A fish farmer releases 200 fish into a pond that originally contained no fish. The fish population, \(P\), grows according to the logistic model, \(\dfrac{d P}{d t}=P\left(1-\dfrac{P}{1000}\right)\) , where \(t\) is the time in years after the release of the 200 fish. --- 2 WORK AREA LINES (style=lined) --- One form of the solution for \(P\) is \(P=\dfrac{1000}{1+D e^{-t}}\ \), where \(D\) is a real constant. --- 2 WORK AREA LINES (style=lined) --- The farmer releases a batch of \(n\) fish into a second pond, pond 2 , which originally contained no fish. The population, \(Q\), of fish in pond 2 can be modelled by \(Q=\dfrac{1000}{1+9 e^{-1.1 t}}\), where \(t\) is the time in years after the \(n\) fish are released. --- 2 WORK AREA LINES (style=lined) --- --- 2 WORK AREA LINES (style=lined) --- --- 4 WORK AREA LINES (style=lined) --- --- 4 WORK AREA LINES (style=lined) --- --- 0 WORK AREA LINES (style=lined) --- The farmer wishes to take 5.5% of the fish from pond 2 each year. The modified logistic differential equation that would model the fish population, \(Q\), in pond 2 after \(t\) years in this situation is \(\dfrac{d Q}{d t}=\dfrac{11}{10}\, Q\left(1-\dfrac{Q}{1000}\right)-0.055Q\) --- 4 WORK AREA LINES (style=lined) --- a. \(A=1, \quad B=\dfrac{1}{1000}\) b. \(\Rightarrow D=4\) c. \(n=\dfrac{1000}{1+9 e^0}=100\) d. \(Q=\dfrac{1000}{1+9 e^{-6.6}} \approx 988\) e.i. \(Q^{\prime}=\frac{121}{100}\, Q\left(1-\frac{Q}{1000}\right)^2-\frac{121}{100\ 000}\left(1-\frac{Q}{1000}\right)\) e.ii. \(t=2\) f. g. \(Q=950\) a. \(\dfrac{dP}{dt}=P\left(1-\dfrac{P}{1000}\right) \ \Rightarrow \ \dfrac{d t}{d P}=\dfrac{1}{P}\left(\dfrac{1}{1-\frac{P}{1000}}\right)\) \(\text{Expand (by CAS):}\) \(\dfrac{1}{P}\left(\dfrac{1}{1-\frac{P}{1000}}\right)=\dfrac{1}{P}-\dfrac{1}{(P-1000)}=\dfrac{1}{P}+\dfrac{1}{1000}\left(\dfrac{1}{1-\frac{P}{1000}}\right)\) \(A=1, \quad B=\dfrac{1}{1000}\) b. \(\text{When}\ \ t=0, P=200 \text{ (given)}\) \(\text{Solve}\ \ 200=\dfrac{1000}{1+D e^0}\ \ \text{for}\ t:\) \(\Rightarrow D=4\ \ \text {(by CAS):}\) c. \(Q=\dfrac{1000}{1+9 e^{-1.1 t}}\) \(\text{At}\ \ t=0, Q=n\): \(n=\dfrac{1000}{1+9 e^0}=100\) d. \(\text{Find } Q \text{ when } t=6:\) \(Q=\dfrac{1000}{1+9 e^{-6.6}} \approx 988\) e.i. \(\dfrac{d Q}{d t}=\dfrac{11}{10}\, Q\left(1-\dfrac{Q}{1000}\right)\) \(\text{Using the product rule:}\) \begin{aligned} e.ii. \(\text{Max } Q^{\prime} \Rightarrow Q^{\prime \prime}=0\) \(\text{Solve } Q^{\prime \prime}=0 \ \ \text {(by CAS):}\) \(\Rightarrow Q=500\) \(\text{Solve } Q=500 \text{ for } t \text{ (by CAS):}\) \(t=1.99 \ldots=2 \ \text{(nearest year)}\) f. g. \(\text{Solve for } Q \ \text{(by CAS):}\) \(\dfrac{d Q}{d t}=\dfrac{11}{10}\, Q\left(1-\dfrac{Q}{1000}\right)-0.055\,Q=0\) \(\Rightarrow Q=950\)
Q^{\prime \prime} & =\frac{11}{10}\left[Q^{\prime}\left(1-\frac{Q}{1000}\right)+Q\left(-\frac{1}{1000}\, Q^{\prime}\right)\right] \\
& =\frac{11}{10}\left[\frac{11}{10}\, Q\left(1-\frac{Q}{1000}\right)^2-\frac{Q}{1000}\left(\frac{11}{10}\, Q\left(1-\frac{Q}{1000}\right)\right)\right] \\
& =\frac{121}{100}\, Q\left(1-\frac{Q}{1000}\right)^2-\frac{121}{100\ 000}\left(1-\frac{Q}{1000}\right)
\end{aligned}
Calculus, SPEC2 2019 VCAA 3
- The growth and decay of a quantity `P` with respect to time `t` is modelled by the differential equation
`qquad qquad(dP)/(dt) = kP`
where `t >= 0`.- Given that `P(a) = r` and `P(b) = s`, where `P` is a function of `t`,
show that `k = 1/(a - b)log_e(r/s)`. (2 marks)
- Specify the condition(s) for which `k >0`. (2 marks)
- Given that `P(a) = r` and `P(b) = s`, where `P` is a function of `t`,
- The growth of another quantity `Q` with respect to time `t` is modelled by the differential equation
`qquad qquad (dQ)/(dt) = e^(t - Q)`
where `t >= 0` and `Q = 1` when `t = 0`.- Express this differential equation in the form `int f(Q)\ dQ = int h(t)\ dt`. (1 mark)
- Hence, show that `Q = log_e(e^t + e - 1)`. (2 marks)
- Show that the graph of `Q` as a function of `t` does not have a point of inflection. (2 marks)
Calculus, SPEC2-NHT 2017 VCAA 3
Bacteria are spreading over a Petri dish at a rate modelled by the differential equation
`(dP)/(dt) = P/2 (1 - P),\ 0 < P < 1`
where `P` is the proportion of the dish covered after `t` hours.
-
- Express `2/(P(1 - P))` in partial fraction form. (1 mark)
- Hence show by integration that `(t - c)/2= log_e(P/(1 - P))`, where `c` is a constant of integration. (2 marks)
- If half of the Petri dish is covered by the bacteria at `t = 0`, express `P` in terms of `t`. (2 marks)
After one hour, a toxin is added to the Petri dish, which harms the bacteria and reduces their rate of growth. The differential equation that models the rate of growth is now
`(dP)/(dt) = P/2 (1 - P) - sqrt P/20` for `t >= 1`
- Find the limiting value of `P`, which is the maximum possible proportion of the Petri dish that can now be covered by the bacteria. Give your answer correct to three decimal places. (2 marks)
- The total time, `T` hours, measured from time `t = 0`, needed for the bacteria to cover 80% of the Petri dish is given by
`qquad qquad T = int_q^r (1/(P/2(1 - P) - sqrt P/20)) dP + s`
where `q, r and s in R`.
Find the values of `q, r` and `s`, giving the value of `q` correct to two decimal places. (2 marks)
- Given that `P = 0.75` when `t = 3`, use Euler’s method with a step size of 0.5 to estimate the value of `P` when `t = 3.5`. Give your answer correct to three decimal places. (3 marks)