You CAN Ace Differential Equations | |
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17calculus > differential equations > exponential growth and decay |
Topics You Need To Understand For This Page
these topics are not required for this page but will help you understand where the equations come from |
Differential Equations Alpha List
Tools
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Related Topics and Links
external links you may find helpful |
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Math Insight - Exponential Growth and Decay: A Differential Equation |
ATTENTION INSTRUCTORS: The new 2018 version of 17calculus will include changes to the practice problem numbering system. If you would like advance information to help you prepare for spring semester, send us an email at 2018info at 17calculus.com. |
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How to Read and Do Proofs: An Introduction to Mathematical Thought Processes |
Exponential Growth and Decay |
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Although this is a differential equation topic, many students come across this topic while studying basic integrals. You do not need to know anything other than integrals to understand where the equations come from. If you are given the equation and not expected to derive it, you need only logarithms and algebra to work many problems. |
quick notes |
rate of change is proportional to quantity \( y' = ky ~~~ \to ~~~ y = Ae^{kt} \) |
You will definitely need to be sharp with your logarithms for this topic. The precalculus logarithms page will help you get up to speed. |
The population dynamics page expands on this discussion of exponential growth and decay applied specifically to population change. |
What Does Exponential Growth/Decay Mean? |
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What does it mean for something to grow or decay exponentially? The idea is that the independent variable is found in the exponent rather than the base. For example, comparing \(f(t)=t^2\) and \(g(t)=2^t\), notice that \(t\) is in the exponent of the \(g(t)\), so \(g(t)\) is considered an example of exponential growth but \(f(t)\) is not (since \(t\) is not in the exponent).
The general form of an exponential growth equation is \(y = a(b^t)\) or \(y=a(1+r)^t\). These equations are the same when \(b=1+r\), so our discussion will center around \(y = a(b^t)\) and you can easily extend your understanding to the second equation if you need to.
When \(b > 1\), we call the equation an exponential growth equation.
When \(b < 1\), it is called exponential decay.
When \(b = 1\), we have \(y=a(1^t)=a\), which is just linear equation and it is not considered an exponential equation.
So it is easy to see that we have the same equation in both cases and the value of \(b\) tells us if the exponential is growing or decaying. Let's watch a quick video before we go on.
PatrickJMT - An Exponential Growth Problem | |
A Special Type of Exponential Growth/Decay |
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A specific type of exponential growth is when \(b=e^r\) and \(r\) is called the growth/decay rate. The equation comes from the idea that the rate of change is proportional to the quantity that currently exists. This produces the autonomous differential equation
autonomous equation |
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A special type of differential equation of the form \(y' = f(y)\) where the independent variable does not explicitly appear in the equation. |
\( y' = ky \) |
where \(k\) is a constant called the growth/decay constant/rate.
When \(k < 0\), we use the term exponential decay.
When \(k > 0\), we use the term exponential growth.
To solve this differential equation, there are several techniques available to us. We will use separation of variables. Although not explicitly written in this equation, the independent variable we usually use in these types of equations is t to represent time.
\(\displaystyle{
\begin{array}{rcl}
y' & = & ky \\
\frac{dy}{dt} & = & ky \\
\frac{dy}{y} & = & k~dt \\
\int{ \frac{dy}{y} } & = & \int{ k~dt } \\
\ln|y| & = & kt+C \\
e^{\ln|y|} & = & e^{kt+C} \\
y & = & Ae^{kt}
\end{array}
}\)
The constant A is usually called the initial amount since at time \(t=0\), we have \(y = Ae^0 = A\).
The key to solving these types of problems usually involves determining \(k\). Depending on what your instructor wants, you can usually just start with the equation \(y = Ae^{kt}\), if you know that you have an exponential decay or growth problem.
Here is a quick video explaining this and showing a graph to give you a feel for these equations.
Brightstorm - The Differential Equation Model for Exponential Growth | |
Half-Life |
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Half-life is the time it takes for half the substance to decay. The idea is to take the equation \(y = Ae^{kt}\), set the left side to \(A/2\) and solve for t. Notice that you don't have to know the initial amount A since in the equation \(A/2 = Ae^{kt}\), the A cancels leaving \(1/2 = e^{kt}\). You can then use basic logarithms to solve for t.
Uniqueness of Solution |
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So it may not have occurred to you but have you thought that maybe it is possible there is another solution to the differential equation \(x'=ax\)? This question is part of a very deep discussion in differential equations involving existence and uniqueness. Here is a really good video that shows that the solution discussed on this page is unique.
Dr Chris Tisdell - Why do initial value problems for x' = ax have exactly one solution? [6min-10secs] | |
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Practice Problems |
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Level A - Basic |
Practice A01 | |
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Suppose that the half-life of a certain substance is 20 days and there are initially 10 grams of the substance. How much of the substance remains after 75 days? | |
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Practice A02 | |
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Suppose that radioactive ion-X decays at a constant annual rate of 4%. What is the half-life of the substance when the initial amount is 100g? | |
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Practice A03 | |
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Suppose a bacteria population starts with 10 bacteria and that they divide every hour. a) What is the population 7 hours later? b) When will there be 1,000,000 bacteria? | |
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Practice A04 | |
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A radioactive material is know to decay at a yearly rate of 0.2 times the amount at each moment. Suppose there are 1000 grams of the material now. What is the amount after 10 years? | |
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Practice A05 | |
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Suppose a material decays at a rate proportional to the quantity of the material and there were 2500 grams 10 years ago. If there are 2400 grams now, what is the half-life? | |
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Practice A06 | |
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Suppose a bacteria population grows at a rate proportional to the population. There were 200 bacteria 3 days ago and 1000 bacteria 1 day ago. How many bacteria will there be tomorrow? | |
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Practice A07 | |
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A bacteria population increases sixfold in 10 hours. Assuming normal growth, how long did it take for their population to double? | |
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Practice A09 | |
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Carbon-14 has a half-life of 5730 years. a) If the initial amount is 300g, how much is left after 2000 years? b) If initial amount is 400 grams, when will there be 350g left? | |
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Practice A10 | |
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Find the half-life of a compound where the decay rate is 0.05. (time is in years) | |
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Practice A11 | |
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After 1000 years, we have 500 grams of a substance with a decay rate of 0.001. What was the initial amount? | |
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Practice A12 | |
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Suppose a radioactive substance decays at a rate of 3.5% per hour. What percent of the substance is left after 6 hours? | |
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Practice A13 | |
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A bacteria culture initially contains 100 cells and grows at a rate proportional to its size. After an hour the population has increased to 420. a) Find an expression for the number of bacteria after t hours. b) Find the number of bacteria after 3 hours. c) Find the rate of growth after 3 hours. d) When will the population reach 10,000? | |
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