# Thread: taking the second derivative of a parametric equation

1. ## taking the second derivative of a parametric equation

I need help with this problem. I have the answer, but i don't know how they got it.

Given the parametric equation x=cost, and y=2t determine dy/dx and the second derivative of this parametric equation.

I understand how they got dy/dx = -2csct, but i don't understand how they got the second derivative. I have the answer, but don't know how to type it on this forum.

2. Originally Posted by kyousuf
I need help with this problem. I have the answer, but i don't know how they got it.

Given the parametric equation x=cost, and y=2t determine dy/dx and the second derivative of this parametric equation.

I understand how they got dy/dx = -2csct, but i don't understand how they got the second derivative. I have the answer, but don't know how to type it on this forum.
$\displaystyle \frac{dy}{dx} = -2\csc{t}$

$\displaystyle \frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right)$

$\displaystyle \frac{d}{dx}\left(-2\csc{t}\right) = 2\csc{t}\cot{t} \cdot \frac{dt}{dx} = 2\csc{t}\cot{t} \cdot (-\csc{t}) = -2\csc^2{t}\cot{t}$