Exponential Models with Differential Equations

#7.8 Exponential Models with Differential Equations

Think of a video going viral and how quickly the number of views it gets increases over time. Or think about the last time who heard a rumor and how fast it spread around. These are real-world scenarios which can be explained through exponential models, a key concept in calculus that helps us understand things that grow or shrink really fast.

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An exponential graph which shows a relationship between popular video “Nyan Cat” and the number of views over time. A number which is steadily increasing.

Image Courtesy of Desmos

#🤔 What is a Differential Equation?

At the heart of exponential models are differential equations.

Imagine these equations as a mystery where you know how fast something is changing (like the speed of a rumor spreading), and you're trying to figure out the entire story (how many people will hear it over time). In calculus, these equations help us map out scenarios of rapid growth or decline.

The following equation is your go-to formula here.

$$dy/dt = ky$$

Rest assured, it's less complex than it looks.

• $dy/dt$ represents the rate of change of something over time, like the number of people watching a video or hearing a rumor.
• The $k$ in the equation is a constant that gives us the rate and nature of this change.
• A positive $k$ means things are growing or spreading (like more people watching the video).
• A negative $k$ means things are declining or fading away.

#✍🏾 Solving the Differential Equation

When the differential equation is solved, it gives us an important equation. Solve the steps below to find out what it is!

1. Start with the Differential Equation: We begin with the equation $dy/dt = ky$, as mentioned above.
2. Separate the Variables: To solve this equation, we first separate the variables $y$ and $t$ . We rearrange the equation to get all the $y$ terms on one side and the $t$ terms on the other: $(1/y)(dy) = k\ dt$ .
3. Integrate Both Sides: Next, we integrate both sides of the equation. The integral of $1/y ; dy$ is $ln |y|$, and the integral of $k\ dt$ is $kt$. After integrating, we have: $ln|y| = ; kt+C$ where $C$ is the constant of integration.
4. Solve for $y$: To make this equation more manageable, we exponentiate both sides to get rid of the natural logarithm. This gives us: $e^{ln|y|} = e^{kt+C}$. After simplifying, we get: $|y| = e^C \cdot e^{kt}$.
5. Determine the Constant $C$: The constant $C$ can be determined using the initial condition of the problem. Suppose we know the value of $y$ at $t = 0$, which we call $y_0$. Plugging these values into our equation gives us: $|y_0| = e^C \cdot e^0$. After we simplify this, we find that $e^C = |y_0|$. Therefore, our equation becomes: $|y| = |y_o| * e^{kt}$.
6. Final Exponential Model: Assuming y is always positive, we can remove the absolute value signs. Our final model is: $y = y_o * e^{kt}$.

This is the exponential model that predicts future patterns. Here, $y_0$ is the starting value, $e$ is Euler's number (important in continuous growth/decay situations), and $kt$ represents how the rate of change and time interact to evolve the initial situation.

#🎯 Walking Through an Exponential Model

Let’s try a practice problem to really show off your newly learned skills.

A small town has a population of 2,000 people. Due to new job opportunities, the population is increasing at a rate proportional to its current size. After 3 years, the population has grown to 3,000 people. Assuming this trend continues, what will the population be after 10 years?

Let’s first identify the information that was given to us:

• Initial population ($y_0$): 2,000 people

• Population after 3 years: 3,000 people

• Time to reach this population: 3 years

• The goal is to find the population after 10 years. Now we can write down the exponential growth equation and substitute any values we know.

• The population growth can be modeled by the equation $y = y_0 \cdot e^{kt}$.

• Initially, $y = 2000$ when $t = 0$. Use the information to find $k$:

• We know that after 3 years $(t = 3)$, $y = 3000$. Plugging these into our equation gives us: $3000 = 2000 \cdot e^{3k}$

• To find k, we first divide both sides by 2000: $$\frac {3000}{2000}=e^{3k}$$

$$1.5=e^{3k}$$

Now, we take the natural logarithm of both sides to solve for k:

$$ln(1.5) = 3k$$

$$k = \frac {ln(1.5)}{3}$$

$$k = \frac {ln(1.5)}{3} \approx .13516$$

Amazing! We solved for $k$. Now, we can apply the value to the problem and predict the population after 10 years.

• Use t = 10 in the equation $y = 2000 \cdot e^{kt}$, making it $y=2000\cdot e^{0.13516×10}$.
• Calculate the final value: $y = 2000 \cdot e^{1.3516}\rightarrow y \approx2000\cdot3.8637\rightarrow y\approx 7727.4$ The population of the town is expected to be approximately 7,727 people after 10 years. Great work!

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#✏️ Practice with Exponential Models

Here are two problems you can try on your own!

1. The rate at which the drug leaves the bloodstream is proportional to the amount in the blood stream. A dose of 200 mg of a drug is administered to a patient. After 3 hours, approximately 127.3 mg remain. The amount of the drug, in milligrams, in the person’s bloodstream after $t$ hours is given by $A(t)$. Write an equation for $A(t)$.
2. Bacteria in a certain culture increase at a rate proportional to the number present. If the number of bacteria doubles in five hours, in how many hours will the number of bacteria quadruple?

Try your best before you scroll and take a look at the way we completed them.

#📌 Solution For Example 1

Let’s again identify the given, relevant information:

• We know that $y_0=200$
• We know that after three hours ($t=3$), $y=127.3$
• Our goal is to find the equation that models this situation Write down the exponential growth equation! The change in the amount of drug can be modeled by the equation $y = y_0 \cdot e^{kt}$. Initially, $y = 200$ when $t = 0$.

Use this information to find $k$:

We know that after 3 hours $(t = 3)$, $y = 127.3$. Plugging these into our equation gives us:

$$127.3 = 200 \cdot e^{3k}$$

To find k, we first divide both sides by 200:

$$\frac {127.3}{200}=e^{3k}$$

$$0.6365=e^{3k}$$

Now, we take the natural logarithm of both sides to solve for k:

$$ln(.6365) = 3k$$

$$k = \frac {ln(.6365)}{3}$$

$$k \approx -0.151$$

Amazing! Now that we have a value for k, we can apply it to our problem. We know that $dy/dt=kA$, and that $A(0)=200$. Thus, we find that $A(t)=200\cdot e^{-0.151t}$.

#📌 Solution For Example 2

This problem seems a little tricky at first glance, since we aren’t given any actual numbers. But, we are given ratios, and we can use that to make up our own numbers.

• We’ll assign the initial value of the bacteria to 100, so $y_0=100$ when $t=0$.
• That means that after 5 hours ($t=5$), $y=200$
• We want to find out how long it will take for the population to reach 400. Writing down the exponential growth equation, we know that the population growth can be modeled by the equation $y = y_0 \cdot e^{kt}$. Initially, $y = 100$ when $t = 0$.

Time to find $k$!

We know that after 5 hours $(t = 5)$, $y = 200$. Plugging these into our equation gives us:

$$200 = 100 \cdot e^{5k}$$

$$\frac {200}{100}=e^{5k}$$

$$2=e^{5k}$$

Now, we take the natural logarithm of both sides to solve for k:

$$ln(2)=5k$$

$$k = \frac {ln(2)}{5}$$

$$k\approx 0.139$$

Almost there! We want to find out how long it will take the population to reach 400, we can plug this information in and solve for $t$:

$$400=100\cdot e^{0.139t}$$

$$4=e^{0.139t}$$

Then, take the natural log of both sides to cancel the $e$:

$$ln(4)=0.139t$$

Finally, just divide by 0.139 and compute the value for $t$!

$$\frac{ln(4)}{0.139}=t\approx9.97 \ \text{hours}$$

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#🥳 Conclusion

Exponential models give us a fantastic tool to understand and predict scenarios of rapid change in our daily lives, from social media trends to the spread of information. By mastering these concepts in AP Calculus, you gain not just the ability to solve math problems, but also a deeper insight into the dynamic world around you.