AP Calculus AB Flashcards: Exponential Models With Differential Equations

What this deck covers

This deck focuses on Exponential Models With Differential Equations, giving you a quick way to review the definitions, rules, and examples that matter most for AP Calculus AB.

How to use these flashcards

Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

Flashcard 1: What initial condition is used for solving $\frac{dy}{dt} = ky$?

Answer: $y(0) = y_0$. This boundary condition determines the constant in the general solution.

Flashcard 2: What is the half-life formula for exponential decay?

Answer: $y(t) = y_0 \left(\frac{1}{2}\right)^{\frac{t}{h}}$. The quantity reduces by half every $h$ time units using base $\frac{1}{2}$.

Flashcard 3: What is the initial value in $N(t) = N_0 e^{-\lambda t}$?

Answer: $N_0$. The initial amount or concentration at time $t=0$.

Flashcard 4: Determine the constant $k$ for the doubling time formula $P(t) = P_0 2^{\frac{t}{d}}$.

Answer: $k = \frac{\ln(2)}{d}$. Since $2^{t/d} = e^{kt}$, we have $k = \ln(2)/d$.

Flashcard 5: State the condition for exponential growth.

Answer: $k > 0$ in $\frac{dy}{dt} = ky$. Positive $k$ means the derivative and function have the same sign.

Flashcard 6: Identify $y(t)$ for $\frac{dy}{dt} = 0.1y$ with $y(1) = 3$.

Answer: $y(t) = 3e^{0.1(t-1)}$. Apply initial condition at $t=1$ with $k=0.1$ and $y_0=3$.

Flashcard 7: Determine $y(t)$ for $\frac{dy}{dt} = -3y$ and $y(0) = 7$.

Answer: $y(t) = 7e^{-3t}$. Apply $k=-3$ and initial value $y_0=7$ to get the solution.

Flashcard 8: State the differential equation model for continuous exponential growth.

Answer: $\frac{dy}{dt} = ky$. The rate of change is proportional to the current value with constant $k$.

Flashcard 9: What does $y_0$ represent in the solution $y(t) = y_0 e^{kt}$?

Answer: Initial value of $y$. The starting value or amount at time $t=0$.

Flashcard 10: What is the $k$ value if a population triples in 6 years?

Answer: $k = \frac{\ln(3)}{6}$. Tripling means $3 = e^{6k}$, so $k = \ln(3)/6$.

Flashcard 11: Determine $y(t)$ for $\frac{dy}{dt} = 2y$ with $y(0) = 8$.

Answer: $y(t) = 8e^{2t}$. Substitute $k=2$ and $y_0=8$ into the general exponential solution.

Flashcard 12: What is the natural base in the exponential model $y(t) = y_0 e^{kt}$?

Answer: $e$. Euler's number $e \approx 2.718$ is the natural exponential base.

Flashcard 13: Find the $y$-intercept for $y(t) = 5e^{3t}$.

Answer: $y$-intercept is 5. At $t=0$, $y(0) = 5e^0 = 5 \cdot 1 = 5$.

Flashcard 14: Calculate $y(t)$ for $\frac{dy}{dt} = 0.2y$ and $y(0) = 7$.

Answer: $y(t) = 7e^{0.2t}$. Substitute $k=0.2$ and $y_0=7$ into the exponential growth formula.

Flashcard 15: Find $y(t)$ for $\frac{dy}{dt} = 0.5y$ if $y(0) = 1$.

Answer: $y(t) = e^{0.5t}$. With $k=0.5$ and $y_0=1$, the solution simplifies to $e^{0.5t}$.

Flashcard 16: Which equation represents a population doubling every period?

Answer: $P(t) = P_0 2^{\frac{t}{d}}$. The base 2 with time divided by doubling period $d$ models doubling.

Flashcard 17: What is the exponential model used for population growth?

Answer: $P(t) = P_0 e^{kt}$. Standard exponential growth model where $P_0$ is initial population.

Flashcard 18: Which function describes a quantity halving every 5 years?

Answer: $y(t) = y_0 \left(\frac{1}{2}\right)^{\frac{t}{5}}$. Half-life of 5 years means the exponent is $t/5$ with base $1/2$.

Flashcard 19: Identify the exponential model for radioactive decay.

Answer: $N(t) = N_0 e^{-kt}$. Standard form for radioactive decay with decay constant $k$.

Flashcard 20: What is the $y$-intercept for $P(t) = 2e^{0.3t}$?

Answer: $y$-intercept is 2. At $t=0$, $P(0) = 2e^0 = 2 \cdot 1 = 2$.

Flashcard 21: For $\frac{dy}{dt} = ky$, determine $y(t)$ if $y_0 = 12$.

Answer: $y(t) = 12e^{kt}$. General solution with specified initial value $y_0=12$.

Flashcard 22: What is the $k$ value for a quantity decreasing by 15% annually?

Answer: $k = -0.15$. A 15% annual decrease corresponds to $k=-0.15$ in the model.

Flashcard 23: Identify the $y$-intercept in $y(t) = 10e^{-t}$.

Answer: $y$-intercept is 10. The $y$-intercept occurs when $t=0$, giving $y(0)=10e^0=10$.

Flashcard 24: Determine $y(t)$ for $\frac{dy}{dt} = -0.3y$ and $y(0) = 9$.

Answer: $y(t) = 9e^{-0.3t}$. Use decay constant $k=-0.3$ with initial value $y_0=9$.

Flashcard 25: State the exponential growth rate if $k = 0.05$.

Answer: 5%. The value $k=0.05$ corresponds to a 5% growth rate per unit time.

Flashcard 26: What is the exponential decay formula for a substance?

Answer: $N(t) = N_0 e^{-\lambda t}$. Standard decay model where $\lambda$ is the decay constant.

Flashcard 27: Express $y(t)$ in terms of $k$ if $\frac{dy}{dt} = ky$ and $y(0) = 6$.

Answer: $y(t) = 6e^{kt}$. General form with initial condition $y_0=6$ and growth constant $k$.

Flashcard 28: Calculate $y(t)$ for $\frac{dy}{dt} = -\frac{1}{2}y$ and $y(0) = 20$.

Answer: $y(t) = 20e^{-\frac{1}{2}t}$. Use $k=-1/2$ and $y_0=20$ in the exponential solution form.

Flashcard 29: Find $y(t)$ if $\frac{dy}{dt} = 4y$ and $y(2) = 16$.

Answer: $y(t) = 16e^{4(t-2)}$. With condition at $t=2$, substitute to get $y=16e^{4(t-2)}$.

Flashcard 30: State the condition for exponential decay.

Answer: $k < 0$ in $\frac{dy}{dt} = ky$. Negative $k$ causes the quantity to decrease exponentially over time.