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

[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?

 

[gabbagabbahey]

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.

Why do you think this?

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

Use the chain rule

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