How do you find the rate at which water is being pumped into the tank in cubic centimeters per minute if water is leaking out of an inverted conical tank at a rate of 12500 cubic cm/min at the same time that water is being pumped into the tank at a constant rate, and the tank has 6m height and the the diameter at the top is 6.5m and if the water level is rising at a rate of 20 cm/min when the height of the water is 1.0m?

Answer 1

Rate of water being pumped in
= Rate of water being pumped out
+ Rate of water increase needed to cause rise in water level.

For all heights within the cone the ratio of radius to height is
#r/h = (3.25)/(600) = (13)/(24)#

The formula for the Volume of a cone dependent on height and radius is
#V(h,r) = (pi r^2 h)/3#

or, with #r = (13)/(24) h#

#V(h) = (169)/(1728) pi h^3# (don't you hate it when the numbers get ugly?)

#(d V(h))/(dh) = (169)/(576) pi h^2#

We're (initially) looking for the rate of input required to raise the water level at a rate of #20 (cm)/(min) when #h = 100 cm#

That is #(d V(h))/(dt)#
which can be calculated as

#(d V(h))/(dt) = (d V(h))/(dh) xx (d h)/(dt)#

# (d V(h=100))/(dt) = (169)/(576) pi * (100)^2#

#= 9,217.52# (using my handy spreadsheet program as a calculator)

Returning to:
Rate of water being pumped in
= Rate of water being pumped out
+ Rate of water increase needed to cause rise in water level.

we get
Rate of water being pumped in
#= 12,500 (cm)^3/(min) + 9,217.52 (cm)^3/(min)#
#= 21,717.52 (cm)^3/(min)#

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Answer 2

To find the rate at which water is being pumped into the tank, use the formula: Rate of change of volume = rate of pumping - rate of leakage.

First, calculate the rate of change of volume of the tank using the given information:

Rate of change of volume = π * r^2 * h

where r is the radius of the water level and h is the height of the water level.

Given:

  • Diameter at the top of the tank = 6.5m, so radius (r) = 6.5/2 = 3.25m
  • Height of the water level (h) = 1.0m

Calculate:

  • Volume = π * (3.25)^2 * 1.0 = approximately 33.18 cubic meters

Given:

  • Rate of change of height (dh/dt) = 20 cm/min
  • Rate of leakage = 12500 cubic cm/min

Calculate the rate of change of volume using the formula:

Rate of change of volume = π * r^2 * (dh/dt)

Substitute the values: Rate of change of volume = π * (3.25)^2 * (20/100) = approximately 20.42 cubic meters/min

Now, find the rate of pumping: Rate of pumping = Rate of change of volume + Rate of leakage Rate of pumping = 20.42 + 12500 = 12520.42 cubic cm/min

So, the rate at which water is being pumped into the tank is approximately 12520.42 cubic cm/min.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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