Chapter 12: Related Rates

Suppose we have two variables and (in most story problems the letters will be different, but for now let's use and ) which are both changing with time. A ``related rates'' problem is a problem where we know one of the rates of change at a given instant --- say, / --- and we want to find the other rate / at that instant.

If is written in terms of , i.e., , then this is easy to do using the chain rule:

=

That is, find the derivative of , plug in the value of x at the instant in question, and multiply by the given value of dx/dt to get dy/dt.

But often and will be related in some other way, for example x = , or , or perhaps , where
and are expressions involving both variables. In all cases, you can solve the related rates problem by taking d/dt of both sides, plugging in all the known values (namely, , and dx/dt), and then solving for dy/dt.

Decide what the two variables are.

Find an equation relating them.

Take d/dt of both sides.

Plug in all known values at the instant in question.

Solve for the unknown rate.

Sometimes it may not be convenient or desirable to find the equation between directly. They might in turn be related to each other by a chain of variables and conditions. We will explain such cases in the examples below.

The radius of a circular puddle is 3 m, and it is increasing at the rate of 1 cm/min. How fast is the puddle's area increasing?

Solution

Here the two things that change with time are: the radius and the area A. We know / = 1 cm/min = 0.01 m/min. We want to find / . Thus, in this problem is playing the role of , and is playing the role of . The relation between the two variables is: .

Taking d/dt of both sides, we obtain

= .

Notice that, since we are taking derivatives with respect to , the derivative of is , by the rule for the derivative of a power of a function (the -rule). In this case we alternately could have used the chain rule for the variables and :

.

We now plug in the information given. At the instant in question, , and / = . Thus, = ( ) ( ) ( ) = / .

Note: A crucial point to notice in this and similar problems is that, even though a numerical value is given for r, r must be regarded as a variable, not a constant. The value of the variable at the instant in question cannot be plugged into the equation until after taking d/dt of the equation .

You are inflating a spherical balloon at the rate of / . How fast is its radius increasing at the instant when ? How fast is the area increasing at the same time?

Solution

First we answer the first question.

Here the variables are the radius and the volume . We know / , and we want / . The two variables are related by means of the equation . Taking / of both sides gives . We now substitute the values we know at the instant in question: . Solving for , we obtain = cm/sec.

Now the second question.

The formula for the area is . Taking / of both sides, we get

= .

Now that we have worked out the answer to / , we can easily deduce that / = = .

Note: If only the second question was asked, then we would have first written down the two well known formulas:

,

and tried to find a direct relation between and . The relation is . Then we would try to solve this for and take derivatives, or take / of both sides and finally arrive at the same result.

The point is that it is not necessary to eliminate right away. It is far easier to take / of the two relations first and get equations

, so or .

Our final answer is now clear!

A plane is flying at 500 mph at an altitude of 3 miles in a direction away from where you're standing (i.e., the point on the ground directly beneath the plane is moving away from you). How fast is the plane's distance from you increasing at the moment when the plane is flying over a point on the ground 4 miles from you?

Solution

To see what's going on, we first draw a triangle whose hypotenuse is the line from you to the plane (see figure). The vertical leg of the triangle is a constant --- 3 mi --- which does not change in the course of the problem. The horizontal leg is the variable whose rate we know --- its rate / is the speed of the plane. The hypotenuse is the variable whose rate we want. The equation relating and is the Pythagorean theorem : . We next take / of both sides, getting / = / . At the instant in question

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Water is poured into a conical container at the rate of /sec. The cone points directly down (see side view at right), and it has a height of and a base radius of . How fast is the water level rising when the water is deep (at its deepest point)?

Solution

In this problem, the water forms a conical shape within the big cone whose height and base radius --- and hence also its volume --- are all increasing as water is poured into the container. This means that we actually have three things varying with time: the water level (the height of the cone of water), the radius of the circular top surface of water (the base radius of the cone of water), and the volume of water V. And we have the relation . We know / , and we want / . At first something seems to be wrong: we have a third variable whose rate we don't know.

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A swing consists of a board at the end of a ft long rope. Think of the board as a point P at the end of the rope, and let Q be the point of attachment at the other end. Suppose that the swing is directly below Q at time , and is being pushed by someone who walks at the speed of 6 ft/sec from left to right. Find (a) how fast the swing is rising after 1 sec; (b) the angular speed of the rope in deg/sec after 1 sec.

Solution

In doing this problem, we must start out by asking: What is the geometric quantity whose rate of change we know, and what is the geometric quantity whose rate of change we're being asked about? Note that the person pushing the swing is moving horizontally at a rate we know. In other words, the horizontal coordinate of P is increasing at 6 ft/sec. In the xy-plane let us make the convenient choice of putting the origin at the location of P at time t = 0, i.e., a distance 10 directly below the point Q of attachment. Then the rate we know is / .

In part (a) the rate we want is dy/dt (the rate at which P is rising).

In part (b) the rate we want is , where theta stands for the angle in radians through which the swing has swung from the vertical. (Actually, since we want our answer in deg/sec, at the end we must convert from rad/sec by multiplying by .)

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A road going North--South crosses a road going East--West at the point . Car is driving North along the first road, and car is driving East along the second road. At time car is to the North of , and car is to the East of . Suppose that at some instant you know the values of , the velocity of car , and the velocity of car . Find a formula in terms of these known values for the rate at which the distance between the two cars is increasing.

Solution

Let be the distance from car to car . By the Pythagorean Theorem, . We now take / of both sides of the Pythagorean Theorem, obtaining + . Dividing by gives the following formula for the unknown rate:

= .

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