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- May 29th 2007, 09:40 PMKimDayLifePhysical Application Of Calculus - Rates Of Change?
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- May 29th 2007, 11:25 PMCaptainBlack
The rate of change of volume is:

$\displaystyle \frac{dV}{dt} = -k$,

then:

$\displaystyle

V(t) = -kt+c

$

Now you are given that at $\displaystyle t=0, V=64 m^3$ and that at $\displaystyle t=9, V=1 m^3$. Use this data to determine the values of $\displaystyle k$ and $\displaystyle c$

RonL - May 29th 2007, 11:28 PMbehemoth100
For queston 1:

Rate of volume change $\displaystyle \frac{dV}{dt}= R $ where R is some constant. Since Rate of change is constant it wil have a constant 'gradient' if graphed, and thus we can do rise over run:

at $\displaystyle t= 0\ x = 4 $ therefore $\displaystyle V= 64$

at $\displaystyle t = 9\ x =1 $ therefore $\displaystyle V = 1 $

thus $\displaystyle \frac{dV}{dt}= R = \frac {64 - 1}{0-9} = -7 $

Thus by integrating

$\displaystyle \int \frac {dV}{dt} = -7t + C = V$

at $\displaystyle t=0; V = C $ and from before at $\displaystyle t=0;\ V = 64=C$

Therefore $\displaystyle V = 64 - 7t$

Can you do the next one on your own? I'll post the answer in 15 minutes.

Edit: Actually after looking at it the second part is ridiculously easy so I don't think you need help with that :) - May 30th 2007, 12:15 AMKimDayLife
Thank you very much behemoth100 and CaptainBlack :D . And yes don't worry behemoth100 and I can work out ii) now :)