Exponential Models with Differential Equations - AP Calculus AB
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What initial condition is used for solving $\frac{dy}{dt} = ky$?
What initial condition is used for solving $\frac{dy}{dt} = ky$?
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$y(0) = y_0$. This boundary condition determines the constant in the general solution.
$y(0) = y_0$. This boundary condition determines the constant in the general solution.
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What is the half-life formula for exponential decay?
What is the half-life formula for exponential decay?
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$y(t) = y_0 \left(\frac{1}{2}\right)^{\frac{t}{h}}$. The quantity reduces by half every $h$ time units using base $\frac{1}{2}$.
$y(t) = y_0 \left(\frac{1}{2}\right)^{\frac{t}{h}}$. The quantity reduces by half every $h$ time units using base $\frac{1}{2}$.
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What is the initial value in $N(t) = N_0 e^{-\lambda t}$?
What is the initial value in $N(t) = N_0 e^{-\lambda t}$?
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$N_0$. The initial amount or concentration at time $t=0$.
$N_0$. The initial amount or concentration at time $t=0$.
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Determine the constant $k$ for the doubling time formula $P(t) = P_0 2^{\frac{t}{d}}$.
Determine the constant $k$ for the doubling time formula $P(t) = P_0 2^{\frac{t}{d}}$.
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$k = \frac{\ln(2)}{d}$. Since $2^{t/d} = e^{kt}$, we have $k = \ln(2)/d$.
$k = \frac{\ln(2)}{d}$. Since $2^{t/d} = e^{kt}$, we have $k = \ln(2)/d$.
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State the condition for exponential growth.
State the condition for exponential growth.
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$k > 0$ in $\frac{dy}{dt} = ky$. Positive $k$ means the derivative and function have the same sign.
$k > 0$ in $\frac{dy}{dt} = ky$. Positive $k$ means the derivative and function have the same sign.
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Identify $y(t)$ for $\frac{dy}{dt} = 0.1y$ with $y(1) = 3$.
Identify $y(t)$ for $\frac{dy}{dt} = 0.1y$ with $y(1) = 3$.
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$y(t) = 3e^{0.1(t-1)}$. Apply initial condition at $t=1$ with $k=0.1$ and $y_0=3$.
$y(t) = 3e^{0.1(t-1)}$. Apply initial condition at $t=1$ with $k=0.1$ and $y_0=3$.
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Determine $y(t)$ for $\frac{dy}{dt} = -3y$ and $y(0) = 7$.
Determine $y(t)$ for $\frac{dy}{dt} = -3y$ and $y(0) = 7$.
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$y(t) = 7e^{-3t}$. Apply $k=-3$ and initial value $y_0=7$ to get the solution.
$y(t) = 7e^{-3t}$. Apply $k=-3$ and initial value $y_0=7$ to get the solution.
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State the differential equation model for continuous exponential growth.
State the differential equation model for continuous exponential growth.
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$\frac{dy}{dt} = ky$. The rate of change is proportional to the current value with constant $k$.
$\frac{dy}{dt} = ky$. The rate of change is proportional to the current value with constant $k$.
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What does $y_0$ represent in the solution $y(t) = y_0 e^{kt}$?
What does $y_0$ represent in the solution $y(t) = y_0 e^{kt}$?
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Initial value of $y$. The starting value or amount at time $t=0$.
Initial value of $y$. The starting value or amount at time $t=0$.
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What is the $k$ value if a population triples in 6 years?
What is the $k$ value if a population triples in 6 years?
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$k = \frac{\ln(3)}{6}$. Tripling means $3 = e^{6k}$, so $k = \ln(3)/6$.
$k = \frac{\ln(3)}{6}$. Tripling means $3 = e^{6k}$, so $k = \ln(3)/6$.
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Determine $y(t)$ for $\frac{dy}{dt} = 2y$ with $y(0) = 8$.
Determine $y(t)$ for $\frac{dy}{dt} = 2y$ with $y(0) = 8$.
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$y(t) = 8e^{2t}$. Substitute $k=2$ and $y_0=8$ into the general exponential solution.
$y(t) = 8e^{2t}$. Substitute $k=2$ and $y_0=8$ into the general exponential solution.
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What is the natural base in the exponential model $y(t) = y_0 e^{kt}$?
What is the natural base in the exponential model $y(t) = y_0 e^{kt}$?
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$e$. Euler's number $e \approx 2.718$ is the natural exponential base.
$e$. Euler's number $e \approx 2.718$ is the natural exponential base.
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Find the $y$-intercept for $y(t) = 5e^{3t}$.
Find the $y$-intercept for $y(t) = 5e^{3t}$.
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$y$-intercept is 5. At $t=0$, $y(0) = 5e^0 = 5 \cdot 1 = 5$.
$y$-intercept is 5. At $t=0$, $y(0) = 5e^0 = 5 \cdot 1 = 5$.
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Calculate $y(t)$ for $\frac{dy}{dt} = 0.2y$ and $y(0) = 7$.
Calculate $y(t)$ for $\frac{dy}{dt} = 0.2y$ and $y(0) = 7$.
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$y(t) = 7e^{0.2t}$. Substitute $k=0.2$ and $y_0=7$ into the exponential growth formula.
$y(t) = 7e^{0.2t}$. Substitute $k=0.2$ and $y_0=7$ into the exponential growth formula.
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Find $y(t)$ for $\frac{dy}{dt} = 0.5y$ if $y(0) = 1$.
Find $y(t)$ for $\frac{dy}{dt} = 0.5y$ if $y(0) = 1$.
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$y(t) = e^{0.5t}$. With $k=0.5$ and $y_0=1$, the solution simplifies to $e^{0.5t}$.
$y(t) = e^{0.5t}$. With $k=0.5$ and $y_0=1$, the solution simplifies to $e^{0.5t}$.
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Which equation represents a population doubling every period?
Which equation represents a population doubling every period?
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$P(t) = P_0 2^{\frac{t}{d}}$. The base 2 with time divided by doubling period $d$ models doubling.
$P(t) = P_0 2^{\frac{t}{d}}$. The base 2 with time divided by doubling period $d$ models doubling.
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What is the exponential model used for population growth?
What is the exponential model used for population growth?
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$P(t) = P_0 e^{kt}$. Standard exponential growth model where $P_0$ is initial population.
$P(t) = P_0 e^{kt}$. Standard exponential growth model where $P_0$ is initial population.
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Which function describes a quantity halving every 5 years?
Which function describes a quantity halving every 5 years?
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$y(t) = y_0 \left(\frac{1}{2}\right)^{\frac{t}{5}}$. Half-life of 5 years means the exponent is $t/5$ with base $1/2$.
$y(t) = y_0 \left(\frac{1}{2}\right)^{\frac{t}{5}}$. Half-life of 5 years means the exponent is $t/5$ with base $1/2$.
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Identify the exponential model for radioactive decay.
Identify the exponential model for radioactive decay.
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$N(t) = N_0 e^{-kt}$. Standard form for radioactive decay with decay constant $k$.
$N(t) = N_0 e^{-kt}$. Standard form for radioactive decay with decay constant $k$.
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What is the $y$-intercept for $P(t) = 2e^{0.3t}$?
What is the $y$-intercept for $P(t) = 2e^{0.3t}$?
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$y$-intercept is 2. At $t=0$, $P(0) = 2e^0 = 2 \cdot 1 = 2$.
$y$-intercept is 2. At $t=0$, $P(0) = 2e^0 = 2 \cdot 1 = 2$.
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For $\frac{dy}{dt} = ky$, determine $y(t)$ if $y_0 = 12$.
For $\frac{dy}{dt} = ky$, determine $y(t)$ if $y_0 = 12$.
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$y(t) = 12e^{kt}$. General solution with specified initial value $y_0=12$.
$y(t) = 12e^{kt}$. General solution with specified initial value $y_0=12$.
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What is the $k$ value for a quantity decreasing by 15% annually?
What is the $k$ value for a quantity decreasing by 15% annually?
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$k = -0.15$. A 15% annual decrease corresponds to $k=-0.15$ in the model.
$k = -0.15$. A 15% annual decrease corresponds to $k=-0.15$ in the model.
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Identify the $y$-intercept in $y(t) = 10e^{-t}$.
Identify the $y$-intercept in $y(t) = 10e^{-t}$.
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$y$-intercept is 10. The $y$-intercept occurs when $t=0$, giving $y(0)=10e^0=10$.
$y$-intercept is 10. The $y$-intercept occurs when $t=0$, giving $y(0)=10e^0=10$.
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Determine $y(t)$ for $\frac{dy}{dt} = -0.3y$ and $y(0) = 9$.
Determine $y(t)$ for $\frac{dy}{dt} = -0.3y$ and $y(0) = 9$.
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$y(t) = 9e^{-0.3t}$. Use decay constant $k=-0.3$ with initial value $y_0=9$.
$y(t) = 9e^{-0.3t}$. Use decay constant $k=-0.3$ with initial value $y_0=9$.
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State the exponential growth rate if $k = 0.05$.
State the exponential growth rate if $k = 0.05$.
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5%. The value $k=0.05$ corresponds to a 5% growth rate per unit time.
5%. The value $k=0.05$ corresponds to a 5% growth rate per unit time.
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What is the exponential decay formula for a substance?
What is the exponential decay formula for a substance?
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$N(t) = N_0 e^{-\lambda t}$. Standard decay model where $\lambda$ is the decay constant.
$N(t) = N_0 e^{-\lambda t}$. Standard decay model where $\lambda$ is the decay constant.
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Express $y(t)$ in terms of $k$ if $\frac{dy}{dt} = ky$ and $y(0) = 6$.
Express $y(t)$ in terms of $k$ if $\frac{dy}{dt} = ky$ and $y(0) = 6$.
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$y(t) = 6e^{kt}$. General form with initial condition $y_0=6$ and growth constant $k$.
$y(t) = 6e^{kt}$. General form with initial condition $y_0=6$ and growth constant $k$.
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Calculate $y(t)$ for $\frac{dy}{dt} = -\frac{1}{2}y$ and $y(0) = 20$.
Calculate $y(t)$ for $\frac{dy}{dt} = -\frac{1}{2}y$ and $y(0) = 20$.
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$y(t) = 20e^{-\frac{1}{2}t}$. Use $k=-1/2$ and $y_0=20$ in the exponential solution form.
$y(t) = 20e^{-\frac{1}{2}t}$. Use $k=-1/2$ and $y_0=20$ in the exponential solution form.
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Find $y(t)$ if $\frac{dy}{dt} = 4y$ and $y(2) = 16$.
Find $y(t)$ if $\frac{dy}{dt} = 4y$ and $y(2) = 16$.
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$y(t) = 16e^{4(t-2)}$. With condition at $t=2$, substitute to get $y=16e^{4(t-2)}$.
$y(t) = 16e^{4(t-2)}$. With condition at $t=2$, substitute to get $y=16e^{4(t-2)}$.
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State the condition for exponential decay.
State the condition for exponential decay.
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$k < 0$ in $\frac{dy}{dt} = ky$. Negative $k$ causes the quantity to decrease exponentially over time.
$k < 0$ in $\frac{dy}{dt} = ky$. Negative $k$ causes the quantity to decrease exponentially over time.
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