Modeling Situations with Differential Equations

#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! ⬇️

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#⛷️ 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 $\frac{dy}{dx} = 5x$. In this equation, $\frac{dy}{dx}$ represents the derivative of the function $y$ with respect to $x$. The equation is saying that the rate of change of $y$ with respect to $x$ is equal to 5 times $x$.

#🧠 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 $a$ is proportional to $b$, then $a = kb$, where $k$ is a constant.
2. Inversely Proportionality: If $a$ is inversely proportional to $b$, then $a=\frac{k}{b}$, where $k$ is a constant.

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#✏️ 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 $S$ with respect to $t$ is inversely proportional to $x$.

Question 2: The rate of change of $A$ with respect to $t$ is proportional to the product of $B$ and $C$.

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 $\frac{dS}{dt} = \frac{k}{x}$ .
2. Again, this problem has a keyword: (directly) proportional. So, the answer is $\frac{dA}{dt} = kBC$.

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 $k$.
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, $F$, with respect to time that is inversely proportional to $D$, 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 $k$ is a constant of proportionality.

$$\frac{dF}{dt} = \frac{k}{D}$$

🔌 Step 2: Substitute the values and solve for $k$.

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

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

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

🧩 Step 3: Form the differential equation!

So, the differential equation representing the given relationship is:

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

#Real-World Scenarios: Example 2

The rate of change of the volume, $V(t)$, of a right rectangular prism with respect to time (in seconds) is increasing at a rate proportional to the product of its length $L$, width $W$, and height $H$. 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 $\frac{dV}{dt}$ be the rate of change of the volume with respect to time. And noting the keyword (directly) proportional, we describe the relationship as…

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

🔌 Step 2: Substitute the values and solve for $k$.

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 ($\frac{dV}{dt} = 3$), we can substitute these values into the equation:

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

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

🧩 Step 3: Form the differential equation!

So, the differential equation representing the given relationship is:

$$\frac{dV}{dt} = \frac{1}{80}LWH$$

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#🌟 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. ➡️