- 44

- 0

**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