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Where tangent line = 0 (parametric)

1. Homework Statement

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. Homework 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
 

Answers and Replies

tiny-tim
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Homework Helper
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[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.
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:
 
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!
 
tiny-tim
Science Advisor
Homework Helper
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249
ah! …well this bit was wrong:
[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]
[tex]cos^{2}\theta sin\theta[/tex] is zero at any multiple of π/2. :smile:
 
Ah ok, so I think here is what I did wrong.
[tex]
\frac{dx}{d\theta} = -3acos^{2}\theta sin\theta
[/tex]

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.
 
tiny-tim
Science Advisor
Homework Helper
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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.
Absolutely! :smile:
 
Yay, thanks for all the help!
 

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