Study Notes on Population Growth and Parametric Equations

A population model is described by the differential equation:
$\frac{dy}{dt} = 0.2y(1000 - y)$
- Here, (y) represents the population, and (t) represents time measured in years.
This is a Logistic Growth Model which indicates that the population growth rate depends on both the current population size and the carrying capacity of the environment (1000 in this case).

To determine the values of (y) for which the population is increasing at a decreasing rate, we need to analyze the differential equation.
The growth rate is given by (\frac{dy}{dt} = 0.2y(1000-y)).
Population is increasing when (\frac{dy}{dt} > 0), which requires:
 - (0.2y(1000 - y) > 0)
 - This inequality means both (y > 0) and ((1000 - y) > 0) must hold true, leading to the following constraint:
 - (0 < y < 1000)
The rate of population growth decreases when (\frac{d^2y}{dt^2} < 0).
We calculate (\frac{d^2y}{dt^2}) based on the first derivative:
  - Using the product rule from the first derivative (\frac{dy}{dt} = 0.2y(1000-y)):
    - Let (P(y) = 0.2y(1000 - y))
    - Differentiate:
       $\frac{d^2y}{dt^2} = 0.2(1000-2y)$
Setting (\frac{d^2y}{dt^2} < 0) gives: - 1000 - 2y < 0 - Solving yields: - 2y > 1000
- y > 500

Therefore, the population is increasing at a decreasing rate for (500 < y < 1000).
Thus, the correct choice is:
- (C) 500 < y < 1000 only.

For the parametric equations defined by:
- (x(t) = 2 + 3t)
- (y(t) = 1 + t^2)
To find the length of the path described by these equations from (t = 0) to (t = 1), we use the formula for the length of a parametric curve:
$L = \int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt$
First, we calculate (\frac{dx}{dt}) and (\frac{dy}{dt}):
- (\frac{dx}{dt} = 3)
- (\frac{dy}{dt} = 2t)
Next, we substitute into the length formula:
  - $L = \int_{0}^{1} \sqrt{(3)^2 + (2t)^2} \, dt$
  - Which simplifies to:
  - $L = \int_{0}^{1} \sqrt{9 + 4t^2} \, dt$
The evaluation of this integral will give the total length of the path from (t=0) to (t=1).
Thus, the correct approach yields the integral that provides the length of the parametric curve described.

To differentiate the expression $\frac{dy}{dt} = 0.2y(1000 - y)$ , follow these steps:

Identify the function: The function can be represented as $y = f(t)$ , which is a relationship between $y$ and $t$ .
Write the equation clearly: The given equation states that the rate of change of population over time is proportional to the current population size and the difference between the carrying capacity and the current population size.
Differentiate using the product rule: Apply the product rule of differentiation because the right-hand side is a product of two functions: $0.2$ , $y$ , and (1000 - y)$. The product rule states that if you have two functions u $and$ v $, their derivative is given by$ \frac{d(uv)}{dt} = u\frac{dv}{dt} + v\frac{du}{dt} $.<ul><li>Here, let$ u = 0.2y $and$ v = (1000 - y) $.</li><li>Now differentiate:<ul><li>$ \frac{du}{dt} = 0.2\frac{dy}{dt} $</li><li>$ \frac{dv}{dt} = -\frac{dy}{dt} $</li></ul></li></ul></li><li>Apply the product rule: Substituting into the product rule gives: $ \frac{d}{dt}(0.2y(1000 - y)) = 0.2(1000 - y)\frac{dy}{dt} - 0.2y\frac{dy}{dt} $.</li><li>Setting the equation to zero: After differentiation, you set the equation equal to zero to find critical points $ 0.2(1000 - y)\frac{dy}{dt} - 0.2y\frac{dy}{dt} = 0$$
This confirms where population growth is stable or at a rate of change.
Interpretation: The resulting equations from differentiation help identify conditions under which the population is increasing or decreasing, providing insights into the behavior of the population under the modeled conditions.