Finding Vertical Tangent of Parametric Curve

**joatmon** · November 26th 2011, 09:43 AM

I know from graphing this that there are 6 vertical tangents to the following curve, but I can only determine two of them:

$x = \cos(3\theta)$ $y = 3 sin \theta$

Clearly, $\frac{dx}{d\theta} = -3\sin(3\theta)$ , so I have to find the points at which this value is equal to zero. As I see it, this happens at each multiple of $\pi$ .

As I plug multiples of $\pi$ into the original equation, I get back either (1,0) or (-1,0), so I know that these points are have vertical tangents. But, I can see from the graph that there are vertical tangents at each multiple of $\frac {\pi}{3}$ rather than simply $\pi$ .

Obviously, I am doing something wrong, but I don't know what. Isn't the idea to simply look at where $\frac{dx}{d\theta} = 0$ ? What am I missing?

Thanks!

**Soroban** · November 26th 2011, 09:54 AM

Hello, joatmon!

Your algebra/trig is off . . .

We have: . $\begin{Bmatrix}x &= &\cos(3\theta) \\ y &=& 3\sin\theta \end{Bmatrix}$

We want: . $\frac{dx}{d\theta} \,=\,0$

Hence: . $-3\sin(3\theta) \:=\:0$

. . . . . . . . $\sin(3\theta) \:=\:0$

. . . . . . . . . . . $3\theta \:=\:\pi n$

. . . . . . . . . . . . $\theta \:=\:\frac{\pi}{3}n$

**joatmon** · November 26th 2011, 10:04 AM

Thanks. I had a feeling that I was missing something obvious.

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