SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

  • Login
  • Get Help
  • About

Calculus, EXT1 C3 2024 HSC 13a

In an experiment, the population of insects, \(P(t)\), was modelled by the logistic differential equation

\(\dfrac{d P}{d t}=P(2000-P)\)

where \(t\) is the time in days after the beginning of the experiment.

The diagram shows a direction field for this differential equation, with the point \(S\) representing the initial population.
 

  1. Explain why the graph of the solution that passes through the point \(S\) cannot also pass through the point \(T\).   (1 mark)

    --- 2 WORK AREA LINES (style=lined) ---

  2. Clearly sketch the graph of the solution that passes through the point \(S\).   (1 mark)

    --- 5 WORK AREA LINES (style=blank) ---

  3. Find the predicted value of the population, \(P(t)\), at which the rate of growth of the population is largest.   (2 marks)

    --- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

 i.    \(\text{Any solution that passes through}\ S\ \text{will follow the}\)

\(\text{slope field (not crossing any lines) and approach a horizontal}\)

\(\text{asymptote at}\ P=2000\ \text{from the lower side}\ (P<2000).\)

ii.    
       

iii.  \(P= 1000\)

Show Worked Solution

 i.    \(\text{Any solution that passes through}\ S\ \text{will follow the}\)

\(\text{slope field (not crossing any lines) and approach a horizontal}\)

\(\text{asymptote at}\ P=2000\ \text{from the lower side}\ (P<2000).\)

ii.    
       

iii.  \(\dfrac{d P}{d t}=P(2000-P)\)

\(\text {Find \(P\) where  \(\dfrac{d P}{d t}\)  is a maximum.}\)

\(\text{Consider the graph}\ \ y=P(2000-P): \)

\(\Rightarrow \ \text {Graph is a concave down quadratic cutting at}\ \ P=0\ \ \text{and}\ \ P=2000\)

\(\Rightarrow \ \text{Max value of}\ \ P(2000-P)\ \ \Big(\text{i.e.}\ \dfrac{dP}{dt}\Big)\ \ \text{occurs at}\ \ P=1000\ \text{(axis).}\)

♦ Mean mark (iii) 51%.

Filed Under: Applications of Differential Equations Tagged With: Band 3, Band 4, Band 5, smc-1198-30-Quantity

Calculus, EXT1 C3 EQ-Bank 8

A researcher estimates the number of brumbies in a National Park after `t` years can be modelled by the equation

`B(t)=(18\ 000)/(1+4e^(-t))`

  1. Sketch the function `B(t)` over the first four years of the research.   (2 marks)

    --- 8 WORK AREA LINES (style=lined) ---

  2. Calculate when the brumby population should reach 13 000, giving your answer to 2 decimal places.   (1 mark)

    --- 6 WORK AREA LINES (style=lined) ---

  3. Show that  `B^(′)(t)=(72\ 000e^t)/(e^t+4)^2`   (2 marks)

    --- 6 WORK AREA LINES (style=lined) ---

  4. What is the maximum growth rate of the brumby population?   (3 marks)

    --- 10 WORK AREA LINES (style=lined) ---

Show Answers Only
  1.  
     
  2. `2.34\ text{years}`
  3. `text{Proof (See Worked Solutions}`
  4. `4500\ text{brumbies per year}`
Show Worked Solution

i.   

ii.   `text {Find}\ t\ text{when}\ \ B(t)=13\ 000`

`13\ 000` `=(18\ 000)/(1+4e^(-t))`  
`13\ 000(1+4e^(-t))` `=18\ 000`  
`1+4e^(-t)` `=18/13`  
`4e^(-t)` `=5/13`  
`e^(-t)` `=5/52`  
`-t` `=ln(5/52)`  
`t` `=2.34\ text{years (2 d.p.)}`  

 

 iii.   `text{Show}\ \ B^(′)(t)=(72\ 000e^t)/(e^t+4)^2`

`B(t)=18\ 000(1+4e^(-t))^(-1)`

`B^(′)(t)` `=-1*-1*4e^(-t)*18\ 000(1+4e^(-t))^-2`  
  `=(72\ 000)/(e^t(1+4e^(-t))^2)`  
  `=(72\ 000)/(e^t(1+4/e^t)^2)`  
  `=(72\ 000)/(e^t((e^t+4)/e^t)^2)`  
  `=(72\ 000)/(e^t/(e^t)^2*(e^t+4)^2)`  
  `=(72\ 000e^t)/(e^t+4)^2\ \ text{… as required}`  

 

iv.  `B^(′)(t)=72\ 000e^t(e^t+4)^(-2)`

`text{Using product rule:}`

`B^(′′)(t)` `=72\ 000e^t(e^t+4)^(-2)+(-2e^t)(e^t+4)^(-3)72\ 000e^t`  
  `=72\ 000e^t(1/(e^t+4)^2-(2e^t)/(e^t+4)^3)`  
  `=72\ 000e^t((e^t+4-2e^t)/(e^t+4)^3)`  
  `=72\ 000e^t((4-e^t)/(e^t+4)^3)`  

 
`text{Find}\ t\ text{when}\ \ B^(′′)(t)=0:`

`4-e^t` `=0`  
`e^t` `=4`  
`t` `=ln4`  
  `=1.386…\ text{years}`  

 
`text{Checking concavity changes:}`

`text{Since}\ e^t>0, (e^t+4)^3>0\ \ text{for all}\ t:`

`text{At}\ t=1, 4-e^1=1.28>0\ \ =>\ \ B^(′′)(1)>0`

`text{At}\ t=2, 4-e^2=-3.4<0\ \ =>\ \ B^(′′)(2)<0`

 
`B^(′)(ln4)=\ text{Max growth rate}`

`B^(′)(ln4)` `=(72\ 000e^(ln4))/(e^(ln4)+4)^2`  
  `=(72\ 000xx4)/((4+4)^2)`  
  `=4500\ text{brumbies per year}`  

Filed Under: Applications of Differential Equations Tagged With: Band 3, Band 4, Band 5, smc-1198-30-Quantity

Calculus, EXT1 C3 EQ-Bank 6

The population of Myna birds in a national park is decreasing at a rate proportional to the population at that time.

  1. Write a differential equation that describes the situation.   (1 mark)

    --- 2 WORK AREA LINES (style=lined) ---

  2. If the population was originally 1300 and decreased to 1040 after 5 years, find the expected population after 10 years.  (3 marks)

    --- 8 WORK AREA LINES (style=lined) ---

Show Answers Only
  1. `(dN)/(dt) = -kN`
  2. `832`
Show Worked Solution

i.   `text{Let}\ (dN)/(dt)=\ text{rate of change of bird population at time}\ t\ text{years.}`

`(dN)/(dt) prop N`

`(dN)/(dt) = -kN\ \ (k>0,\ text{decreasing population)}`
 

ii.    `(dN)/(dt)` `=-kN`
  `1/N* (dN)/(dt)` `=-k`
  `int1/N\ dN` `=-intk\ dt`
  `ln absN` `=-kt+c`
  `N` `=e^(-kt+c)`
    `=e^(-kt)*e^c`
    `=Ae^(-kt)\ \ \ text{(where}\ A=e^c)`

 
`text{When}\ \ t=0, N=1300:`

`1300=Ae^0\ \ =>\ \ A=1300`

`N=1300e^(-kt)`
 

`text{When}\ \ t=5, N=1040:`

`1040` `=1300e^(-5k)`  
`e^(-5k)` `=1040/1300`  
`-5k` `=ln(0.8)`  
`k` `=-(ln(0.8))/(5)`  
  `=0.04462…`  

 
`text{Find}\ N\ text{when}\ \ t=10:`

`N` `=1300e^(-0.04462 xx 10)`  
  `=832\ \ text{myna birds}`  

Filed Under: Applications of Differential Equations Tagged With: Band 3, Band 4, smc-1198-30-Quantity

Calculus, EXT1 C3 EQ-Bank 4

A varroa virus is infecting commercial beehives in a regional NSW town.

All infected hives detected so far lie within a circular region with radius 16 km and researchers believe that the increase of the radius `r` km can be modelled by a differential equation, where `(dr)/(dt)=2/5sqrtr` where `t` denotes the time in months.

What does this model predict for the radius of the region affected by the pest after `t` months?   (3 marks)

--- 8 WORK AREA LINES (style=lined) ---

Show Answers Only

`r=((t+20)/5)^2`

Show Worked Solution
`(dr)/(dt)` `=2/5sqrtr`  
`5/(2sqrtr)\ dr` `=1\ dt`  
`5/2r^(-1/2)\ dr` `=1\ dt`  
`5/2intr^(-1/2)\ dr` `=int1\ dt`  
`5r^(1/2)` `=t+C`  

 

`text{When}\ \ t=0, r=16`

`5sqrt16=C\ \ =>\ \ C=20`

`5r^(1/2)` `=t+20`  
`r^(1/2)` `=(t+20)/5`  
`:.r` `=((t+20)/5)^2`  

Filed Under: Applications of Differential Equations Tagged With: Band 4, smc-1198-30-Quantity

Calculus, EXT1 C3 2021 HSC 14b

In a certain country, the population of deer was estimated in 1980 to be 150 000.

The population growth is given by the logistic equation  `(dP)/(dt) = 0.1P((C - P)/C)`  where `t`  is the number of years after 1980 and `C` is the carrying capacity.

In the year 2000, the population of deer was estimated to be 600 000.

Use the fact that  `C/(P(C - P)) = 1/P + 1/(C - P)`  to show that the carrying capacity is approximately 1 130 000.  (4 marks)

Show Answers Only

`text(See Worked Solution)`

Show Worked Solution

`(dP)/(dt) = 0.1P((C – P)/C)`

♦ Mean mark 45%.
`(dt)/(dP)` `= 10/P(C/(C – P))`
  `= 10 xx C/(P(C – P))`
  `= 10(1/P + 1/(C – P))`

 

`t` `= 10 int 1/P + 1/(C – P)\ dP`
  `= 10[ln P – ln(C – P)] + c`
  `= 10 ln(P/(C – P)) + c`

 

`text(When)\ \ t= 0, P = 150\ 000:`

`0` `= 10 ln((150\ 000)/(C – 150\ 000)) + c`
`c` `= -10 ln((150\ 000)/(C – 150\ 000))`
  `= 10 ln((C – 150\ 000)/(150\ 000))`

 

`text(When)\ \ t = 20, P = 600\ 000:`

`20` `= 10 ln((600\ 000)/(C – 600\ 000)) + 10 ln((C – 150\ 000)/(150\ 000))`
`2` `= ln((600\ 000)/(C – 600\ 000) xx (C – 150\ 000)/(150\ 000))`
`2` `= ln((4C – 600\ 000)/(C – 600\ 000))`
`e^2` `= (4C – 600\ 000)/(C – 600\ 000)`
`e^2(C – 600\ 000)` `= 4C – 600\ 000`
`e^2C – 4C` `= e^2* 600\ 000 – 600\ 000`
`C(e^2 – 4)` `= 600\ 000(e^2 – 1)`
`C` `= (600\ 000(e^2 – 1))/(e^2 – 4)`
  `~~ 1\ 131\ 121`

 
`:.\ text(Carrying capacity)\ ~~1\ 130\ 000`

Filed Under: Applications of Differential Equations Tagged With: Band 5, smc-1198-30-Quantity

Calculus, EXT1 C3 SM-Bank 5

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.

Given  `2/(P(1 - P)) = 2/P + 2/(1 - P),`

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

    --- 5 WORK AREA LINES (style=lined) ---

  2. If half of the Petri dish is covered by the bacteria at  `t = 0`, express  `P`  in terms of  `t`.  (2 marks)

    --- 6 WORK AREA LINES (style=lined) ---

Show Answers Only
  1. `text(Proof)\ text{(See Worked Solutions)}`
  2. `P = e^(t/2)/(1 + e^(t/2))`
Show Worked Solution

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

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

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


b.
  `text(When)\ \ t=0,\ P=0.5`

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

 

`t/2` `= ln (P/(1 – P))`
`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))`

Filed Under: Applications of Differential Equations Tagged With: Band 4, smc-1198-30-Quantity

Copyright © 2014–2025 SmarterEd.com.au · Log in