Homework Help: Where tangent line = 0 (parametric)

  1. Mar 29, 2008 #1
    1. The problem statement, all variables and given/known data

    At which point is the tanget line to the following curve horizontal?
    [tex] y= a sin^{3}\theta [/tex]
    [tex] x = acos^{3}\theta [/tex]

    2. Relevant equations

    3. The attempt at a solution
    [tex] \frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}[/tex]

    When [tex]\frac{dy}{dx} = 0 [/tex], this means that that the tanget line is horizontal.

    [tex] \frac{dy}{dx}=0[/tex] when [tex]\frac{dy}{d\theta} = 0 [/tex]

    [tex]\frac{dy}{d\theta} = 3asin^{2}\thetacos\theta [/tex]

    [tex] 0=3asin^{2}\thetacos\theta [/tex]
    [tex] 0 = sin^{2}\thetacos\theta [/tex]
    [tex] 0 = (1-cos^{2}\theta)(cos\theta) [/tex]
    [tex] cos\theta = 0 [/tex] when [tex] \theta = \pi/2 [/tex] and [tex]\theta = 3\pi/2[/tex]
    [tex]1-cos^{2}\theta = 0 [/tex] when [tex]\theta = 0 [/tex]
    We must also find where [tex]dx/d\theta =0 [/tex] since if both [tex]dy/d\theta [/tex] and [tex]dx/d\theta [/tex] = 0 at the same [tex]\theta[/tex] then we can't use that value
    [tex] \frac{dx}{d\theta} = -3acos^{2}\theta sin\theta[/tex]

    [tex] dx/d\theta = 0[/tex] at [tex]\theta = 0, \pi, \pi/2 [/tex]

    So therefore we can't use [tex] \theta = \pi/2 [/tex] and [tex] \theta = 0 [/tex] from the [tex] dy/d\theta [/tex] expression. Thus we are left with only [tex]\theta = 3\pi/2[/tex]

    If we sub [tex]\theta = 3\pi/2[/tex] back into the expression for x and y, we see that this corresponds to (0, -a), which means at (0, -a) the tangent line is horizontal.

    However, this answer is wrong, the curve is an astroid and the astroid have horizontal tangets at (+/- a, 0).

    I don't get what I did wrong, appreciate any help. Thanks
  2. jcsd
  3. Mar 29, 2008 #2


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    Hi motornoob! :smile:

    Why didn't you stop there? :confused:

    You needed [tex]\frac{dy}{dx} = 0[/tex] , but you started examining [tex]\frac{dy}{d\theta}[/tex] instead.

    Just do dy/dx = 3sin^2.cos/-3cos^2.sin = … ? :smile:
  4. Mar 29, 2008 #3
    Well yeah of course I could have figured out what dy/dx is which is just [tex] -tan\theta [/tex] and go from there, but I want to know why the method where I figure out where [tex] dy/d\theta = 0 [/tex] doesn't work. It bugs me!
  5. Mar 29, 2008 #4


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    ah! …well this bit was wrong:
    [tex]cos^{2}\theta sin\theta[/tex] is zero at any multiple of π/2. :smile:
  6. Mar 29, 2008 #5
    Ah ok, so I think here is what I did wrong.
    \frac{dx}{d\theta} = -3acos^{2}\theta sin\theta

    I set it to on the left side

    [tex] 0 = cos^{2}\theta sin\theta [/tex]
    [tex] 0= (1-sin^{2}\theta)sin\theta [/tex]

    [tex] sin\theta = 0 [/tex] when [tex] \theta = 0, \pi [/tex]

    [tex] 1-sin^{2}\theta =0[/tex]
    [tex] sin^{2}\theta =1[/tex]

    I think here I made my mistake, thinking that I can just write [tex] sin\theta = 1 [/tex] but it turns out [tex] sin\theta = +/- 1 [/tex] so [tex] \theta [/tex] here equals to [tex] \pi/2, 3\pi/2 [/tex]

    So then with this corrected mistake, I see that [tex] dy/d\theta [/tex] equals to 0 at the same places that [tex] dx/d\theta[/tex] goes to 0, so therefore I must use the [tex] dy / dx [/tex] expression directly rather than playing with [tex] dy/d\theta [/tex] or [tex] dx/d\theta [/tex]

    Would this reasoning be right? Thanks.
  7. Mar 29, 2008 #6


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    Absolutely! :smile:
  8. Mar 29, 2008 #7
    Yay, thanks for all the help!
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