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October 6th, 2013, 01:45 AM  #1 
Newbie Joined: Aug 2013 Posts: 24 Thanks: 0  Finding an expression for Instantaneous rate of change
The question is  v(t)=250(160080t+t^2) Find an expression for the instantaneous rate at which water is leaving a pool at any given time, t minutes. This is my working  V'(t)=lim(h0) 400,00020,000t+t^2/h V'(t)=lim(h0)40000020000t+h+(t+h)^2(40000020000t+t^2)/h V'(t)=lim(h0)40000020000t+h+t^2+h^2+2th400000+20000t^2/h V'(t)=lim(h0)h^2+2th/h V'(t)=h+2t V'(t)=2t Is this correct? Thanks 
October 6th, 2013, 02:08 AM  #2 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 522 Math Focus: Calculus/ODEs  Re: Finding an expression for Instantaneous rate of change
Now that's not correct. I would work the problem, assuming we are to use first principles, as follows: Now, let's rewrite the volume function a bit: Let's next make the substitution: and so we have: And so, we may write: Now, backsubstitute for 
October 6th, 2013, 02:14 AM  #3 
Newbie Joined: Aug 2013 Posts: 24 Thanks: 0  Re: Finding an expression for Instantaneous rate of change
Thanks, I see where I went wrong, I didn't factorise the equation. Using differentiation, how would I do this?

October 6th, 2013, 02:21 AM  #4 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 522 Math Focus: Calculus/ODEs  Re: Finding an expression for Instantaneous rate of change
If you use the rules of differentiation, you could write: 

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