A cylindrical tank with height 15m and diameter 2m is being

[kimmy_koo51]

A cylindrical tank with height 15m and diameter 2m is being filled with gasoline at a rate of 500L/min.
a) At what rate is the fluid level in the tank rising? (1 L = 1000 cm cubed)
b) About how long will it take to fill the tank?

 

[galactus]

Convert the volume from liters to cubic meters.

There is \(\displaystyle 0.5\;\ m^{3}\) in 500L

Because it's a cylinder, the change in height will be constant if the volume coming in remains the same.

The volume of the cylinder is \(\displaystyle 15{\pi}\approx{47.124}\;\ m^{3}\)

The gas is entering at \(\displaystyle 0.5\;\ \frac{m^{3}}{min}\)

Can you take it from here?. How many minutes 'til its filled?.

What is the height of the liquid level when the volume is \(\displaystyle 0.5\;\ m^{3}\)?.

 

[kimmy_koo51]

Woah...sorry but to be rather honest I have no idea what you did. Does it help you if I let you know that I am doing related rates?

This is what I have so far.

h= 15m
dh/dt = 0

d (for diameter) = 2, therefore the radius is 1

V= 15pie
dV/dt= 500 L

The equation for volume of a cylinder is V = pie r^2h

 

[skeeter]

\(\displaystyle \L V = \pi r^2 h\)

the radius has a constant value of 1 m ...

\(\displaystyle \L V = \pi h\)

\(\displaystyle \L \frac{d}{dt}[V = \pi h]\)

\(\displaystyle \L \frac{dV}{dt} = \pi \frac{dh}{dt}\)

you were given dV/dt = 500 L/min = 0.5 m³/min, find dh/dt.
time to fill the tank will be h/(dh/dt).

btw, pi is the lower case Greek letter that represents the ratio of the circumference of a circle to its diameter ... pie is something you eat.

 

[galactus]

No doubt, skeeter's method is what you need if you are studying related rates. What I was getting at is that you can do it without related rates and see if you get the same thing.

Since the radius is constant, you can find the entire volume of the tank and divide by the inflow rate to find how long it'll take to fill.

Since there is \(\displaystyle 0.5\;\ m^{3}\) coming in per minute and the total volume is \(\displaystyle 15{\pi}\) m^3, you have:

\(\displaystyle \L\\\frac{15{\pi}}{0.5}\)

To find the change in depth you can find the height of the gas during the first minute.

\(\displaystyle 0.5={\pi}h\)

Solve for h.

That is the same thing as the related rates method skeeter showed you.

You know the height of the tank is 15m. Divide by the h you just found to find the time to fill. It should be the same as the way I showed at the top.

Please, no more 'pie' unless there's ice cream with it.