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7.1 Modeling Situations with Differential Equations

2 min readjune 18, 2024

Zaina Siddiqi

Zaina Siddiqi

Zaina Siddiqi

Zaina Siddiqi

7.1 Modeling Situations with Differential Equations

Welcome back to AP Calculus with Fiveable! We are now diving into one of the most valuable concepts in calculus to simulate real-life scenarios. Our focus in this section is modeling situations with differential equations. These equations involve rates of change and help us understand how a quantity changes with respect to another. Let’s get into it! ⬇️


⛷️ Differential Equations

Differential equations involve derivatives and represent the relationship between a function and its rate of change. They help us understand how functions change with respect to the independent variable in real-world scenarios.

For example, a differential equation can look like dydx=5x\frac{dy}{dx} = 5x. In this equation, dydx\frac{dy}{dx} represents the derivative of the function yy with respect to xx. The equation is saying that the rate of change of yy with respect to xx is equal to 5 times xx.

🧠 Understanding Proportionality

Proportionality, which is the concept that two quantities vary in a consistent way with respect to each other, forms the basis of many differential equations! You can have two values be directly proportional to each other, or indirectly proportional to each other.

  1. Directly Proportionality: If aa is proportional to bb, then a=kba = kb, where kk is a constant.
  2. Inversely Proportionality: If aa is inversely proportional to bb, then a=kba=\frac{k}{b}, where kk is a constant.

✏️ Working With Differential Equations

Let’s go through some of the common problems you’ll see!

🔍 Describing A Relationship

For the following two phrases, write the corresponding differential equation!

Question 1: The rate of change of SS with respect to tt is inversely proportional to xx.

Question 2: The rate of change of AA with respect to tt is proportional to the product of BB and CC.

Think about the basics of differential equations and whether there is direct or inverse proportionality.

⚡ Here are the answers:

  1. Note that the problem has a keyword: inversely proportional! So, the answer is dSdt=kx\frac{dS}{dt} = \frac{k}{x} .
  2. Again, this problem has a keyword: (directly) proportional. So, the answer is dAdt=kBC\frac{dA}{dt} = kBC.

Now let’s take a look at how to take this concept a few steps further and apply them to real-life scenarios.

🌎 Modeling Real-World Scenarios

When taking a look at the following questions, it may be helpful to follow these three steps:

  1. 👀 Identify the keyword to describe the relationship.
  2. 🔌 Substitute the values and solve for kk.
  3. 🧩 Form the differential equation!

Real-World Scenarios: Example 1

Mrs. May is an amateur singer. Her voice change can be modeled by the rate of change of frequency, FF, with respect to time that is inversely proportional to DD, the decibel level of her voice. If the frequency changes by 4 vibrations per second when she is projecting at 60 decibels, find the differential equation that describes this relationship.

👀 Step 1: Identify the keyword to describe the relationship.

Like we did before, we note the keyword inversely proportional and describe the relationship as the following, where kk is a constant of proportionality.

dFdt=kD\frac{dF}{dt} = \frac{k}{D}

🔌 Step 2: Substitute the values and solve for kk.

Given the frequency changes by 4 vibrations per second (dFdt=4\frac{dF}{dt} = 4) when Mrs. May is projecting at 60 decibels (D=60)D = 60), we can substitute these values into the equation…

4=k604 = \frac{k}{60}

Solving for kk, we get k=240.k=240.

🧩 Step 3: Form the differential equation!

So, the differential equation representing the given relationship is:

dFdt=240D\frac{dF}{dt} = \frac{240}{D}

Real-World Scenarios: Example 2

The rate of change of the volume, V(t)V(t), of a right rectangular prism with respect to time (in seconds) is increasing at a rate proportional to the product of its length LL, width WW, and height HH. Find the differential equation if the prism has a length of 10 units, width of 4 units, height of 6 units, and the volume is changing by 3 cubic units per second.

👀 Step 1: Identify the keyword to describe the relationship.

We let dVdt\frac{dV}{dt} be the rate of change of the volume with respect to time. And noting the keyword (directly) proportional, we describe the relationship as…

dVdt=kLWH\frac{dV}{dt} = kLWH

🔌 Step 2: Substitute the values and solve for kk.

Given that the prism has a length of 10 units, a width of 4 units, a height of 6 units, and the volume is changing by 3 cubic units per second (dVdt=3\frac{dV}{dt} = 3), we can substitute these values into the equation:

3=k10463 = k \cdot 10 \cdot 4 \cdot 6

Solving for kk, we get k=180k=\frac{1}{80}.

🧩 Step 3: Form the differential equation!

So, the differential equation representing the given relationship is:

dVdt=180LWH\frac{dV}{dt} = \frac{1}{80}LWH

🌟 Closing

Great work! You made it to the end of the first key topic in unit seven. In the next key topic, we’ll get into verifying solutions for differential equations. ➡️