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Ripple on a Calm Lake

As we move into applications of derivatives, the first applications we will see is that of related rates. These examples are difficult for students to get a grasp on, so we will take the time to go through and see why they are difficult, what can help the process and finally find a solution to the example we are given.

Our process

The largest difficulty students have with related rates problems is that they are word problems. At the point in time you are taking Calculus, it is difficult to see the connection between the equations we see so often in class and the ideas and topics we see in life. In order to help facilitate learning to make these connections, I generally give the following outline for approaching related rates problems.

  1. Read the problem carefully.
    • While this may seem evident, we too often read through the problem quickly and try to guess what we are supposed to find or what is given to us.
    • Instead, read the problem multiple times, piece by piece and keep careful track of everything that is being given to you or asked.
    • Drawing a picture to help visualize the situation can be extremely helpful.
    • When drawing a picture or writing details, label things so that you understand.
  2. Once you have read the problem and written down everything that is given, you now need to define any variables.
    • Anything that will change needs to be assigned a variable.
    • Using \(x\) and \(y\) is not a great choice for variables. Instead, use letters or symbols that remind you of what the variable represents.
  3. For all the variables, write down the rates given for them or ask about them. That is, you should find the derivative of any variable with respect to some other variable as described in the problem.
  4. Find an equation relating the variables, that is an equation that has the variables in it.
  5. Implicitly differentiate the equation you came up with in order to get an equation involving rates.
  6. Substitute in what you know and solve for what you are looking for based on the information given for the problem.
  7. Write your answer in terms of the question.

Note that what we are doing is taking an example of some situation. We then convert that example into mathematical notation, thus giving a mathematical model. If is often helpful to think of this as translating into another language. We do this so that we can use the mathematical notation to facilitate the manipulation of the information in such a way that we will arrive at a solution to the model. However, getting a solution to the model isn’t enough. Instead, we have to remember to answer the original question, that is, we have to translate the model back into English.

Make sure to follow the steps as we go through the example.

Find how quickly a ripple is growing

Suppose that you are standing on the shore of a calm lake on a beautiful spring day. Without the wind blowing, you look out onto the lake as it sits completely still. You cast a fishing line out into the lake as you try to enjoy a day of fishing. As your lure hits the water, it causes a ripple to expand from the point of impact, creating a circle. If the radius of the ripple is increasing by 3 inches per second, how fast is the area of the ripple growing when the radius is 24 inches?

Read problem

As we read through the problem a few times, we want to pay particular attention to anything that will be helpful. While the first three sentences help to set a scene, we don’t get to anything that will help us with the mathematical model until the fourth sentence. As we find this, we will keep track by drawing pictures along the way.

  • The ripple is a circle.

    circle

  • The ripple is increasing by 3 inches per second.

    radius movement

  • We want to know how fast the area of the ripple is growing.
    looking for
  • At the current point in time, the radius is 24 inches.
    radius

Variables

We now have to find all of our variables. As we do this, we want to think about what is changing in this problem. Here, we note that the size of the ripple be is increasing. Therefore, the radius of the ripple is changing. We need to assign this a variable, so \(r=\)radius would be a good descriptive choice. Also, the area of the ripple is changing, we therefore assign \(A=\)area. One further not is that these are changing with respect to time, therefore we need to assign \(t=\)time.

I would like to note that there may be other things changing in the problem. For example, the water or air temperature, the humidity, or our happiness level may be changing. One or more of these may helpful to know, but we don’t need to worry about them at this point. That is, for the time being, we only want to include variables that are mentioned within the problem. Hopefully this will be enough to solve our model, but, if not, we will find other variables later.

If is also helpful to note that, currently, \(r=24\). It may be helpful to find the current area or time, but we will refrain from doing so until we know that we will need this. We will, however, update our picture to include the assigned variables.

variables

Rates

Now that we know our variables, we want to find write out the rates that we know.

  • We know that the radius is changing at a rate of \(3\) inches per second.
  • This tells us that the radius is changing with respect to time.
  • That is the rate of change of radius with respect to time is \(3\).
  • Therefore, \(\frac{dr}{dt}=\pm 3\).

Above, I have \(\pm 3\), because it is worth taking an extra step to determine if the derivative is positive or negative. For this, we recall that the derivative is positive if the function is increasing and negative if the function is decreasing. Since the radius is increasing, this would tell us that \(\frac{dr}{dt}=3\).

In addition to \(\frac{dr}{dt}\), we also see that we want to find the rate of change of area with respect to time. That is, we need to find \(\frac{dA}{dt}\). Since this is what we want to find with our model, we can denote this as \(\frac{dA}{dt}=?\) where we use to \(?\) to show that this is the goal. Again, we will update our picture to show the rates.

rates

Equation

We now have to find an equation relating the variables in our problem. Since we have radius and area of a circle, we would like to find an equation with each of these in it. This is often the most difficult part of a related rates problem. This is because the equation you use is different for nearly every example. Therefore, you have to use your background knowledge and previous experience to figure this out.

In this case, because there is a circle, we would try things like circumference or area. Here we get that \(A=\pi r^{2}\) is the area of a circle. Since both \(A\) and \(r\) are in this equation, it is the one we want to use.

Implicitly Differentiate

Now that we have an equation, we can implicitly differentiate. Recall that we want to find \(\frac{dA}{dt}\). Therefore, we will take the derivative of both sides of our equation with respect to \(t\). This will give us
\begin{align*}
\frac{d}{dt}A=\frac{d}{dt}\pi r^{2}.
\end{align*}

Now, note that since \(A\) is a function of \(t\), the best we can do for \(\frac{d}{dt}A\) is to say that it is the derivative of \(A\) with respect to \(t\), that is \(\frac{dA}{dt}\). When taking the derivative of \(\pi r^{2}\) we can factor out the \(\pi\), but we have to remember that \(r\) is a function of \(t\). We will then have to use a chain rule with the outside function being \(i^{2}\) and the inside function being \(r\). Using this, we find

\begin{align*}
\frac{d}{dt}A&=\frac{d}{dt}\pi r^{2} \\
\frac{dA}{dt}&=2\pi r \frac{dr}{dt}.
\end{align*}

Substitute

Note that by implicitly differentiating, we have found an equation involving \(\frac{dA}{dt}\) and \(\frac{dr}{dt}\). This was the goal of using this model. Now that we have done this, we can now substitute in what we know and solve for what we want. We then have, at the current point in time,
\begin{align*}
\frac{dA}{dt}&=2\pi r \frac{dr}{dt} \\
\frac{dA}{dt}&=2 \pi (24) (3) \\
\frac{dA}{dt}&=144 \pi \\
\frac{dA}{dt}& \approx 452.2.
\end{align*}

Solution

Here we note that we have found the solution to the model we created of the problem, but we haven’t given the solution to the original question. Therefore, we have to convert this back into the original question. In order to do this, we have to determine what is going on.

  • Note \(\frac{dA}{dt}=452.2\)
  • Since radius is in inches, and \(A= \pi r^{2}\), \(A=\pi (inches)^{2}= (inches)^{2}\) since \(\pi\) has no units.
  • We furthermore recall that time is in seconds.
  • Therefore, \(\frac{dA}{dt}\) is inches squared divided by seconds, that is square inches per second.
  • Also, \(\frac{dA}{dt}\) is positive, so the area is increasing.

If we combine all of this together, we find that the area is increasing at a rate of 452.2 square inches per second. Now, to write that in more colloquial terms, we get “The area of the ripple is getting bigger by 452.2 square inches per second.” We can also give a more complete answer and say that, “As we cast our lure out onto the calm lake it created a ripple. The ripple got bigger over time, and, when the radius of the ripple was 24 inches, the area of the ripple was getting bigger by 452.2 square inches per second.”

Conclusion

Note that, as is a common theme in Calculus, we really tried to break this problem into smaller pieces. We focused on one thing at a time and combined our work in order to get answer a more complicated problem. While each step will look different in each related rate problem, we can use the general outline used here in any of the situations that will arise. I hope seeing this helped you get more comfortable with relates rates. If it did, make sure to like the post and share on Social Media. You can also find more Calculus help here.

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