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Calculus, SPEC2 2023 VCAA 4

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.

  1. The above logistic differential equation can be expressed as
  2. \(\displaystyle \int \frac{A}{P}+\dfrac{B}{1-\dfrac{P}{1000}} d P=\int dt \text {, where } A, B \in R .\)
  3. Find the values of \(A\) and \(B\).  (1 mark)

One form of the solution for \(P\) is  \(P=\dfrac{1000}{1+D e^{-t}}\ \),  where \(D\) is a real constant.

  1. Find the value of \(D\).  (1 mark)

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.

  1. Find the value of \(n\).  (1 mark)

  2. Find the value of \(Q\) when \(t=6\).
  3. Give your answer correct to the nearest integer.  (1 mark)

  4.  i. Given that  \(\dfrac{dQ}{dt}=\dfrac{11}{10} Q\left(1-\dfrac{Q}{1000}\right)\),  express  \(\dfrac{d^2 Q}{d t^2}\)  in terms of \(Q\).  (1 mark)

  5. ii. Hence or otherwise, find the size of the fish population in pond 2 and the value of \(t\) when the rate of growth of the population is a maximum. Give your answer for \(t\) correct to the nearest year.  (2 marks)

  6. Sketch the graph of \(Q\) versus \(t\) on the set of axes below. Label any axis intercepts and any asymptotes with their equations.  (2 marks)
     

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

\(\dfrac{d Q}{d t}=\dfrac{11}{10}\, Q\left(1-\dfrac{Q}{1000}\right)-0.055Q\)

  1. Find the maximum number of fish that could be supported in pond 2 in this situation.  (1 mark)

Show Answers Only

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.  \(Q=950\)

Show Worked Solution

a.    \(\dfrac{dP}{dt}=P\left(1-\dfrac{P}{1000}\right) \ \Rightarrow \ \dfrac{d t}{d P}=\dfrac{1}{P}\left(\dfrac{1}{1-\frac{P}{1000}}\right)\)

\(\text{Expand (by CAS):}\)

\(\dfrac{1}{P}\left(\dfrac{1}{1-\frac{P}{1000}}\right)=\dfrac{1}{P}-\dfrac{1}{(P-1000)}=\dfrac{1}{P}+\dfrac{1}{1000}\left(\dfrac{1}{1-\frac{P}{1000}}\right)\)

\(A=1, \quad B=\dfrac{1}{1000}\)
 

♦ Mean mark (a) 48%.

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}
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}

 

♦♦♦ Mean mark (e)(i) 21%.

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.  \(\text{Solve for } Q \ \text{(by CAS):}\)

\(\dfrac{d Q}{d t}=\dfrac{11}{10}\, Q\left(1-\dfrac{Q}{1000}\right)-0.055\,Q=0\)

\(\Rightarrow Q=950\)

♦ Mean mark (g) 40%.

Calculus, SPEC2 2019 VCAA 3

  1. 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`.

    1. 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)

    2. Specify the condition(s) for which  `k >0`.   (2 marks)

  2. 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`.

    1. Express this differential equation in the form  `int f(Q)\ dQ = int h(t)\ dt`.   (1 mark)

    2. Hence, show that  `Q = log_e(e^t + e-1)`.   (2 marks)

    3. Show that the graph of `Q` as a function of `t` does not have a point of inflection.   (2 marks)

Show Answers Only
    1. `text(See Worked Solutions)`
    2. `a-b\ text(and)\ s > r > 0`
    1. `int e^Q\ dQ =  int e^t\ dt`
    2. `text(See Worked Solutions)`
    3. `text(See Worked Solutions)`
Show Worked Solution
a.i.    `(dP)/(dt)` `= kP`
  `(dt)/(dP)` `= 1/(kP)`

`t = 1/k int 1/P\ dP = 1/k log_e P + c`

`P(a) = r`

`a = 1/k · log_e r + c`

`P(b) = r`

`b = 1/k · log_e s + c`

`a-b` `= 1/k(log_e r + c-log_e s + c)`
`:.k` `= 1/(a-b)log_e (r/s)`

 

a.ii.   `text(Conditions for)\ \ k > 0,`

`text(Scenario 1:)`

`a-b > 0\ \ text(and)\ \ log_e(r/s) > 0`

`=> a > b\ \ text(and)\ \ r > s > 0`
 

`text(Scenario 2:)`

`a-b < 0\ \ text(and)\ \ log_e(r/s) < 0`

`=> a < b\ \ text(and)\ \ s > r > 0`
 

b.i.    `(dQ)/(dt)` `= (e^t)/(e^Q)`
  `int e^Q dQ` `= int e^t\ dt`

 

b.ii.    `int e^Q dQ` `= int e^t dt`
  `e^Q` `= e^t + c`

 
`text(When)\ \ Q = 1,  t = 0`

`e` `= e^0 + c`
`c` `= e-1`
`e^Q` `= e^t + e-1`

 
`:. Q = log_e(e^t + e-1)`

 

b.iii.   `Q = log_e(e^t + e-1)`

`(dQ)/(dt)` `= (e^t)/(e^t + e-1)`
`(d^2Q)/(dt^2)` `= (e^t(e-1))/((e^t + e-1)^2)`

 
`(d^2Q)/(dt^2) > 0\ text(for)\ t >= 0`

`:.\ text(No POI)`

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.

    1. Express  `2/(P(1-P))`  in partial fraction form.   (1 mark)

    2. Hence show by integration that  `(t-c)/2= log_e(P/(1-P))`, where  `c`  is a constant of integration.   (2 marks)

    3. 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`

  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)

  2. 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)

  1. 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)

Show Answers Only
    1. `2/(P(1-P)) = 2/P + 2/(1-P)`
    2. `text(Proof)\ text{(See Worked Solutions)}`
    3. `P = e^(t/2)/(1 + e^(t/2))`
  2. `P ~~ 0.894`
  3. `r = 0.8`
    `s = 1`
    `q ~~ 0.62`
  4. `~~ 0.775`
Show Worked Solution

a.i.   `2/(P(1-P)) = A/P + B/(1 ⋅ P)`

`A(1-P) + BP = 2`

`text(If)\ \ P = 0\ \ =>\ \ A = 2`

`text(If)\ \ P = 1\ \ =>\ \ B = 2`

`:. 2/(P(1-P)) = 2/P + 2/(1-P)\ \ \ text{(can also solve by CAS)}`

 

a.ii.  `(dt)/(dP) = 2/(P(1-P)) = 2/P + 2/(1-P)`

`t` `= int 2/P + 2/(1-P)\ dP`
  `= 2 ln |P|-2ln|1-P| + c`
`(t-c)/2` `=ln|P|-ln|1-P|`
  `=ln |(P)/(1-P)|`
  `= ln (P/(1-P))`

  
`text(S)text(ince)\ \ 0 < P < 1 :\ |P| = P\ and\ |1-P| = 1-P`


a.iii.  `text(When)\ \ t=0,\ P=0.5`

`(-c)/2` `= ln (0.5/0.5)`
`c` `= ln (1) = 0`

 

`t/2` `= ln (P/(1-P))\ \ \ text{(solve manually or by CAS)}`
`e^(t/2)` `= P/(1-P)`
`e^(t/2) (1-P)` `= P`
`e^(t/2)-Pe^(t/2)` `= P`
`e^(t/2)` `= P(1 + e^(t/2))`
`:. P` `= e^(t/2)/(1 + e^(t/2))`

 

b.   `(dP)/(dt) = P/2 (1-P)-sqrt P/20`

`text(Limiting value occurs when)\ \ (dP)/(dt) = 0,`

`P ~~ 0.894\ \ \ text{(by CAS)}`
 

`=>\ text(Lower solution values at levels already exceeded)`

 `text(are ignored.)`

 

c.   `(dt)/(dP) = 1/(P/2 (1-P)-sqrt P/20)\ \ text(for)\ \ t>=1`

`text(When)\ \ t=1\ \ =>\ \ P=0.622`
 

`T = int_0.62^0.8 1/(P/2 (1-P)-sqrt P/20) dP + 1`
 

`:. r = 0.8,\ s = 1 and q = 0.62`

 

d.   `P(3.5) ~~ P(3) + h* (dP)/(dt)|_(P= 0.75)`

   `= 0.75 + 0.5 (0.75/2 (1-0.75)-sqrt 0.75/20)`

   `~~ 0.775`
 

`:. P~~0.775\ \ text(when)\ \ t=3.5`