Posted by Subhotosh Khan on April 30, 2001 at 09:04:29:
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
No!
Re-write the dy/dt as
dy/dt = (3/2)[t^2/{sqrt(t^3+1)}]
Now you have function of the form U(t)/V(t) where
U(t) = t^2
and
V(t) = sqrt(t^3+1)
Now you know
d/dt(U/V) = (U'V - V'U)/(V^2) where U' and V' are the first derivatives of U and V. Go through this step by step - there is lot of algebra to get to the final answer.