# rate of change, swimming pool

• Nov 21st 2011, 01:25 PM
stefanhinote
rate of change, swimming pool
Swimming pool is 10m long, 6m wide, depth is 1m at one end and 1.5m at the other end.

It's being filled with water at a rate of 50,000 cm^3 / minute.

After 225 minutes after the hose is turned on, the water is rising at a rate of _____cm per second.

Here's what I did:

dv/dt = 833 1/3 cm^3 / second

dv/dt = wl(dD/dt)

wl = 60,000cm

so dD/dt = dv/dt / 60,000
dD/dt = (833 1/3) / 60,000

dD/dt = 0.014 cm/sec for the answer.

Does the 225 minutes even come into play? I'm thinking it would since the depth isn't constant?
• Nov 21st 2011, 02:07 PM
TheEmptySet
Re: rate of change, swimming pool
Quote:

Originally Posted by stefanhinote
Swimming pool is 10m long, 6m wide, depth is 1m at one end and 1.5m at the other end.

It's being filled with water at a rate of 50,000 cm^3 / minute.

After 225 minutes after the hose is turned on, the water is rising at a rate of _____cm per second.

Here's what I did:

dv/dt = 833 1/3 cm^3 / second

dv/dt = wl(dD/dt)

wl = 60,000cm

so dD/dt = dv/dt / 60,000
dD/dt = (833 1/3) / 60,000

dD/dt = 0.014 cm/sec for the answer.

Does the 225 minutes even come into play? I'm thinking it would since the depth isn't constant?

Some care must be taken as the volume of the pool is a piecewise defined function

$V=\begin{cases}30h, \quad 0 \le h \le .5 \\ 15+60h, \quad h > .5 \end{cases}$

This happens becuase the bottom of the pool is NOT flat.

so when you take the derivative you get that

$\frac{dV}{dt}=\begin{cases}30 \frac{dh}{dt}, \quad 0 < h < .5 \\ 60 \frac{dh}{dt}, \quad h > .5 \end{cases}$

You need to use the 225 seconds so you know which one of the above to use.

Draw a diagram to help illustrate this point.
• Nov 21st 2011, 02:12 PM
stefanhinote
Re: rate of change, swimming pool
Thank you, makes more sense now.

Greetings from Tucson ;)
• Nov 21st 2011, 02:17 PM
skeeter
Re: rate of change, swimming pool
Quote:

Originally Posted by stefanhinote
Swimming pool is 10m long, 6m wide, depth is 1m at one end and 1.5m at the other end.

It's being filled with water at a rate of 50,000 cm^3 / minute.

After 225 minutes after the hose is turned on, the water is rising at a rate of _____cm per second.

Here's what I did:

dv/dt = 833 1/3 cm^3 / second

dv/dt = wl(dD/dt)

wl = 60,000cm

so dD/dt = dv/dt / 60,000
dD/dt = (833 1/3) / 60,000

dD/dt = 0.014 cm/sec for the answer.

Does the 225 minutes even come into play? I'm thinking it would since the depth isn't constant?

10^6 cm^3 = 1 m^3

(50000 cm^3/min)(225 min) = 11.25 m^3

the lower section of the pool (the triangular prism) has a volume of

V = (0.5 m)(5 m)(6 m) = 15 m^3

so, the water level has not reached the 0.5 m mark at the deep end yet. dh/dt is variable until it reaches that point, then the water level rises at a constant rate.

you'll need to get the volume of water in the pool below that point as a function of depth at the deep end, then take the time derivative.