# Finding d^2y/dx^2 for a parametrized curve

1. Mar 23, 2010

### IntegrateMe

Find d2y/dx2 as a function of t if x = t - t2 and y = t - t3

Ok, so this basically means we're going to take the second derivative of y with respect to t over the derivative of x with respect to t.

So the second derivative of y would be.

[2 - 6t + 6t2]/(1 - 2t)2

Now, we just put that over the first derivative of x which is 1 - 2t so we get:

[[2 - 6t + 6t2]/(1 - 2t)2]/(1 - 2t)

=

(2 - 6t + 6t2)/(1 - 2t)3

Correct or incorrect?

2. Mar 23, 2010

### gabbagabbahey

Why do you think this?

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

Use the chain rule

$$\frac{d}{dx}=\frac{dt}{dx}\frac{d}{dt}=\frac{1}{x'(t)}\frac{d}{dt}$$