--- title: "AP Calculus 7.8: Exponential Models with Differential Equations" description: "Review AP Calc 7.8 exponential models with differential equations, including dy/dt = ky, y = y0e^(kt), separation of variables, k, growth, and decay." canonical: "https://fiveable.me/ap-calc/unit-7/exponential-models-with-differential-equations/study-guide/fzn9urmywhPSE4KUxVcq" type: "study-guide" subject: "AP Calculus AB/BC" unit: "Unit 7 – Differential Equations" lastUpdated: "2026-06-11" --- # AP Calculus 7.8: Exponential Models with Differential Equations ## Summary Review AP Calc 7.8 exponential models with differential equations, including dy/dt = ky, y = y0e^(kt), separation of variables, k, growth, and decay. ## Guide Exponential models come from the [differential equation](/ap-calc/unit-7/verifying-solutions-for-differential-equations/study-guide/s2nX7AhIBDxGIWwlL82x "fv-autolink") $\frac{dy}{dt}=ky$, which says the [rate of change](/ap-calc/key-terms/rate-of-change "fv-autolink") of a quantity is proportional to the size of the quantity. Solving it by separation of variables gives $y=y_0e^{kt}$, where $y_0$ is the starting amount and $k$ is the growth ($k>0$) or decay ($k<0$) constant. For AP Calculus, interpret $k$, $y_0$, and the units in context. ## Why This Matters for the AP Calculus Exam This topic connects the differential equation skills from earlier in [Unit 7](/ap-calc/unit-7 "fv-autolink") (separation of variables and [initial conditions](/ap-calc/key-terms/initial-condition "fv-autolink")) to real-world growth and decay. On the AP Calculus exam you may need to: - Translate a verbal statement like "the rate of change is proportional to the amount" into $\frac{dy}{dt} = ky$. - Solve a separable equation and apply an initial condition to find the [particular solution](/ap-calc/key-terms/particular-solution "fv-autolink"). - Use the model $y = y_0 e^{kt}$ to predict future values or solve for time. - Interpret what the variables and the constant $k$ mean in context. You can recognize the exponential form and use it directly without re-deriving it every time, but you should still be ready to show the separation-of-variables steps when a problem asks for them. Writing clear setup, correct notation, and labeled units makes your work easier to follow and check. ## Key Takeaways - The phrase "rate of change is proportional to the size of the quantity" always means $\frac{dy}{dt} = ky$. - The [general solution](/ap-calc/key-terms/general-solution "fv-autolink") is $y = y_0 e^{kt}$, where $y_0 = y(0)$ is the initial value. - A positive $k$ models growth; a negative $k$ models decay. - Find $k$ by plugging a second known data point into $y = y_0 e^{kt}$ and using natural logs. - Keep the [constant of integration](/ap-calc/key-terms/constant-of-integration "fv-autolink") when you integrate, and separate variables before integrating. - Track units on $t$ and $k$ so your final answer makes sense in context. ## What Is a Differential Equation Here? A differential equation relates a quantity to its own rate of change. In exponential models, you know how fast something is changing relative to how much is present, and you want the full [function](/ap-calc/unit-1/defining-continuity-at-point/study-guide/JbsR9iQfAzCznNOCG6JK "fv-autolink") that describes the quantity over time. The model you use is: $$ \frac{dy}{dt} = ky $$ - $\frac{dy}{dt}$ is the rate of change of the quantity over time (people, milligrams, bacteria, etc.). - $k$ is the proportionality constant that sets the rate and direction of change. - $k > 0$ means the quantity is growing. - $k < 0$ means the quantity is decaying. ### Solving the Differential Equation You can derive $y = y_0 e^{kt}$ from $\frac{dy}{dt} = ky$ using separation of variables. 1. **Start with the differential equation:** $\frac{dy}{dt} = ky$. 2. **Separate the variables:** Move all $y$ terms to one side and all $t$ terms to the other: $\frac{1}{y}\,dy = k\,dt$. 3. **Integrate both sides:** The integral of $\frac{1}{y}\,dy$ is $\ln|y|$, and the integral of $k\,dt$ is $kt$. This gives $\ln|y| = kt + C$, where $C$ is the constant of integration. 4. **Solve for $y$:** Exponentiate both sides to undo the natural log: $e^{\ln|y|} = e^{kt+C}$, which simplifies to $|y| = e^C \cdot e^{kt}$. 5. **Use the initial condition:** Suppose $y = y_0$ when $t = 0$. Plugging in gives $|y_0| = e^C \cdot e^0$, so $e^C = |y_0|$. Then $|y| = |y_0| \cdot e^{kt}$. 6. **Final exponential model:** Assuming $y$ stays positive, drop the absolute values to get $y = y_0 e^{kt}$. Here $y_0$ is the starting value, $e$ is Euler's number, and $kt$ combines the rate constant and time. ## Worked Example: Population Growth A small town has a population of 2,000 people. The population increases at a rate proportional to its current size. After 3 years, the population grows to 3,000 people. What will the population be after 10 years? Identify the given information: - Initial population $y_0 = 2000$ - Population after 3 years: $3000$ - Goal: find the population after 10 years. Set up the model $y = y_0 e^{kt}$ with $y = 2000$ when $t = 0$. Find $k$ using the point at $t = 3$, $y = 3000$: $$ 3000 = 2000 \cdot e^{3k} $$ Divide both sides by 2000: $$ \frac{3000}{2000} = e^{3k} $$ $$ 1.5 = e^{3k} $$ Take the natural log of both sides: $$ \ln(1.5) = 3k $$ $$ k = \frac{\ln(1.5)}{3} \approx 0.13516 $$ Now use $t = 10$ in $y = 2000 \cdot e^{kt}$: $$ y = 2000 \cdot e^{0.13516 \times 10} $$ $$ y = 2000 \cdot e^{1.3516} \approx 2000 \cdot 3.8637 \approx 7727.4 $$ The population is expected to be about 7,727 people after 10 years. ## How to Use This on the AP Calculus Exam ### Problem Solving Use this routine for any exponential growth or decay problem: 1. Recognize the phrase "proportional to the amount" and write $\frac{dy}{dt} = ky$. 2. Write the solution $y = y_0 e^{kt}$ and substitute the initial value $y_0$. 3. Plug in a second known point to solve for $k$ using natural logs. 4. Answer the actual question (a future value, or a time you solve for) and include units. ### Free Response If a problem asks you to find a general or particular solution, show the separation-of-variables steps: separate, integrate both sides, add the constant of integration, and use the initial condition to solve for the constant. Skipping the separation step or dropping the constant of integration [limits](/ap-calc/unit-10-infinite-sequences-and-series-bc-only-/defining-convergent-divergent-infinite-series/study-guide/CIVFHStGQM90EJ4GtIDB "fv-autolink") the points you can earn. ### Common Trap Not every differential equation with a fraction leads to a log or exponential answer. Only equations in the form $\frac{dy}{dt} = ky$ (rate proportional to amount) give exponential solutions. Check the structure before assuming the answer is exponential. ## Practice Problems Try these before checking the solutions. 1. The rate at which a drug leaves the bloodstream is proportional to the amount in the bloodstream. A dose of 200 mg is given to a patient. After 3 hours, approximately 127.3 mg remain. The amount of the drug in milligrams after $t$ hours is $A(t)$. Write an equation for $A(t)$. 2. Bacteria in a 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 quadruple? ### Solution for Problem 1 Identify the given information: - $y_0 = 200$ - After 3 hours ($t = 3$), $y = 127.3$ - Goal: find the equation that models the situation. Use $y = y_0 e^{kt}$ with $y = 200$ when $t = 0$. Find $k$ using the point at $t = 3$, $y = 127.3$: $$ 127.3 = 200 \cdot e^{3k} $$ Divide both sides by 200: $$ \frac{127.3}{200} = e^{3k} $$ $$ 0.6365 = e^{3k} $$ Take the natural log of both sides: $$ \ln(0.6365) = 3k $$ $$ k = \frac{\ln(0.6365)}{3} \approx -0.151 $$ Since $\frac{dy}{dt} = kA$ and $A(0) = 200$, the model is: $$ A(t) = 200 \cdot e^{-0.151t} $$ The negative $k$ confirms decay, which fits a drug leaving the bloodstream. ### Solution for Problem 2 There are no exact numbers, only ratios, so pick convenient starting values. - Let $y_0 = 100$ when $t = 0$. - After 5 hours ($t = 5$), $y = 200$ (doubled). - Goal: find how long until $y = 400$ (quadrupled). Use $y = y_0 e^{kt}$ with $y = 100$ when $t = 0$. Find $k$ from the point at $t = 5$, $y = 200$: $$ 200 = 100 \cdot e^{5k} $$ $$ \frac{200}{100} = e^{5k} $$ $$ 2 = e^{5k} $$ Take the natural log of both sides: $$ \ln(2) = 5k $$ $$ k = \frac{\ln(2)}{5} \approx 0.139 $$ Now solve for the time when $y = 400$: $$ 400 = 100 \cdot e^{0.139t} $$ $$ 4 = e^{0.139t} $$ Take the natural log of both sides: $$ \ln(4) = 0.139t $$ $$ t = \frac{\ln(4)}{0.139} \approx 9.97 \ \text{hours} $$ ## Common Misconceptions - **"Any differential equation with a fraction gives a log or exponential answer."** Only the form $\frac{dy}{dt} = ky$ (rate proportional to amount) produces exponential solutions. Look at the actual structure first. - **"You can skip separating variables."** You need to move all $y$ terms to one side and all $t$ terms to the other before integrating. Skipping this leads to incorrect work and lost points. - **"The constant of integration does not matter."** Forgetting $C$ when you integrate means you cannot correctly use the initial condition to find the particular solution. - **"$k$ is always positive."** A positive $k$ models growth and a negative $k$ models decay. Solving for $k$ tells you which is happening; do not assume the sign. - **"$y_0$ is just any data point."** $y_0$ is specifically the value of $y$ when $t = 0$. Use a different known point to solve for $k$. - **"Units do not matter."** Track the units on time and on $k$ so your prediction makes sense in context. ## Related AP Calculus Guides - [Unit 7 Overview: Differential Equations](/ap-calc/unit-7/review/study-guide/iNRxaToienfCUUDM9YGi) - [7.2 Verifying Solutions for Differential Equations](/ap-calc/unit-7/verifying-solutions-for-differential-equations/study-guide/s2nX7AhIBDxGIWwlL82x) - [7.3 Sketching Slope Fields](/ap-calc/unit-7/sketching-slope-fields/study-guide/eVAoF3mU4CepiUyAqltB) - [7.6 Finding General Solutions Using Separation of Variables](/ap-calc/unit-7/finding-general-solutions-using-separation-variables/study-guide/qYWqPrBHjoXf0x451c3H) - [7.5 Approximating Solutions Using Euler’s Method](/ap-calc/unit-7/approximating-solutions-using-eulers-method/study-guide/XZF01jg29LPjZaV7jKjE) - [7.7 Finding Particular Solutions Using Initial Conditions and Separation of Variables](/ap-calc/unit-7/finding-particular-solutions-using-initial-conditions-separation-variables/study-guide/v0tgJcQJwgznHGMkq2Uy) ## Vocabulary - **differential equation**: An equation that relates a function to its derivatives, describing how a quantity changes in relation to one or more variables. - **exponential decay**: A process in which a quantity decreases at a rate proportional to its current size, modeled by dy/dt = ky where k < 0. - **exponential growth**: A process in which a quantity increases at a rate proportional to its current size, modeled by dy/dt = ky where k > 0. - **exponential growth and decay model**: A differential equation of the form dy/dt = ky that models quantities that increase or decrease at a rate proportional to their current amount. - **general solution**: The complete family of solutions to a differential equation, containing arbitrary constants that represent all possible particular solutions. - **initial condition**: Specified values of a function at particular points that determine which particular solution to a differential equation is selected. - **particular solution**: A specific solution to a differential equation obtained by using initial conditions to determine the values of arbitrary constants. - **proportional**: A relationship between two quantities where one is a constant multiple of the other. - **rate of change**: The measure of how quickly a quantity changes with respect to another variable, often time. ## FAQs ### What is an exponential model with a differential equation? An exponential model comes from a differential equation where the rate of change is proportional to the current amount. In AP Calculus, that usually appears as dy/dt = ky, and its solution is y = y0e^(kt). ### How do you solve dy/dt = ky? Separate the variables to get (1/y) dy = k dt, integrate both sides, and solve for y. After applying the initial condition y(0) = y0, the particular solution is y = y0e^(kt). ### What does k mean in exponential growth and decay? The constant k tells you the rate and direction of change. If k is positive, the quantity grows; if k is negative, the quantity decays. Its units are the reciprocal of the time unit, such as per year or per hour. ### How do you find k from a data point? Plug the initial value and a second known point into y = y0e^(kt). Divide by y0, take the natural log of both sides, and solve for k. ### When do you use exponential growth and decay on AP Calc? Use it when a problem says a rate is proportional to the current amount. Common contexts include population growth, radioactive decay, drug concentration, cooling, or any quantity changing at a rate tied to its size. ### How is AP Calc 7.8 tested? AP Calc 7.8 can ask you to translate a verbal model into dy/dt = ky, solve for a general or particular solution, use an initial condition, and interpret y, t, and k in context. ## Structured Data ```json {"@context":"https://schema.org","@type":"FAQPage","inLanguage":"en","mainEntity":[{"@type":"Question","@id":"https://fiveable.me/ap-calc/unit-7/exponential-models-with-differential-equations/study-guide/fzn9urmywhPSE4KUxVcq#what-is-an-exponential-model-with-a-differential-equation","name":"What is an exponential model with a differential equation?","acceptedAnswer":{"@type":"Answer","text":"An exponential model comes from a differential equation where the rate of change is proportional to the current amount. In AP Calculus, that usually appears as dy/dt = ky, and its solution is y = y0e^(kt)."}},{"@type":"Question","@id":"https://fiveable.me/ap-calc/unit-7/exponential-models-with-differential-equations/study-guide/fzn9urmywhPSE4KUxVcq#how-do-you-solve-dydt-ky","name":"How do you solve dy/dt = ky?","acceptedAnswer":{"@type":"Answer","text":"Separate the variables to get (1/y) dy = k dt, integrate both sides, and solve for y. After applying the initial condition y(0) = y0, the particular solution is y = y0e^(kt)."}},{"@type":"Question","@id":"https://fiveable.me/ap-calc/unit-7/exponential-models-with-differential-equations/study-guide/fzn9urmywhPSE4KUxVcq#what-does-k-mean-in-exponential-growth-and-decay","name":"What does k mean in exponential growth and decay?","acceptedAnswer":{"@type":"Answer","text":"The constant k tells you the rate and direction of change. If k is positive, the quantity grows; if k is negative, the quantity decays. Its units are the reciprocal of the time unit, such as per year or per hour."}},{"@type":"Question","@id":"https://fiveable.me/ap-calc/unit-7/exponential-models-with-differential-equations/study-guide/fzn9urmywhPSE4KUxVcq#how-do-you-find-k-from-a-data-point","name":"How do you find k from a data point?","acceptedAnswer":{"@type":"Answer","text":"Plug the initial value and a second known point into y = y0e^(kt). Divide by y0, take the natural log of both sides, and solve for k."}},{"@type":"Question","@id":"https://fiveable.me/ap-calc/unit-7/exponential-models-with-differential-equations/study-guide/fzn9urmywhPSE4KUxVcq#when-do-you-use-exponential-growth-and-decay-on-ap-calc","name":"When do you use exponential growth and decay on AP Calc?","acceptedAnswer":{"@type":"Answer","text":"Use it when a problem says a rate is proportional to the current amount. Common contexts include population growth, radioactive decay, drug concentration, cooling, or any quantity changing at a rate tied to its size."}},{"@type":"Question","@id":"https://fiveable.me/ap-calc/unit-7/exponential-models-with-differential-equations/study-guide/fzn9urmywhPSE4KUxVcq#how-is-ap-calc-78-tested","name":"How is AP Calc 7.8 tested?","acceptedAnswer":{"@type":"Answer","text":"AP Calc 7.8 can ask you to translate a verbal model into dy/dt = ky, solve for a general or particular solution, use an initial condition, and interpret y, t, and k in context."}}]} ```