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 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 2: 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 3: 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 4: 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 5: 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 6: 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 7: 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 8: Find the solution to $\frac{dy}{dt} = 3y$ given $y(0) = 5$.

Answer: $y(t) = 5e^{3t}$. Substitute $k=3$ and $y_0=5$ into the general solution $y=y_0e^{kt}$.

Flashcard 9: 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 10: 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 11: Calculate the decay constant $k$ if the half-life is 10 years.

Answer: $k = \frac{\ln(2)}{10}$. For half-life decay, $k = \ln(2)/t_{1/2} = \ln(2)/10$.

Flashcard 12: For $\frac{dy}{dt} = -2y$, what is $y(t)$ if $y(0) = 4$?

Answer: $y(t) = 4e^{-2t}$. Use $k=-2$ and $y_0=4$ in the exponential decay formula.

Flashcard 13: What is the general solution form for a differential equation $\frac{dy}{dt} = ky$?

Answer: $y(t) = y_0 e^{kt}$. This is the general exponential form where $k$ determines growth/decay rate.

Flashcard 14: 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 15: 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 16: 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 17: What is the solution to $\frac{dy}{dt} = -y$ given $y(0) = 15$?

Answer: $y(t) = 15e^{-t}$. Apply $k=-1$ and initial condition $y_0=15$ to the general form.

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: 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 20: 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 21: 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 22: 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 23: 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 24: 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 25: 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 26: 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 27: 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 28: 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 29: Solve for $k$ in $y = y_0 e^{kt}$ if $y(2) = 8$ and $y_0 = 2$.

Answer: $k = \frac{\ln(4)}{2}$. From $8 = 2e^{2k}$, get $4 = e^{2k}$, so $k = \ln(4)/2$.

Flashcard 30: 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 31: 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 32: 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 33: State the condition for exponential decay.

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

Flashcard 34: What is the solution for $\frac{dy}{dt} = ky$ if $y(0) = y_0$?

Answer: $y(t) = y_0 e^{kt}$. Separating variables and integrating yields this exponential solution.

Flashcard 35: For $\frac{dy}{dt} = -2y$, what is $y(t)$ if $y(0) = 4$?

Answer: $y(t) = 4e^{-2t}$. Use $k=-2$ and $y_0=4$ in the exponential decay formula.

Flashcard 36: 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 37: What is the solution for $\frac{dy}{dt} = ky$ if $y(0) = y_0$?

Answer: $y(t) = y_0 e^{kt}$. Separating variables and integrating yields this exponential solution.

Flashcard 38: What is the specific solution if $\frac{dy}{dt} = 5y$ and $y(1) = 10$?

Answer: $y(t) = 10e^{5(t-1)}$. Use initial condition at $t=1$ to find $C=10e^{-5}$, so $y=10e^{5(t-1)}$.

Flashcard 39: 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 40: 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 41: What is the solution to $\frac{dy}{dt} = -y$ given $y(0) = 15$?

Answer: $y(t) = 15e^{-t}$. Apply $k=-1$ and initial condition $y_0=15$ to the general form.

Flashcard 42: 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 43: 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 44: 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 45: What is the general solution form for a differential equation $\frac{dy}{dt} = ky$?

Answer: $y(t) = y_0 e^{kt}$. This is the general exponential form where $k$ determines growth/decay rate.

Flashcard 46: 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 47: 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 48: 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 49: 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 50: 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 51: 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 52: 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 53: Calculate the decay constant $k$ if the half-life is 10 years.

Answer: $k = \frac{\ln(2)}{10}$. For half-life decay, $k = \ln(2)/t_{1/2} = \ln(2)/10$.

Flashcard 54: 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 55: 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 56: 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 57: 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 58: Identify the differential equation for exponential decay.

Answer: $\frac{dy}{dt} = -ky$. Negative $k$ indicates the quantity decreases over time exponentially.

Flashcard 59: Identify the differential equation for exponential decay.

Answer: $\frac{dy}{dt} = -ky$. Negative $k$ indicates the quantity decreases over time exponentially.

Flashcard 60: Find the solution to $\frac{dy}{dt} = 3y$ given $y(0) = 5$.

Answer: $y(t) = 5e^{3t}$. Substitute $k=3$ and $y_0=5$ into the general solution $y=y_0e^{kt}$.

Flashcard 61: Solve for $k$ in $y = y_0 e^{kt}$ if $y(2) = 8$ and $y_0 = 2$.

Answer: $k = \frac{\ln(4)}{2}$. From $8 = 2e^{2k}$, get $4 = e^{2k}$, so $k = \ln(4)/2$.

Flashcard 62: 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 63: 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 64: 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 65: 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 66: 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 67: 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 68: Identify the exponential model for radioactive decay.

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

Flashcard 69: 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 70: 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 71: 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 72: Identify the exponential model for radioactive decay.

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

Flashcard 73: What is the specific solution if $\frac{dy}{dt} = 5y$ and $y(1) = 10$?

Answer: $y(t) = 10e^{5(t-1)}$. Use initial condition at $t=1$ to find $C=10e^{-5}$, so $y=10e^{5(t-1)}$.

Flashcard 74: State the condition for exponential decay.

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

Flashcard 75: 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 76: 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 77: 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 78: 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.