Why Study Calculus? - Related Rates

One of the more interesting applications of the calculus is in related rates problems.

Problems such as these demonstrate the sheer power of this branch of mathematics to answer questions that would seem unanswerable.

Here we examine a specific problem in related rates and show how the calculus allows us to come up with the solution quite easily.

Any quantity which increases or decreases with respect to time is a candidate for a related rates problem. It should be noted that all functions in related rates problems are dependent on time.
Since we are trying to find an instantaneous rate of change with respect to time, the process of differentiation (taking derivatives) comes into play and this is done with respect to time.

Once we map out the problem, we can isolate the rate of change we are looking for, and then solve using differentiation.

A specific example will make this procedure clear. (Please note I have taken this problem from Protter/Morrey, "College Calculus," Third Edition, and have expanded upon the solution and application of such.
) Let us take the following problem: Water is flowing into a conical tank at the rate of 5 cubic meters per minute.
The cone has altitude 20 meters and base radius 10 meters (the vertex of the cone is facing down).
How fast is the water level rising when the water is 8 meters deep? Before we solve this problem, let us ask why we might even need to address such a problem. Well suppose the tank serves as part of an overflow system for a dam. When the dam is overcapacity because of flooding as a result of, let us say, excessive rain or river drainage, the conical tanks serve as outlets to release pressure on the dam walls, preventing damage to the overall dam structure. This whole system has been designed so that there is an emergency procedure which kicks in when the water levels of the conical tanks reach a certain level.
Before this procedure is implemented a certain amount of preparation is necessary.

The workers have taken a measurement of the depth of the water and find that it is 8 meters deep. The question becomes how long do the emergency workers have before the conical tanks reach capacity? To answer this question, related rates come into play. By knowing how fast the water level is rising at any point in time, we can determine how long we have until the tank is going to overflow.
To solve this problem, we let h be the depth, r the radius of the surface of the water, and V the volume of the water at an arbitrary time t.

We want to find the rate at which the height of the water is changing when h = 8.
This is another way of saying we wish to know the derivative dh/dt.

We are given that the water is flowing in at 5 cubic meters per minute.

This is expressed as dV/dt = 5. Since we are dealing with a cone, the volume for the water is given by V = (1/3)(pi)(r^2)h, such that all quantities depend on time t.
We see that this volume formula depends on both variables r and h.
We wish to find dh/dt, which only depends on h.

Thus we need to somehow eliminate r in the volume formula.
We can do this by drawing a picture of the situation.

We see that we have a conical tank of altitude 20 meters, with a base radius of 10 meters.
We can eliminate r if we use similar triangles in the diagram.

(Try to draw this out to see this.

) We have 10/20 = r/h, where r and h represent the constantly changing quantities based on the flow of water into the tank.
We can solve for r to get r = 1/2h.
If we plug this value of r into the formula for the volume of the cone, we have V = (1/3)(pi)(.

5h^2)h.

(We have replaced r^2 by 0.5h^2).

We simplify to get V = (1/3)(pi)(h^2/4)h or (1/12)(pi)h^3. Since we want to know dh/dt, we take differentials to get dV = (1/4)(pi)(h^2)dh.
Since we want to know these quantities with respect to time, we divide by dt to get (1) dV/dt = (1/4)(pi)(h^2)dh/dt. We know that dV/dt is equal to 5 from the original statement of the problem.

We want to find dh/dt when h = 8.

Thus we can solve equation (1) for dh/dt by letting h = 8 and dV/dt = 5.
Inputting we get dh/dt = (5/16pi)meters/minute, or 0.

099 meters/minute.
Thus the height is changing at a rate of less than 1/10 of a meter every minute when the water level is 8 meters high.
The emergency dam workers now have a better assessment of the situation at hand.

For those who have some understanding of the calculus, I know you will agree that problems such as these demonstrate the awesome power of this discipline.

Before calculus, there would never have been a way to solve such a problem, and if this were a real world impending disaster, no way to avert such a tragedy. This is the power of mathematics.