Exponential decay describes a quantity decreasing at a rate proportional to its current size, modeled by the differential equation dy/dt = ky with k < 0, whose solution is y = y₀e^(kt). It's the core application of Topic 7.8 in AP Calculus Unit 7.
Exponential decay is what happens when a quantity shrinks at a rate proportional to how much of it is left. The more you have, the faster you lose it. Less stuff means slower loss. That sentence in math form is the differential equation dy/dt = ky, where k is negative for decay (FUN-7.F.2). Radioactive substances, cooling coffee, and drug concentrations in the bloodstream all follow this pattern.
Solve that differential equation (separation of variables works) and you get y = y₀e^(kt), where y₀ is the amount you start with at t = 0 (FUN-7.G.1). When k < 0, the e^(kt) factor shrinks toward zero as t grows, so the quantity decays. Here's the key mental model. Exponential decay isn't just "it goes down fast then slow." It's the unique behavior you get when the rate of change is proportional to the amount. That proportionality statement is what the AP exam wants you to recognize and translate into dy/dt = ky.
Exponential decay lives in Topic 7.8 (Exponential Models with Differential Equations) in Unit 7, and it supports two learning objectives directly. 7.8.A asks you to interpret a differential equation in context, meaning you read "a substance decays at a rate proportional to the amount present" and write dy/dt = ky. 7.8.B asks you to actually solve it, finding the general solution y = Ce^(kt) and then using an initial condition to pin down the particular solution y = y₀e^(kt). This is one of the two named applications of differential equations in the CED (the other is motion along a line, per FUN-7.F.1), so it's a reliable source of multiple-choice questions. It also pulls together half of the course in one problem type, since solving it uses separation of variables, integration, and properties of e^x and ln(x).
Keep studying AP Calculus Unit 7
Visual cheatsheet
view galleryExponential Growth (Unit 7)
Growth and decay are the same model, dy/dt = ky, with one sign flipped. Positive k means growth, negative k means decay. On the exam, the sign of k is often the entire question, so check it before anything else.
Half-Life (Unit 7)
Half-life is the time it takes a decaying quantity to drop to half its starting value. It's a fixed number for any given k, and you find it by solving y₀/2 = y₀e^(kt), which gives t = ln(2)/|k|. Half-life problems are just exponential decay problems wearing a costume.
Initial Condition (Unit 7)
The general solution y = Ce^(kt) describes infinitely many decay curves. An initial condition like y(0) = 100 picks out exactly one of them, turning C into y₀. Without an initial condition you can't answer "how much is left at time t."
Separation of Variables (Unit 7)
y = y₀e^(kt) isn't a formula handed down from above. You derive it by separating dy/dt = ky into dy/y = k dt and integrating both sides. Knowing the derivation matters because FRQs can ask you to show the work, not just quote the answer.
Exponential decay shows up most often in multiple choice, and the questions follow a few predictable patterns. One type asks you to identify the correct differential equation for a decay scenario, where the answer is dy/dt = ky with negative k (or written as dy/dt = -ky with positive k). Another gives you something like dy/dx = -0.1y with y(0) = 100 and asks for the particular solution, which is y = 100e^(-0.1x). A third type is a word problem, like a radioactive substance decaying at 6% per year from an initial 200 grams, where you build y = 200e^(-0.06t) yourself. No released FRQ has centered on this exact phrase, but exponential models are fair game in any FRQ about solving differential equations in context, so be ready to set up the equation, separate variables, integrate, apply the initial condition, and interpret your answer with units.
Linear decay loses the same fixed amount every time period (constant rate, dy/dt = -c), while exponential decay loses the same fixed percentage of what's left (rate proportional to size, dy/dt = ky). A linear model hits zero and keeps going negative; an exponential model approaches zero forever without reaching it. If the problem says the rate is "proportional to the amount present," that's exponential, full stop.
Exponential decay means the rate of change of a quantity is proportional to the quantity itself, written as dy/dt = ky with k < 0.
Solving dy/dt = ky with the initial condition y(0) = y₀ gives the particular solution y = y₀e^(kt), per FUN-7.G.1.
The phrase "proportional to the amount present" in a word problem is your signal to write dy/dt = ky immediately.
Growth and decay use the identical model, so always check the sign of k before answering.
An exponentially decaying quantity approaches zero as t increases but never actually reaches it.
You should be able to derive y = y₀e^(kt) by separation of variables, not just memorize it, since FRQs can ask for the work.
It's the model for a quantity that decreases at a rate proportional to its current size, expressed as the differential equation dy/dt = ky where k is negative. Its solution is y = y₀e^(kt), where y₀ is the initial amount. It's covered in Topic 7.8 of Unit 7.
No. Since y = y₀e^(kt) is a positive number times an exponential, it gets arbitrarily close to zero as t grows but never equals zero. The x-axis is a horizontal asymptote, not a finish line.
Negative, if the equation is written dy/dt = ky. Some problems write it as dy/dt = -ky with k > 0 instead, which means the same thing. Either way, the rate of change must be negative for the quantity to decrease.
Same equation, opposite sign. Both come from dy/dt = ky and both solve to y = y₀e^(kt). With k > 0 the quantity grows without bound; with k < 0 it decays toward zero. On the exam, identifying the correct sign is often the whole question.
Separate variables to get dy/y = -0.5 dx, integrate both sides to get ln|y| = -0.5x + C, then exponentiate to get y = Ce^(-0.5x). The initial condition y(0) = 8 gives C = 8, so the particular solution is y = 8e^(-0.5x).