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

  • #1
217
0
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?
 

Answers and Replies

  • #2
gabbagabbahey
Homework Helper
Gold Member
5,002
7
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?

[tex]\frac{d^2y}{dx^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)[/tex]

Use the chain rule

[tex]\frac{d}{dx}=\frac{dt}{dx}\frac{d}{dt}=\frac{1}{x'(t)}\frac{d}{dt}[/tex]
 

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