Logistic Models with Differential Equations - AP Calculus BC
Card 1 of 30
What does $P(0)$ represent in the context of logistic growth?
What does $P(0)$ represent in the context of logistic growth?
Tap to reveal answer
Initial population size. $P(0)$ is the starting population at time $t = 0$.
Initial population size. $P(0)$ is the starting population at time $t = 0$.
← Didn't Know|Knew It →
What property of logistic growth makes it realistic for modeling populations?
What property of logistic growth makes it realistic for modeling populations?
Tap to reveal answer
Incorporates carrying capacity. Unlike exponential growth, logistic models have a maximum capacity.
Incorporates carrying capacity. Unlike exponential growth, logistic models have a maximum capacity.
← Didn't Know|Knew It →
How is the constant $A$ related to initial population in logistic growth?
How is the constant $A$ related to initial population in logistic growth?
Tap to reveal answer
$A = \frac{K - P(0)}{P(0)}$. Derived from solving $P(0) = \frac{K}{1 + A}$ for $A$.
$A = \frac{K - P(0)}{P(0)}$. Derived from solving $P(0) = \frac{K}{1 + A}$ for $A$.
← Didn't Know|Knew It →
What are the units of $r$ in the logistic model $\frac{dP}{dt} = rP(1 - \frac{P}{K})$?
What are the units of $r$ in the logistic model $\frac{dP}{dt} = rP(1 - \frac{P}{K})$?
Tap to reveal answer
Inverse time. Since $\frac{dP}{dt}$ has units of population per time.
Inverse time. Since $\frac{dP}{dt}$ has units of population per time.
← Didn't Know|Knew It →
State the solution for $P(t)$ when $P(0) = \frac{K}{2}$ in logistic growth.
State the solution for $P(t)$ when $P(0) = \frac{K}{2}$ in logistic growth.
Tap to reveal answer
$P(t) = \frac{K}{1 + e^{-rt}}$. When $P(0) = \frac{K}{2}$, then $A = 1$ in the solution.
$P(t) = \frac{K}{1 + e^{-rt}}$. When $P(0) = \frac{K}{2}$, then $A = 1$ in the solution.
← Didn't Know|Knew It →
What characterizes the initial phase of logistic growth?
What characterizes the initial phase of logistic growth?
Tap to reveal answer
Exponential-like growth. Early phase resembles exponential growth when $P << K$.
Exponential-like growth. Early phase resembles exponential growth when $P << K$.
← Didn't Know|Knew It →
In logistic growth, what happens as $P$ approaches $K$?
In logistic growth, what happens as $P$ approaches $K$?
Tap to reveal answer
Growth rate decreases. Factor $ (1 - \frac{P}{K}) $ approaches zero as $P$ nears $K$.
Growth rate decreases. Factor $ (1 - \frac{P}{K}) $ approaches zero as $P$ nears $K$.
← Didn't Know|Knew It →
What is the impact of a negative $r$ in the logistic model?
What is the impact of a negative $r$ in the logistic model?
Tap to reveal answer
Population declines. Negative $r$ reverses the direction of population change.
Population declines. Negative $r$ reverses the direction of population change.
← Didn't Know|Knew It →
What is logistic growth's behavior at $t \to \text{infinity}$?
What is logistic growth's behavior at $t \to \text{infinity}$?
Tap to reveal answer
$P(t) \to K$. As $t \to \infty$, the exponential term vanishes leaving $P = K$.
$P(t) \to K$. As $t \to \infty$, the exponential term vanishes leaving $P = K$.
← Didn't Know|Knew It →
Find the value of $A$ if $P(0) = 5$, $K = 10$, and $r = 0.3$. Use $P(t) = \frac{K}{1 + Ae^{-rt}}$.
Find the value of $A$ if $P(0) = 5$, $K = 10$, and $r = 0.3$. Use $P(t) = \frac{K}{1 + Ae^{-rt}}$.
Tap to reveal answer
$A = 1$. Using $A = \frac{K - P(0)}{P(0)} = \frac{10 - 5}{5} = 1$.
$A = 1$. Using $A = \frac{K - P(0)}{P(0)} = \frac{10 - 5}{5} = 1$.
← Didn't Know|Knew It →
Compute $P(t)$ for $r = 1$, $K = 50$, and $P(0) = 25$. Use logistic solution.
Compute $P(t)$ for $r = 1$, $K = 50$, and $P(0) = 25$. Use logistic solution.
Tap to reveal answer
$P(t) = \frac{50}{1 + e^{-t}}$. With $P(0) = 25 = \frac{K}{2}$, we have $A = 1$.
$P(t) = \frac{50}{1 + e^{-t}}$. With $P(0) = 25 = \frac{K}{2}$, we have $A = 1$.
← Didn't Know|Knew It →
Find $P(t)$ when $r = 0.5$, $K = 100$, and $P(0) = 10$.
Find $P(t)$ when $r = 0.5$, $K = 100$, and $P(0) = 10$.
Tap to reveal answer
$P(t) = \frac{100}{1 + 9e^{-0.5t}}$. Using $A = \frac{K - P(0)}{P(0)} = \frac{90}{10} = 9$.
$P(t) = \frac{100}{1 + 9e^{-0.5t}}$. Using $A = \frac{K - P(0)}{P(0)} = \frac{90}{10} = 9$.
← Didn't Know|Knew It →
Determine the point of inflection for $P(t) = \frac{K}{1 + Ae^{-rt}}$.
Determine the point of inflection for $P(t) = \frac{K}{1 + Ae^{-rt}}$.
Tap to reveal answer
$P = \frac{K}{2}$. Inflection occurs at half the carrying capacity in logistic growth.
$P = \frac{K}{2}$. Inflection occurs at half the carrying capacity in logistic growth.
← Didn't Know|Knew It →
What is the effect of a larger $r$ on the logistic growth curve?
What is the effect of a larger $r$ on the logistic growth curve?
Tap to reveal answer
Faster growth towards $K$. Higher $r$ means the population reaches $K$ more quickly.
Faster growth towards $K$. Higher $r$ means the population reaches $K$ more quickly.
← Didn't Know|Knew It →
Explain the impact of $K$ in the logistic growth equation.
Explain the impact of $K$ in the logistic growth equation.
Tap to reveal answer
Sets upper limit for population. $K$ represents the environmental limit on population size.
Sets upper limit for population. $K$ represents the environmental limit on population size.
← Didn't Know|Knew It →
What is the behavior of the logistic model when $P > K$?
What is the behavior of the logistic model when $P > K$?
Tap to reveal answer
Population decreases. When $P > K$, the factor $(1 - \frac{P}{K})$ becomes negative.
Population decreases. When $P > K$, the factor $(1 - \frac{P}{K})$ becomes negative.
← Didn't Know|Knew It →
State the logistic growth model in terms of $y(t)$ if $y(0) = 1$ and $K = 10$.
State the logistic growth model in terms of $y(t)$ if $y(0) = 1$ and $K = 10$.
Tap to reveal answer
$y(t) = \frac{10}{1 + 9e^{-rt}}$. With $y(0) = 1$ and $K = 10$, we get $A = 9$.
$y(t) = \frac{10}{1 + 9e^{-rt}}$. With $y(0) = 1$ and $K = 10$, we get $A = 9$.
← Didn't Know|Knew It →
Calculate $P(t)$ when $K = 200$, $r = 0.2$, and $P(0) = 100$. Use logistic solution.
Calculate $P(t)$ when $K = 200$, $r = 0.2$, and $P(0) = 100$. Use logistic solution.
Tap to reveal answer
$P(t) = \frac{200}{1 + e^{-0.2t}}$. With $P(0) = 100 = \frac{K}{2}$, we get $A = 1$.
$P(t) = \frac{200}{1 + e^{-0.2t}}$. With $P(0) = 100 = \frac{K}{2}$, we get $A = 1$.
← Didn't Know|Knew It →
What condition leads to logistic growth equilibrium?
What condition leads to logistic growth equilibrium?
Tap to reveal answer
$P = K$. At equilibrium, $\frac{dP}{dt} = 0$, which occurs when $P = K$.
$P = K$. At equilibrium, $\frac{dP}{dt} = 0$, which occurs when $P = K$.
← Didn't Know|Knew It →
Identify the phase where logistic growth is approximately linear.
Identify the phase where logistic growth is approximately linear.
Tap to reveal answer
Near point of inflection. Around the inflection point, growth rate is roughly constant.
Near point of inflection. Around the inflection point, growth rate is roughly constant.
← Didn't Know|Knew It →
What happens to $P(t)$ when $r$ is zero in the logistic model?
What happens to $P(t)$ when $r$ is zero in the logistic model?
Tap to reveal answer
Population remains constant. Zero growth rate means $\frac{dP}{dt} = 0$ for all $t$.
Population remains constant. Zero growth rate means $\frac{dP}{dt} = 0$ for all $t$.
← Didn't Know|Knew It →
Describe the role of $e^{-rt}$ in logistic growth solution.
Describe the role of $e^{-rt}$ in logistic growth solution.
Tap to reveal answer
Determines rate of approach to $K$. The exponential decay controls how quickly $P$ approaches $K$.
Determines rate of approach to $K$. The exponential decay controls how quickly $P$ approaches $K$.
← Didn't Know|Knew It →
Describe the asymptotic behavior of $P(t)$ in logistic growth.
Describe the asymptotic behavior of $P(t)$ in logistic growth.
Tap to reveal answer
Approaches $K$ as $t \to \text{infinity}$. The exponential term vanishes as $t$ increases without bound.
Approaches $K$ as $t \to \text{infinity}$. The exponential term vanishes as $t$ increases without bound.
← Didn't Know|Knew It →
What happens to growth rate as $P$ approaches $K$ in logistic growth?
What happens to growth rate as $P$ approaches $K$ in logistic growth?
Tap to reveal answer
Growth rate approaches zero. The term $(1 - \frac{P}{K})$ approaches zero as $P \to K$.
Growth rate approaches zero. The term $(1 - \frac{P}{K})$ approaches zero as $P \to K$.
← Didn't Know|Knew It →
Identify the inflection point time $t$ for logistic model $P(t) = \frac{K}{1 + Ae^{-rt}}$.
Identify the inflection point time $t$ for logistic model $P(t) = \frac{K}{1 + Ae^{-rt}}$.
Tap to reveal answer
$t = \frac{\text{ln}(A)}{r}$. Inflection occurs when the exponential term $Ae^{-rt} = 1$.
$t = \frac{\text{ln}(A)}{r}$. Inflection occurs when the exponential term $Ae^{-rt} = 1$.
← Didn't Know|Knew It →
What is the logistic growth rate at $P = \frac{K}{2}$?
What is the logistic growth rate at $P = \frac{K}{2}$?
Tap to reveal answer
Maximum growth rate. At the inflection point, $\frac{dP}{dt}$ reaches its maximum value.
Maximum growth rate. At the inflection point, $\frac{dP}{dt}$ reaches its maximum value.
← Didn't Know|Knew It →
Identify the type of growth when $P$ is much less than $K$ in the logistic model.
Identify the type of growth when $P$ is much less than $K$ in the logistic model.
Tap to reveal answer
Approximately exponential growth. When $P << K$, the factor $(1 - \frac{P}{K}) \approx 1$.
Approximately exponential growth. When $P << K$, the factor $(1 - \frac{P}{K}) \approx 1$.
← Didn't Know|Knew It →
Explain the impact of $K$ in the logistic growth equation.
Explain the impact of $K$ in the logistic growth equation.
Tap to reveal answer
Sets upper limit for population. $K$ represents the environmental limit on population size.
Sets upper limit for population. $K$ represents the environmental limit on population size.
← Didn't Know|Knew It →
State the logistic growth model in terms of $y(t)$ if $y(0) = 1$ and $K = 10$.
State the logistic growth model in terms of $y(t)$ if $y(0) = 1$ and $K = 10$.
Tap to reveal answer
$y(t) = \frac{10}{1 + 9e^{-rt}}$. With $y(0) = 1$ and $K = 10$, we get $A = 9$.
$y(t) = \frac{10}{1 + 9e^{-rt}}$. With $y(0) = 1$ and $K = 10$, we get $A = 9$.
← Didn't Know|Knew It →
Identify the phase where logistic growth is approximately linear.
Identify the phase where logistic growth is approximately linear.
Tap to reveal answer
Near point of inflection. Around the inflection point, growth rate is roughly constant.
Near point of inflection. Around the inflection point, growth rate is roughly constant.
← Didn't Know|Knew It →