Read this section to learn to connect derivatives to the concept of the rate at which things change. Work through practice problems 1-3.
In Section 2.1, several interpretations were given for the derivative of a function. Here we will examine how the "rate of change of a function" interpretation can be used. If several variables or quantities are related to each other and some of the variables are changing at a known rate, then we can use derivatives to determine how rapidly the other variables must be changing.
Example 1: Suppose we know that the radius of a circle is increasing at a rate of 10 feet each second (Fig. 1), and we want to know how fast the area of the circle is increasing when the radius is 5 feet. What can we do?
We could get a better approximation by calculating over a shorter time interval, say seconds. Then the original area was , the new area is feet (Why is the new radius 6 feet?) so . This is
the slope of the secant line through the point and in Fig. 3, and it is a much better approximation of the slope of the tangent line at , but it is still only an approximation. Using derivatives, we can
get an exact answer without doing very much work.
We know that the two variables in this problem, the radius and the area , are related to each other by the formula , and we know that both and are changing over time so each of them is a function of an additional variable time. We will continue to write the radius and area variables as and , but it is important to remember that each of them is really a function of and . The statement that "the radius is increasing at a rate of 10 feet each second" can be translated into a mathematical statement about the rate of change, the derivative of with respect to time: . The question about the rate of change of the area is a question about . Collecting all of this information, we have
Finally, we are ready to find - we just need to differentiate each side of the equation with respect to the independent variable .
. The last piece, , appears in the derivative because is a function of and we must use the differentiation rule for a function to a power (or the Chain Rule):
We know from the problem that so . This answer tells us that the rate of increase of the area of the circle, , depends on the value of the radius as well as on the value of . Since , the area of the circle will be increasing at a rate of ( 5 feet) square feet per second.
The key steps in finding the exact rate of change of the area of the circle were to:
Example 2: Divers lives depend on understanding situations involving related rates. In water, the pressure at a depth of feet is approximately pounds per square inch (compared to approximately 15 pounds per square inch at sea level ). Volume is inversely proportional to the pressure, , so doubling the pressure will result in half the original volume. Remember that volume is a function of the pressure: .
(a) Suppose a diver's lungs, at a depth of 66 feet, contained 1 cubic foot of air, and the diver ascended to the surface without releasing any air, what would happen?
(b) If a diver started at a depth of 66 feet and ascended at a rate of 2 feet per second, how fast would the pressure be changing?
(Dives deeper than 50 feet also involve a risk of the "bends" or decompression sickness if the ascent is too rapid. Tables are available which show the safe rates of ascent from different depths).
Solution: (a) The diver would risk getting ruptured lungs. The 1 cubic foot of air at a depth of 66 feet would be at a pressure of pounds per square inch (psi). Since the pressure at sea level, , is only 1/3 as great, each cubic foot of air would expand to 3 cubic feet, and the diver's lungs would be in danger. Divers are taught to release air as they ascend to avoid this danger.
(b) The diver is ascending at a rate of 2 feet/second so the rate of change of the diver's depth is . The pressure, , is a function of (or so .
Example 3: The height of a cylinder is increasing at 7 meters per second and the radius is increasing at 3 meters per second. How fast is the volume changing when the cylinder is 5 meters high and has a radius of 6 meters? (Fig. 4)
Solution: First we need to translate our known information into a mathematical format. The height and radius are given: height and radius . We are also told how fast and are changing: and . Finally, we are asked to find , and we should expect the units of to be the same as which are .
We also need an equation which relates the variables and (all of which are functions of time ) to each other:Then, differentiating each side of this equation with respect to (remembering that and are functions), we have
by the Power Rule for functions.
The rest is just substituting values and doing some arithmetic:
The volume of the cylinder is increasing at a rate of 1357.2 cubic meters per second. (It is always encouraging when the units of our answer are the ones we expect).
Practice 1: How fast is the surface area of the cylinder changing in the previous example?
(Assume that , and have the same values as in the example and use Fig. 5 to help you determine an equation relating the surface area of the cylinder to the variables
and . The cylinder has a top and bottom).
Practice 2: How fast is the volume of the cylinder in the previous example changing if the radius is decreasing at a rate of 3 meters per second? (The height, radius and rate of change of the height are the same as in the previous example: and respectively).
Usually, the most difficult part of Related Rate problems is to find an equation which relates or connects all of the variables. In the previous problems, the relating equations required a knowledge of geometry and formulas for areas and volumes (or knowing where to find them). Other Related Rates problems may require other information about similar triangles, the Pythagorean formula, or trigonometry - it depends on the problem.
It is a good idea, a very good idea, to draw a picture of the physical situation whenever possible. It is also a good idea, particularly if the problem is very important (your next raise depends on getting the right answer), to calculate at least one approximate answer as a check of your exact answer.
Example 4: Water is flowing into a conical tank at a rate of . If the radius of the top of the cone is (see Fig. 6), the height is , and the depth of the water is
, then how fast is the water level rising?
Solution: Lets define our variables to be height (or depth) of the water in the cone and the volume of the water in the cone. Both and are changing, and both of them are functions of time . We are told in the problem that and , and we are asked to find We expect that the units of will be the same as which are meters/second.
Unfortunately, the equation for the volume of a cone, , also involves an additional variable , the radius of the cone at the top of the water. This is a situation in which the picture can be a great help by suggesting that we have a pair of similar triangles so top radius total height and . Then we can rewrite the volume of the cone of water, , as a function of the single variable :
The rest of the solution is straightforward.
We know that and so it is easy to solve for
This example was a little more difficult than the others because we needed to use similar triangles to get an equation relating to and because we eventually needed to do a little arithmetic to solve for .
Practice 3: A rainbow trout has taken the fly at the end of a 60 foot line, and the line is being reeled in at a rate of 30 feet per minute. If the tip of the rod is 10 feet above the water and the trout is at the surface of the water, how fast is the trout being pulled toward the angler? (Suggestion: Draw a picture and use the Pythagorean formula).
Example 5: When rain is falling vertically, the amount (volume) of rain collected in a cylinder is proportional to the area of the opening of the cylinder. If you place a narrow cylindrical glass and a wide cylindrical glass out in the rain (Fig. 7),
(a) which glass will collect water faster, and
(b) in which glass will the water level rise faster?
Solution: Let's assume that the smaller glass has a radius of and the larger has a radius of , so the areas of their openings are and respectively.
(a) The smaller glass will collect water at the rate , and the larger at the rate so , and the larger glass will collect water faster than the smaller glass.
(b) The volume of water in each glass is a function of the radius of the glass and the height of the water in the glass: and where and are the heights of the water levels in the smaller and larger glasses, respectively. The heights and vary with (are functions of ) so
Then so the water level in each glass is rising at the same rate. In one minute period, the larger glass will collect more rain, but the larger glass also requires more rain to raise its water level by each inch. How do you think the volumes and water levels would change if we placed a small glass and a large plastic box side by side in the rain?