A water tank has the shape of an upside-down cone with radius 2 m and height 5 m. The water is running out of the tank through a small hole at the bottom. What is the rate of change of the water level when flow-out reads #3 m^3/min# at height, #h= 4 m#?

Question

Charles Aeschliman · Accepted Answer

rate of change is #6.2times 10^-3ms^-1=0.37m/ min#

We are given the rate of change of volume #(dV)/dt=3m^3/min=0.05m^3/s# and are asked to find the rate of change of water level , #(dh)/dt#.

In terms of a differential equation we can express this as #(dV)/dt=(dV)/(dh)*(dh)/dt=0.05# which we will call equation1

we want to find #(dh)/dt# so we will need to find #(dV)/(dh)#.

The volume of a cone is given by #V=pir^2h/3#

And in this case #r/h=2/5# so #r=2/5h#

Substituting this into the Volume equation gives:

#V=pir^2h/3=pi(2/5h)^2(h/3)=pi4/75h^3#

We can now differentiate:

#(dV)/(dh)=pi12/75h^2#

When h = 4, then #(dV)/(dh)=pi12/75h^2=2.56pi#

So if we substitute this back into equation 1:

#(2.56pi)*((dh)/dt)=0.05#

#(dh)/dt=0.05/(2.56pi)=6.2times 10^-3ms^-1=0.37m/ min#

Elena Albarran · Answer

undefined To find the rate of change of the water level, we can use the formula for the volume of a cone: V = (1/3)πr^2h, where V is the volume, r is the radius, and h is the height. Then, we can differentiate the volume equation with respect to time (t) to find the rate of change of the volume. Finally, we can use the relationship between the volume and the height of the water to find the rate of change of the water level.

Given:
- Radius (r) = 2 m
- Height (h) = 5 m
- Volume flow rate (dV/dt) = 3 m^3/min
- Height at which the rate of change of water level is to be found (h) = 4 m

First, we need to find the rate of change of the volume with respect to time (dV/dt).

dV/dt = (dV/dh) * (dh/dt)

To find dV/dh, differentiate the volume equation with respect to height:

V = (1/3)πr^2h

dV/dh = (1/3)π * 2^2

Now, we need to find dh/dt. We are given that dV/dt = 3 m^3/min and h = 4 m.

dh/dt = (dV/dt) / (dV/dh)

Finally, substitute the values into the equation:

dh/dt = 3 / ((1/3)π * 2^2)

Evaluate dh/dt to find the rate of change of the water level at h = 4 m.