# Thread: Parametric Equations Horzontal Tangents

1. ## Parametric Equations Horzontal Tangents

I have a question that is confusing me.

Suppose the curve is traced by the parametric equations:
x = 3sin(t)
y = 25-6cos^2(t)-12sin(t)
At what point (x,y) on this curve is the tangent line horizontal??

I know that I need to find (dy/dt)/(dx/dt)
I think that the derivative of y is messing me up to start with.
Then I know I need to find the points for t and sub them into the x and y equations given but I am not getting the correct answers.
Can anyone help?

2. Hello,

A tangent line is horizontal when the derivative is 0 (the derivate number is the slope of the tangent).

So you have to solve for t as you said.

For the derivative of y, let's do it slightly :

$y = 25-6\cos^2(t)-12\sin(t)$

We know that :

- $(u+v)'=u'+v'$

- the derivative of a constant is 0.

- $\forall \lambda \in \mathbb{R}, \ (\lambda u(x))'=\lambda u'(x)$

- $(u^n(x))'=n \ u'(x) \ u^{n-1}(x)$ or here, what's going to interest you is : $(u^2(x))'=2u'(x)u(x)$

- the derivate of cos is -sin and the derivate of sin is cos

I think you got all you need here =)
If you still have difficulties show where there is a problem