📝 Logistic growth curves calculus (21 MCQs)

📖 Explanation: Because the inflection time is $t_{\text{inf}}=\frac{1}{k}\ln A$. When $k$ grows, the denominator becomes larger, making the whole expression smaller, so the inflection occurs at a smaller (earlier) time.

📖 Explanation: The inflection time is $t_{\text{inf}}=\frac{\ln A}{k}$. A larger $A$ produces a larger numerator, so $t_{\text{inf}}$ increases, meaning the inflection point moves to a later time.

📖 Explanation: The second derivative is $k^{2}L^{2}y(L-y)(L-2y)$. Since $y=0.4L<L/2$, the factor $(L-2y)$ is positive, making the second derivative positive, which indicates the curve is concave up.

📖 Explanation: The horizontal asymptote of the logistic function is $y=L$. Decreasing $L$ therefore lowers the asymptote, shifting it downward on the $y$-axis.

📖 Explanation: Both models converge to the same limiting value $L$; however, the larger growth constant $k_{2}$ accelerates the approach, so the population with $k_{2}$ reaches values near $L$ more quickly.

📖 Explanation: All factors in the expression are positive: $k>0$, $L>0$, $y>0$, and $L-y>0$. The product of positive numbers is positive, so $\frac{dy}{dt}>0$ for any admissible $y$.

📖 Explanation: The exponential function continues to increase indefinitely, whereas the logistic function has a horizontal asymptote at $y=L$; thus, for large $t$ the logistic curve levels off while the exponential keeps rising.

📖 Explanation: The factor $k$ appears in the exponent and also multiplies the derivative. A larger $k$ accelerates the transition from low to high values, sharpening the S‑shape and increasing the slope at the inflection point.

📖 Explanation: At $t=0$, $y(0)=\frac{L}{1+A}$. With $0<A<1$, the denominator lies between 1 and 2, so $y(0)$ lies between $L/2$ and $L$, i.e., it is greater than $L/2$.

📖 Explanation: The sign change occurs when the factor $(L-2y)$ equals zero, which gives $y=\frac{L}{2}$. For $y<L/2$ the second derivative is positive (concave up), and for $y>L/2$ it becomes negative (concave down).

📖 Explanation: As $t\to\infty$, the term $e^{-kt}$ tends to zero, so the denominator approaches $1$. The limit therefore simplifies to $\frac{L}{1}=L$.

📖 Explanation: Setting $\frac{L}{1+Ae^{-kt}}=\frac{3L}{4}$ gives $Ae^{-kt}=1/3$. Solving for $t$ yields $t=\frac{1}{k}\ln(3A)$, which is the expression in option D.

📖 Explanation: When $A<1$ the denominator $1+Ae^{-kt}$ is smaller for early times, giving larger values of $y$ and moving the curve leftward. Conversely, $A>1$ yields a smaller early $y$ and shifts the curve right.

📖 Explanation: In the logistic model, $L$ is the horizontal asymptote and denotes the carrying capacity—the largest population that the environment can sustain over the long term.

📖 Explanation: The term $(L-2y)$ in $\frac{d^{2}y}{dt^{2}}$ determines the sign of the curvature. When $y=L/2$, this factor equals zero, causing the second derivative to change sign, which defines an inflection point.

📖 Explanation: The factor $(L-y)$ becomes smaller as the population nears $L$, diminishing the overall growth rate. When $y$ equals $L$, the factor is zero, stopping growth entirely and thus enforcing the limit.

📖 Explanation: The derivative $\frac{dy}{dt}=kL\,y(L-y)$ is a quadratic in $y$ that attains its maximum at the vertex, which occurs at $y=\frac{L}{2}$.

📖 Explanation: The inflection time is $t_{\text{inf}}=\frac{\ln A}{k}$. Doubling $k$ divides the denominator by two, so the time to reach $L/2$ is reduced by a factor of two.

📖 Explanation: Compute $y(5)=\frac{1000}{1+4e^{-0.3\cdot5}}=\frac{1000}{1+4e^{-1.5}}\approx\frac{1000}{1+0.8925}=528.6$, which rounds to 529.

📖 Explanation: The logistic curve starts with near‑exponential increase when the population is small, then environmental limits cause the growth rate to decline, producing the familiar S‑shaped pattern.

📖 Explanation: The classic logistic function is written as $y(t)=\frac{L}{1+Ae^{-kt}}$, where $L$ is the carrying capacity, $A$ reflects the initial condition, and $k$ is the growth rate constant.