Related Rates: The Draining Tank Problem

Related Rates: The Draining Tank Problem
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  • 0:06 Draining Tank Problem
  • 1:51 Related Rates
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Lesson Transcript
Instructor: Erin Monagan

Erin has been writing and editing for several years and has a master's degree in fiction writing.

Grab an empty cup and pour some water into it. In this lesson we will watch how the height of the water changes as we learn about related rates of change and learn how to solve the draining tank problem.

The Tank

Diagram of the draining tank
Draining Tank Diagram

Let's say you go to a party, where they have a big box full of liquid. This could be punch, water or anything you want. This box is about 50 cm deep, 10 cm wide and about 20 cm tall. The flow of rate coming out of the box when you open the spigot is about 50 cubic centimeters per minute (which is about 50 ml/sec).

The Draining Tank Problem

How fast is the level of water (or drink) in the tank dropping? To figure this out, let's first determine how much liquid is in the tank. I can write the volume in the tank as V = (area of the bottom) * (height of the liquid in the tank), so V = 10 * 50 * h, so it's 500 cubic centimeters.

The height of the liquid in the tank is changing, so we'll just call it h. We now have V = 500h. Let's say I differentiate both sides of this equation with respect to dt. So I get dV/dt (how the volume of liquid in the tank is changing over time) = 500 (dh/dt)(how the height is changing as a function of time).

Related Rates

This related rate shows how height changes in time to the volume
drinking tank

This is called a related rate. We're relating the height and how it changes in time to the volume and how it changes in time. We did that by taking the derivative of a relationship we know (in this case, the volume as a function of height).

We have dV/dt = 500 (dh/dt). Well, dV/dt is the flow rate coming out of this tank or box. We know that liquid is coming out at a rate of 50 cubic cm/sec. That gives us -50 (because the volume is decreasing). Plugging that in, we have -50 = 500 (dh/dt). I can solve that for dh/dt and I get that the height is changing at a rate of -0.1 cm/sec. This means the height is dropping 1 mm/sec.

Let's say, as you're thinking of this, you're filling up a little paper cup that you made in the shape of a cone. You want to know how quickly the height of liquid is changing in your cone. You know that the volume of liquid in your cone is equal to 1/3 the height of the liquid in your cone times the largest base area of liquid in your cone.

Let's take a look at a cross-section of this cone, which is a circle with a radius. You know that the cone you made had a radius that is 1/2 of the height. So, at the top the height of the cone may be 10 cm, the radius is then 5 cm and the diameter is 10 cm. You can write V = 1/3 h (pi (h/2) ˆ2). You could simplify this so that V = pi/12 *hˆ3. Now you have a relationship between the height and the volume.

Solving for dh/dt shows that the height is dropping 1 mm/sec
Draining Tank Height Change

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