Find the rate At which water

Question

Water is leaking out of an inverted conical tank at a rate of 7000.07000.0 cubic centimeters per min at the same time that water is being pumped into the tank at a constant rate. The tank has height 14.014.0 meters and the diameter at the top is 4.04.0meters. If the water level is rising at a rate of 28.028.0 centimeters per minute when the height of the water is 4.54.5 meters, find the rate at which water is being pumped into the tank in cubic centimeters per minute. 

Answer 1

Start by pulling out the unknowns and knowns from the problem. IMPORTANT: be sure to pay attention to what units the problems asks us to put our final answer we're looking for in. We want cubic centimeters per minute, so we will have to take some of the information we were given and convert the values to centimeters accordingly.

We know for the conical tank that:

The top has a radius of 2.0 meters.

The height of the tank is 14.0 meters

Water is coming out at 7000.0 cubic centimeters per minute

When converting all of our info to proper units, we have:

R = 200 cm

H = 1400 cm

dv / dt = 7000 cm³ / min

For the portion of the tank full of water, let's consider it as a tinier cone that is part of the larger cone that makes up the tank. From the info we're given:

We don't know the radius, r of the filled portion of the tank

We know the height of the filled portion of the tank is 4.5 m

Water is raising in the tank at a rate of 28.0 cm / min

Let's put this in variable and rates form:

r = ?

h = 450 cm

dh / dt = 28.0 cm / min

Alright, now we need to understand an additional piece of information. We want to find the rate of water being pumped into the tank. We already know 7000 cubic centimeters are coming out per minute. The rate of change of water in the tank is equal to the difference of the amount of water coming in and the amount of water coming out:

dV / dt = (water in) - (water out)

So, by some basic algebra, we know that if we add the amount of water coming out as a rate to the change in volume of water of the tank over time, we will find the amount of water coming in per minute:

water in = (dV / dt) + water out

Let's find r first. We can do this by relating the proportional relationship between the radius and height of the filled portion of the tank and the tank itself:

(R / H) = (r / h)

We are doing this to relate the radius of the filled part of the tank as a function of height. When isolating for r, we get:

r = (R / H) * h

Or when adding in values, r = (200 / 1400) * h

When simplified:

r = (1 / 7) * h

Now, we want to use a function of volume that we can find the derivative of to get the rate of change of water volume over time for our tank. Since our tank is a cone, we can use the volume for a cone:

V = (π / 3) * r² * h

We already put r in terms of h previously, so we can plug that into this equation now:

V = (π / 3) * ( (1/7) * h)² * h

Simplifying this, we get:

V = (π / 3) * (1 / 49) * h³

Now we can derive both sides with respect to time:

(dV / dt) = (π / 3) * (1 / 49) * (3h²) * (dh / dt)

Simplifying more, we get the following:

(dV / dt) = (π)* (1/ 49) * h² * (dh/dt)

This is going to give us our rate of change of volume over time for the tank. Let's plug in the height of the water we were given as well as the rate of change of height of the water level over time:

(dV / dt) = (π)* (1 / 49) * (450)² * (28.0)

dV / dt = 363527.15 cm³ / min

Now, to find the amount of water coming into the tank, simply add this amount we calculated to the amount of water coming out over time (the 7000 cubic centimeters per minute):

water in = 363527.15 + 7000

water in = 370527.15

Let's make this a smaller number to write down using significant figures and rounding. I am going to leave my answer using 3 significant figures. Always check that you have your exponential right too for this.

water in ≈ 3.71 * 10⁵ cm³ / min

Always be careful with your units and calculations. Make sure to look over your rates of change and give them different variables too depending on the context of the problem. Hope this helps!

Answer 2

Hi Nowshin:

Let's do this one! Not sure why all the numbers are duplicated but that's OK. Remember to convert all units to cm and cm³.

The shape of the tank is an inverted cone, that is, it looks like a water fountain paper drinking cup. Be sure to draw a diagram.

The volume of a cone, V = (1/3) ∏ r² h where r = radius of the base, h = height from the round base

But, in this case, the cone is inverted (upside down)

so we need the height from the pointy end, which = 1400 - h => V = (1/3) ∏ r² (1400 - h) . . . . . Eqn (1)

This equation has 2 variables, h and r, so we must eliminate one by using another piece of information.

Now, in your diagram, you should have the cone seen from the side as a triangle with base (at the top) of radius 200 cm (half of he diameter of 400 cm) and height 1400 cm.

Draw a horizontal somewhere in the middle to represent the water when its height is 450 cm. The water, seen from the side, is also a triangle with radius, r, and height, h (from the pointy end)

By similar triangles, we can write: 1400 / 200 = h / r

So, cross-multiplying the equation: 1400 r = 200 h

=> r = 200 h / 1400 = h / 7

Substitute into Eqn (1): V = (1/3) ∏ (h/7)² (1400 - h)

= (1/3) ∏ (1/49) h² (1401 - h) . . . . . . . . . Eqn (2)

Since the water level is changing, we need the change in volume per minute, namely, dV/dt

Differentiate Eqn (2), using the product rule: dV/dt = (1/3) ∏ (1/49) [2h (1400 - h) + h² (-1)] dh/dt

= (1/3) ∏ (1/49) [2800h - 2h² - h²] dh/dt

= (1/3) ∏ (1/49) [2800h - 3h² ] dh/dt

So, when h = 450 cm and dh/dt = 28 cm/min, dV/dt = (1/3) ∏ / 49 (2800 x 450 - 3 x 450²⁾) 28 cm³/min

This is the net rate at which the volume is increasing.

But, since water is also leaking out at 7000 cm³/min, we must add 7000 to the above value for dV/dt.

I leave the arithmetic to you!