Related rate problems were the bane of my existence back when I took Calculus. So I will be going over the strategy for approaching these problems, followed by a couple examples. If you have a specific example in mind you want me to go through, feel free to message me!
Hopefully this guide will be able to help you understand it better!
So what exactly are related rates?
Basically, given two or more qualities, the change in one quality will result in the change of another quality.
STEP 1: Find out exactly what the question is asking
- What do you already know?
- What are you trying to find out?
- Drawing a diagram of the situation if it is helps you make sense of the information.
STEP 2: Find the relationship between the qualities given in the problem
STEP 3: Take the derivative of both sides of the equation relative to time
- This will involve using the chain rule, as illustrated in Example 1.
STEP 4: Plug and chug
- You might need to use the information given by the problem to help find the vales of variables in the relationship, as seen in Example 1 and 2.
Time for examples!
The sides of a square are increasing at a rate of 2ft/s. What is the area of the square when the area is 64ft2?
STEP 1: Letting s equal the sides of a square, and a equal the area of a square, we know the rate of change for the sides. We are trying to find the rate of change for the area when the area of the square is 642. So…
STEP 2: Find the relationship between the side and area of the square. Of course, the area of a square is the side of a square squared.
STEP 3: Take the derivative of both sides relative to time. This gives us…
STEP 4: Because the area at which we want to find da/dt is 642, s must equal 8. Plug s=8 and a=64 into the derivative equation above and find da/dt
This means that when the area of the square is 64ft2 and the sides of the square are increasing at 2ft/s, the area of the square is increasing at 32ft2/s
Wait, chain rule??
Let’s take a step back and see how we got the derivative. Recall the formula for using the chain rule.
Using the equation of example 1 as an example, let’s take the derivative of…
Using the chain rule to find ds/dt…
Therefore, we get the differential equation…
Let’s see the whole thing again through another example.
Find dz/dt of the following equation when x = 1 and y = 3. dx/dt = 1 and dy/dt = 2.
STEP 1: Because we already know the relationship, we can skip straight to the derivative. Using the chain rule…
STEP 2: Find z, then find dz/dt.
So dz/dt = 1 when x = 1 and y = 3.
THAT’S THE GIST OF IT!
Click here for examples of more complicated related rate problems, including ones involving similar triangles and distance between two points.
Alternatively, click here for worksheets and extra practice! Solutions to the worksheets will be linked on respective worksheets.