Chapter 2:

Related Rates

First notice that it is not usually the case that a constant change in an input variable will produce a corresponding constant change in the output.  That is, suppose a circle's radius is increasing at a constant 0.1 cm/sec.  Its area will not be increasing at that rate.  A small increase in radius will produce a rather large increase in area, once the radius is large.  That's really just another way of saying that the area function does not have constant slope.  Consider this simple applet:

Applet:

So, if we want to realistically model the changes in a complex system, with interrelated variables changing through time, we will need to think of the chain of relationships between them.  This involves derivatives and the chain rule:

Using the Chain Rule to make still pictures into motion pictures:

Any relationship that is "always true" can be turned from a static relationship to a motion relationship, by simply examining what happens if the variables are allowed to vary through time.

That is, for example, if you think of the base and the height of a triangle as being functions of time:

base = b(t) and height = h(t)

Then the old tried and true function for the area of a triangle,

A = (base*height)/2

can be turned into the "motion equation" for a triangle area, by applying the differential operator, with respect to time:

D_t [ A(t) = (b(t)*h(t))/2 ]

Notice that we had to use the product rule, since b and h are both functions of t.

So, consider the applet below, where we have an oscillating base and an oscillating height, and we are concerned with what's happening to the area:

Applet:

If the base of the triangle at any time, t, is given by: 

b(t) = 2 + sin(pi*t/2), and the height is given by

h(t) = 2 - cos(pi*t/3), then we could form the area function, and examine its derivatives, or we could simply apply the "motion formula" for the area:

b'(t) = pi/2*cos(pi*t/2)

h'(t) = pi/3*sin(pi*t/3)

At time t = 5,

b(t) = 2 + sin(pi*5/2) = 3

h(t) = 2 - cos(pi*5/3) = 1.5

b'(t) = pi/2*cos(pi*5/2) = 0

h'(t) = pi/3*sin(pi*5/3) = -pi*sqrt(3)/6 = -0.9069

A(t) = (3*1.5)/2 = 2.25

A'(t) = (0*1.5 + 3*(-0.9069))/2 = -1.3603

and so, as we see in the graph, the area is decreasing when t = 5, at a rate of about -1.36 square units per second.

The beauty of it is that we didn't even have to know the actual function b(t) or h(t) (although this time we did know them); we only need to know the values of b, h, b' and h' at a particular time, t.

Example:  Suppose at a particular time, the base of a triangle is 8 cm and is decreasing at 3 cm/sec, while the height of a triangle is 4 cm and is increasing at 6 cm/sec.

At what rate is the area of the triangle changing?  Is it increasing or decreasing?

Ans: 

b = 8, db/dt = -3 (since it's decreasing!), h = 4, dh/dt = 6:

dA/dt = ( (-3)*4 + 8*6)/2 = (-12 + 48)/2 = 18 cm²/sec.  Since it's positive, the area is increasing.

The two steps we need to practice, evidently are:

A)  Finding a static equation that is true for all time.

B)  Deriving the motion equation from the static equation. 

After that, everything's just fairly straight-forward algebra.

Look at the following examples:

The first example is a simplified version of a common phenomenon in rural Minnesota.  Picture a grain truck unloading overflow soybeans in a conical pile (see p. 154 #23):

Applet:

Example #2 from the book is a famous problem, found in almost every Calculus textbook of the past 100 years (p. 155 #27):

Applet:

Example #3 is usually, in most books, for some reason I don't know, an ostrich, rather than a person, walking either toward or away from a light pole. (Book p. 156 #35)

At first glance, "A" and "B" seem to be asking the same question, but see the diagram for the difference.

Applet:

The fourth example takes a little outside information about falling objects, although it is not mentioned in the problem (see the book, p. 157 #54).

Applet:

Next Time:  Chapter 3.

Also, you can start work on Project #3 and on Section 2.6 Homework.