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October 6th, 2013, 01:45 AM   #1
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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
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October 6th, 2013, 02:08 AM   #2
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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, back-substitute for

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October 6th, 2013, 02:14 AM   #3
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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?
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October 6th, 2013, 02:21 AM   #4
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Re: Finding an expression for Instantaneous rate of change

If you use the rules of differentiation, you could write:

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