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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(1600-80t+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(h-0) 400,000-20,000t+t^2/h V'(t)=lim(h-0)400000-20000t+h+(t+h)^2-(400000-20000t+t^2)/h V'(t)=lim(h-0)400000-20000t+h+t^2+h^2+2th-400000+20000-t^2/h V'(t)=lim(h-0)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: $v'(t)=\lim_{h\to0}\frac{v(t+h)-v(t)}{h}$ Now, let's rewrite the volume function a bit: $v(t)=250$$1600-80t+t^2$$=250(t-40)^2$ Let's next make the substitution: $u=t-40$ and so we have: $v(u)=250u^2$ And so, we may write: $v'(u)=\lim_{h\to0}\frac{v(u+h)-v(u)}{h}=250\lim_{h\to0}\frac{(u+h)^2-u^2}{h}=250\lim_{h\to0}\frac{(h$$2u+h$$}{h}=250\li m_{h\to0}$$2u+h$$=500u$ Now, back-substitute for $u$ $v'(t)=500(t-40)=500t-20000$
 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: $v'(t)=250\frac{d}{dt}$$t^2-80t+1600$$=250$$2t-80$$=500t-20000$

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