Related Rates

Secant lines \u0026 average rate of change | Derivatives introduction | AP Calculus AB | Khan Academy
Secant lines \u0026 average rate of change | Derivatives introduction | AP Calculus AB | Khan Academy

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Related Rates

In this section, we will learn how to solve problems about related rates – these are questions in which there are two or more related variables that are both changing with respect to time.

There are many different applications of this, so I’ll walk you through several different types. They can all be solved by following several general steps:

Example 1 – Ripples in a Pool

A rock is dropped into a pool of water, causing ripples to form in expanding outward circles. The radius r of a ripple is increasing at a rate of 1 foot per second. When the radius is 6 feet, at what rate is the area A of the water inside the ripple changing?

As a note, remember that a derivative is a rate of change. If we have a variable, the derivative of that variable is the rate that it is changing. For example, in this problem, we have the variable r; r is the radius of the ripple. dr/dt is the rate at which the ripple is changing – in this example, it is increasing at 1 foot per second. The same with A; A is the area, while dA/dt is the rate at which the area is changing.

To do this problem, we first need to identify our variables – r is given, dr/dt is given, and A is to be determined. Here is a picture of this scenario:

Now that we have identified our variables and drawn a sketch, let’s find an equation that involves both r and A of a circle. Since r is a variable, dr/dt will be included once the equation is differentiated. These variables can be related by the equation for the area of a circle,

A = π r2

Differentiation with respect to t will obtain the related rate equation that we need to plug our information into:

When the radius is 6 feet, the area is changing at a rate of 12π ft2/second, which is about 37.7 ft2/second

Example 2 – Ripples in a Pool

Yes, we’re going to do another problem with ripples in a pool; however, this time we will be solving for something other than the rate at which the area is expanding. This is to show how we can start with the same formula, but solve for a different rate of change.

A rock is dropped into a pool of water, creating ripples which move outward from it. One of these ripples creates a circle with an area increasing at a rate of 30 ft2/sec. When the area is 25π square feet, at what rate is the radius expanding?

As before, we need to make a sketch of the problem, listing and labeling our knowns and unknowns:

Now, we need an equation that relates A and r of a circle – good thing we already know that from the last problem: A = π r2! Even though we’ve already done it, let’s differentiate it again, just for practice:

This was fine for the last problem, but notice that we are missing a variable; we have A and dA/dt, but we’re missing r. Fortunately, we have a formula which relates the two right in front of us; since we know the A at the specific time that we need, we can use it to solve for r, using A.

Using A = 25π, the radius is 5. Now we have all the information we need to solve for dr/dt!

When the area is 25π ft2, the radius is expanding at 3/π, which is about .95, ft/s.

Example 3 – An Inflating Balloon

A spherical balloon is being inflated at a rate of 10 cubic feet per minute. Find how fast the radius is expanding when the radius is 1 foot.

We need to name our knowns and unknowns, and label them on a drawing:

Now we need an equation which relates A and r for a sphere; the volume of a sphere is

Differentiating this equation, we get:

Finally, we can solve for dr/dt and plug in our variables:

When the radius is 1 foot, it is expanding at a rate of approximately .78 feet per minute.

Example 4 – An Airplane Overhead

An observer sees an airplane at a height of 5 miles moving toward him. If the distance s between him and the airplane is decreasing at a rate of 300 miles per hour when s is 12 miles, what is the speed of the airplane?

With a problem such as this, a drawing is usually given, since the wording can be a bit confusing. Still, we need to list our knowns and unknowns, and also assign some variables of our own:

Remember that since x in the above image is the horizontal distance, dx/dt is the rate at which the distance is changing, or the speed. Since we need the horizontal speed, dx/dt, a right triangle gives us the variable x to work with. First, let’s use the Pythagorean Theorem to solve for x:

Now we have all of our variables, and we also have an equation which relates s and x, so let’s differentiate it and solve for the plane’s speed, dx/dt:

We got a negative number for the velocity; this is just because of the direction the airplane is moving, based on how we defined our variables in the beginning. Since we are only finding the speed, we don’t need to worry about the negative sign; when s is 12 miles, and s is decreasing at 300 miles per hour, the speed of the airplane is 330.3 miles per hour.

Example 6 – A Changing Angle of Elevation

A camera is on the ground, filming a rocket launch. The rocket is rising according to the position function s = 30t2, where y is measured in feet and t in seconds. Find the rate of the change of the camera’s angle at 15 seconds after the rocket initially launches. The camera is 1000 feet away from the rocket.

First, let’s list our variables and label them on a drawing; as with the last example, we’ll need to use a right triangle to help us solve it:

We need to solve for s. Since s is the height of the rocket, we can use the position function that we were given for it, and evaluate at 15 seconds:

We also need ds/dt; this is the derivative of s, which is 60t.

Now, using this information, we need to find the change in the angle of elevation of the camera when t = 15 and s = 6750. We can relate these using the trigonometric equation in our picture above:

We are almost done, however we need to use our right triangle, along with some new terms, to find an expression for the cosine of our angle:

Finally, we can substitute cosine for our new expression, and evaluate the problem:

The angle of the camera at time t = 15 seconds is changing at approximately .02 radians per second.

Hello! My name is Alex, and I am the creator of I’m not a genius or a math guru; in fact, I struggled with it for several years before becoming proficient. Eventually, I passed the AP Calculus exam in high school, and Calculus II and III in college; because I struggled with it, I understand how necessary clear, concise explanations are. That is why I created this site, and am working on a book to go along with it – to help you cope with calculus!

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