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

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!