Posted by kevin on April 29, 2001 at 21:32:16:
In Reply to: Re: find acceleration posted by dan on April 29, 2001 at 19:19:36:
: The problem reads: A particle moves along the curve given by y=sqrt(t^3+1). Find the acceleration when t=2.
: Kevin wrote:second derivative again. first derivate would be 1/2(t^3+1)^(-1/2)(3t^2).
: Would this workout to be 1/(3t)^(-1/2)(6t)?
: I have tried this a few different ways but have not gotten the correct answer yet (multi choice)
: Thank you very much, I have been working on this all day
I may have written is confusing. It should look more like (3t^2)/[2(t^3+1)^(1/2)]
A website gave me [3t(t^3+4)]/[4(t^3+1)^(3/2)]
If that is the correct 2nd derivative f''(2) equals 13/27. I may have made a mistake somewhere but hopefully you will get the concept