Posted by **Lucy** on Sunday, September 27, 2009 at 12:09pm.

As water drains out of a 2000L hot tub, the amount of water remaining in the tub can be modelled by the function V=0.00002(100-t)^4 , where t is the time, in minutes (t is greater than or equal to 0 and less than or equal to 100), and V is the volume of water, in litres, remaining in the tub at time t.

Determine the average rate of cahnge of the volume of water during

- the entire 100 min

- the first 30 min

- the last 30 min

I'm really confused as to what I actually have to do. I'd really appreciate some help.

- Maths -
**DrBob222**, Sunday, September 27, 2009 at 12:31pm
I believe what you are asked to do is something like this.

For the 100 min period--

V=0.00002(100-100)^4 = 0 L remaining.

So all 2000 L have drained; therefore, the rate of change is 2000L/100 min = 20L/min.

For the first 30 min--

V=0.00002(100-30)^4 = 480.2 L remaining or 2000-480.2 =1519.8 L drained.

rate of change is 1519.8/30 = 50.7 L/min.

Etc. Check my thinking. Check my work.

- Maths -
**Lucy**, Sunday, September 27, 2009 at 12:36pm
oh okay, thank you.

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