Physical Application Of Calculus - Rates Of Change?

**KimDayLife** · May 29th 2007, 09:40 PM

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**CaptainBlack** · May 29th 2007, 11:25 PM

Originally Posted by KimDayLife

Today in class our teacher gave us a *fun* take home question and normally I'm fine with them except this time (maybe because physical applications is hard for me), it's been wracking my brain, I've looked in my text books and can't seem to find any real help so any help would be very appreciated

The Question

A block of ice in the form of a cube melts in such a way that the block remains cubic. After a time t hours the black is 3cm^3. Initially the cube has edge 4m and after 9 hours the cube has an end of 1m.
i) If the rate of change of volume. Rm^3 is given by R=-k for some constant k>0 show that V=64-7t.
ii) Hence find when the cube has an edge of 2m.

I intergrated (V=64-7t) and got 64t - 7 x(t^2/2). Is this the right approach or am I going about it all wrong? Thanks, Kim

The rate of change of volume is:

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

then:

$<br /> V(t) = -kt+c<br />$

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

RonL

**behemoth100** · May 29th 2007, 11:28 PM

Originally Posted by KimDayLife

Today in class our teacher gave us a *fun* take home question and normally I'm fine with them except this time (maybe because physical applications is hard for me), it's been wracking my brain, I've looked in my text books and can't seem to find any real help so any help would be very appreciated

The Question

A block of ice in the form of a cube melts in such a way that the block remains cubic. After a time t hours the black is 3cm^3. Initially the cube has edge 4m and after 9 hours the cube has an end of 1m.
i) If the rate of change of volume. Rm^3 is given by R=-k for some constant k>0 show that V=64-7t.
ii) Hence find when the cube has an edge of 2m.

I intergrated (V=64-7t) and got 64t - 7 x(t^2/2). Is this the right approach or am I going about it all wrong? Thanks, Kim

For queston 1:
Rate of volume change $\frac{dV}{dt}= R$ 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 $t= 0\ x = 4$ therefore $V= 64$
at $t = 9\ x =1$ therefore $V = 1$
thus $\frac{dV}{dt}= R = \frac {64 - 1}{0-9} = -7$
Thus by integrating
$\int \frac {dV}{dt} = -7t + C = V$
at $t=0; V = C$ and from before at $t=0;\ V = 64=C$
Therefore $V = 64 - 7t$
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

**KimDayLife** · May 30th 2007, 12:15 AM

Thank you very much behemoth100 and CaptainBlack

. And yes don't worry behemoth100 and I can work out ii) now

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