Where tangent line = 0 (parametric)

[motornoob101]

Homework Statement

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

Homework Equations

The Attempt at a Solution

\frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}

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

\frac{dy}{dx}=0 when \frac{dy}{d\theta} = 0

\frac{dy}{d\theta} = 3asin^{2}\thetacos\theta

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

dx/d\theta = 0 at \theta = 0, \pi, \pi/2

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

If we sub \theta = 3\pi/2 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

 

[tiny-tim]

\frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}

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

Hi motornoob!

Why didn't you stop there?

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

Just do dy/dx = 3sin^2.cos/-3cos^2.sin = … ?

 

[motornoob101]

Well yeah of course I could have figured out what dy/dx is which is just -tan\theta and go from there, but I want to know why the method where I figure out where dy/d\theta = 0 doesn't work. It bugs me!

 

[tiny-tim]

ah! …well this bit was wrong:

\frac{dx}{d\theta} = -3acos^{2}\theta sin\theta

dx/d\theta = 0 at \theta = 0, \pi, \pi/2

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

cos^{2}\theta sin\theta is zero at any multiple of π/2.

 

[motornoob101]

Ah ok, so I think here is what I did wrong.
<br /> \frac{dx}{d\theta} = -3acos^{2}\theta sin\theta<br />

I set it to on the left side

0 = cos^{2}\theta sin\theta
0= (1-sin^{2}\theta)sin\theta

sin\theta = 0 when \theta = 0, \pi

1-sin^{2}\theta =0
sin^{2}\theta =1

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

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

Would this reasoning be right? Thanks.

 

[tiny-tim]

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

Would this reasoning be right? Thanks.

Absolutely!

 

[motornoob101]

Yay, thanks for all the help!