Caution: In order to solve related-rates problems, you really need to know how to differentiate implicitly. See the page on implicit differentiation to learn how.
Well, let's use calculus to solve some problems, shall we? There is a class of problems in one-variable called related rates problems. As the name suggests, the rate of one thing is related through some function to the rate of change of another. We can take advantage of that relationship and the fact that calculus is the mathematics of change to solve a whole bunch of new problems.
The general approach to a related-rates problem will be to identify the two things that are changing, to find some sort of relationship between them – often it's geometric, to equate them and then take the derivative (implicit) with respect to time of both sides.
The best way to learn related-rates problems is by doing examples, so here are a few
This is a classic problem in calculus. What happens at the top and bottom of a ladder that is sliding down the wall (and thus away from the wall at the bottom) is a little surprising: While the bottom of the ladder travels at a constant speed, the top accelerates downward. Weird.
The problem: A 6-meter ladder leans against a wall. The bottom of the ladder is 2 m from the wall at time t=0, and it slides away from the wall at the rate of 0.5 m/s. Find the downward speed of the top of the ladder at any time, t, and at t = 1 second.
Notice that while the bottom of the ladder traces out constant intervals, the intervals against the wall increase between each equal time step. Here's the basic geometry:
We are given that
where t = time. We can relate x and h through the Pythagorean triangle (left):
Now we take the derivative of each side with respect to time:
Note that the derivative of the square of the ladder height is zero so height of the ladder doesn't matter in this equation. Now rearrange to isolate dh/dt:
Finally, substitute for dx/dt to get the final related-rates equation:
Now to apply our equation at time t = 1s, we have to know x and h at t = 1. The bottom moves at 0.5 ms-1, so after 1s it is 0.5 meter away. Now when the bottom is 1m away, the Pythagorean relationship tells us that h is √(62 - 2.52) = 5.45 m, and the rate of drop of the top of the ladder is (-0.5)(2.5/5.45) = -0.23 ms-1. The negative sign is a result of our coordinate system, in which height was measured upward from the ground. The graph of the solution below shows the velocity of the top of the ladder as a function of the distance of the base from the wall. The curve in the graph implies acceleration.
The problem: A conical tank with the dimensions shown ( → ) is filled with liquid at a rate of 1.5 m3·min-1. At what rate is the water level rising when it passes a height of 5 meters? Sketch a graph of the rate as a function of time.
What we know and don't know:
This problem is typical of a great many others in two ways. First, two rates are related (duh). In this case the rate of filling (or draining) is proportional in some way to the change in height of the liquid, measured as h, as shown. Second, while at first glance it's a two-variable problem [V = f(r, h)], those two variables are related by an equation, thus one variable can be eliminated to reduce the problem to a single variable one.
The ratio of the radius of the cone to its height is the same at any height, by similar triangles. We use that relationship (first equation → ) to eliminate the variable r from our problem. We could also use it to eliminate h, but we're interested in height of the liquid as a function of time. Here we go.
Then we take the time derivative of both sides, relating the rate of volume change to the rate of change of the height.
Implicit differentiation and the chain rule leads us to an expression for f'(h) that can be used to find the rate of change of height at any time.
At a height of 5 meters, the rate is
The ratio of radius to height of the water cone is equivalent to that ratio of the whole cone, which we know:
Plugging our new expression for the radius into the volume formula gives
Now we take the derivative of both sides with respect to time:
We know dV/dt, so that reduces to:
where we've taken the implicit derivative with respect to time on the right. Then we solve for dh/dt:
and the result is:
The graphs below show the height of liquid as a function of time for this problem, and its derivative with respect to time. Note that each is drawn on its own set of axes. Even though it's often done, it is incorrect to superimpose the curves on the same graph; they have different y-axes. Make sure you understand the relationship between the height vs. time curve and its derivative.
Let's say you track the progress of a rising rocket using a clinometer, an instrument for measuring the angle between the line of sight (to the nose of the rocket here) and the level ground. The angle is labeled θ in the picture → .
The problem: At a given moment, the angle between the nose of the rocket and the ground is 39˚ and is changing at a rate of 25˚·min-1. Calculate the velocity of the rocket at that instant.
Here's what we know and don't know:
Here's a picture of the problem and its geometry. We will assume that the rocket takes off at a right angle to the ground.
The first trick is to find a relationship between θ and y, and we find it in trigonometry: tan(θ) = y/5. From there we differentiate both sides, plug in what we know and solve for the vertical velocity, dy/dt.
Then it's just a matter of calculating that velocity when θ = 39˚:
By now these related rates problems should show a familiar pattern. I think they're fun because there's a definite method, but we still need to find our way through the specific algebra of each kind of problem.
The key relationship between y and θ is
Taking the derivative of both sides with respect to time,
Now we solve for dy/dt to get:
Finally, plugging in our target angle of 39˚ gives us the answer we were seeking:
Not available yet – Stay tuned.
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