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Calculus, EXT1* C1 2019 HSC 12c

The number of leaves, `L(t)`, on a tree `t` days after the start of autumn can be modelled by

`L(t) = 200\ 000e^(-0.14t)`

  1. What is the number of leaves on the tree when  `t = 31`?  (1 mark)

  2. What is the rate of change of the number of leaves on the tree when  `t = 31`?  (2 marks)

  3. For what value of `t` are there 100 leaves on the tree?  (2 marks)

Show Answers Only
  1. `2607\ text(leaves)`
  2. `-365.02…`
  3. `54.3\ text{(1 d.p.)}`
Show Worked Solution

i.  `text(When)\ \ t = 31`

`L(t)` `= 200\ 000 xx e^(-0.14(31))`
  `=2607.305…`
  `= 2607\ text(leaves)`

 

ii.    `L` `= 2000\ 000^(-0.14t)`
  `(dL)/(dt)` `= -0.14 xx 200\ 000 e^(0.14t`
    `= -28\ 000e^(-0.14t)`

 
`text(When)\ \ t = 31,`

`(dL)/(dt)` `= -28\ 000 xx e^(-0.14(31))`
  `= -365.02…`

 
`:.\ text(365 leaves fall per day.)`

 

iii.   `text(Find)\ t\ text(when)\ \ L = 100:`

`100` `= 200\ 000 e^(-0.14t)`
`e^(-0.14t)` `= 0.0005`
`e^(-0.14t)` `= ln 0.0005`
`t` `= (ln 0.0005)/(-0.14)`
  `= 54.292…`
  `= 54.3\ text{(1 d.p.)}`

Calculus, EXT1* C1 2018 HSC 5 MC

The diagram shows the number of penguins, `P(t)`, on an island at time `t`.
  


  

Which equation best represents this graph?

A.     `P(t) = 1500 + 1500e^(-kt)`

B.     `P(t) = 3000 - 1500e^(-kt)`

C.     `P(t) = 3000 + 1500e^(-kt)`

D.     `P(t) = 4500 - 1500e^(-kt)`

Show Answers Only

`A`

Show Worked Solution

`P(0) = 3000\ text{(from graph) → Eliminate B and C}`
 

`text(As),\ t -> oo,\ 1500e^(-kt) -> 0,`

`:. 1500 + 1500e^(-kt) => 1500`

`=>  A`

Calculus, EXT1* C1 2017 HSC 14c

Carbon-14 is a radioactive substance that decays over time. The amount of carbon-14 present in a kangaroo bone is given by

`C(t) = Ae^(kt),`

where `A` and `k` are constants, and `t` is the number of years since the kangaroo died.

  1. Show that `C(t)` satisfies  `(dC)/(dt) = kC`.  (1 mark)

  2. After 5730 years, half of the original amount of carbon-14 is present.

     

    Show that the value of `k`, correct to 2 significant figures, is – 0.00012.  (2 marks)

  3. The amount of carbon-14 now present in a kangaroo bone is 90% of the original amount.

     

    Find the number of years since the kangaroo died. Give your answer correct to 2 significant  figures.  (2 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `870\ text{years (2 sig. fig.)}`
Show Worked Solution
i.   `C` `= Ae^(kt)`
   `(dC)/(dt)` `= k * Ae^(kt)`
    `= kC …\ text(as required)`

 

ii.  `text(When)\ \ t = 5730, qquad A = 0.5 A_0`

`0.5 A_0` `= A_0 * e^(5730 k)`
`e^(5730 k)` `= 0.5`
`text(ln)\ e^(5730 k)` `= text(ln)\ 0.5`
`5730 k` `= text(ln)\ 0.5`
`k` `= {text(ln)\ 0.5}/5730`
  `= -0.0001209…`
  `= -0.00012\ text{(2 sig fig) … as required}`

 

iii.  `text(Find)\ t\ text(when)\ A = 0.9 A_0`

`0.9 A_0` `= A_0 e^(kt)`
`e^(kt)` `= 0.9`
`kt` `= text(ln)\ 0.9`
`t` `= (text(ln)\ 0.9)/k`
  `= (5730 xx text(ln)\ 0.9)/(text(ln)\ 0.5)`
  `= 870.97…`
  `= 870\ text{years (2 sig.fig.)}`

Calculus, EXT1* C1 2016 HSC 13c

A radioactive isotope of Curium has a half-life of 163 days. Initially there are 10 mg of Curium in a container.

The mass `M(t)` in milligrams of Curium, after `t` days, is given by

`M(t) = Ae^(-kt),`

where `A` and `k` are constants.

  1. State the value of `A`.  (1 mark)

  2. Given that after 163 days only 5 mg of Curium remain, find the value of `k`.  (2 marks)

Show Answers Only
  1. `10`
  2. `(ln 2)/163`
Show Worked Solution

i.  `text(When)\ \ t = 0,\ \ M = 10`

`10` `= Ae°`
`:. A` `= 10`

 

ii.  `text(When)\ \ t = 163,\ \ M = 5`

`5` `= 10 e^(-163k)`
`e^(-163k)` `= 1/2`
`e^(163k)` `= 2`
`163 k` `= ln 2`
`:. k` `= (ln 2)/163`

Calculus, EXT1* C1 2015 HSC 15a

The amount of caffeine, `C`, in the human body decreases according to the equation

`(dC)/(dt) = -0.14C,` 

where `C` is measured in mg and `t` is the time in hours.

  1. Show that  `C = Ae^(-0.14t)`  is a solution to  `(dC)/(dt) = -0.14C,` where ` A` is a constant.

     

    When `t = 0`, there are 130 mg of caffeine in Lee’s body.  (1 mark)

  2. Find the value of `A.`  (1 mark)

  3. What is the amount of caffeine in Lee’s body after 7 hours?   (1 mark)

  4. What is the time taken for the amount of caffeine in Lee’s body to halve?  (2 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `130`
  3. `48.8\ text(mg)`
  4. `4.95\ text(hours)`
Show Worked Solution
i. `C` `= Ae^(-0.14t)`
  `(dC)/(dt)` `= d/(dt) (Ae^(-0.14t))`
    `= -0.14 xx Ae^(-0.14t)`
    `= -0.14\ C`

 
`:.\ C = Ae^(-0.14t)\ \ text(is a solution)`

 

ii.  `text(When)\ \ t = 0,\ C = 130`

`130` `= Ae^(-0.14 xx 0)`
`:.\ A` `= 130`

 

iii.  `text(Find)\ \ C\ \ text(when)\ \ t = 7`

`C` `= 130\ e^(-0.14 xx 7)`
  `= 130\ e^(-0.98)`
  `= 48.79…`
  `= 48.8\ text{mg  (to 1 d.p.)}`

 
`:.\ text(After 7 hours, Lee will have 48.8 mg)`

`text(of caffeine left in her body.)`

 

iv.  `text(Find)\ \ t\ \ text(when caffeine has halved.)`

`text(When)\ \ t = 0,\ \ C = 130`

`:.\ text(Find)\ \ t\ \ text(when)\ \ C = 65`

`65` `= 130 e^(-0.14 xx t)`
`e^(-0.14t)` `= 65/130`
`ln e^(-0.14t)` `= ln\ 65/130`
`-0.14t xx ln e` `= ln\ 65/130`
`t` `= (ln\ 65/130)/-0.14`
  `= 4.951…`
  `= 4.95\ text{hours  (to 2 d.p.)}`

 

`:.\ text(It will take 4.95 hours for Lee’s)`

`text(caffeine to halve.)`

Calculus, EXT1* C1 2006 HSC 6b

A rare species of bird lives only on a remote island. A mathematical model predicts that the bird population, `P`, is given by

`P = 150 + 300 e^(-0.05t)`

where `t` is the number of years after observations began.

  1. According to the model, how many birds were there when observations began?  (1 mark)

  2. According to the model, what will be the rate of change in the bird population ten years after observations began?  (2 marks)

  3. What does the model predict will be the limiting value of the bird population?  (1 mark)

  4. The species will become eligible for inclusion in the endangered species list when the population falls below `200`. When does the model predict that this will occur?  (2 marks)

Show Answers Only
  1. `450\ text(birds)`
  2. `-9.1\ text{(to 1 d.p.)}`
  3. `150\ text(birds)`
  4. `text(After 35.83… years)`
Show Worked Solution

i.  `P = 150 + 300 e^(-0.05t)`

`text(When)\ \ t = 0`

`P` `= 150 + 300 e^0`
  `= 450`

 
`:.\ text(There were 450 birds when observations began.)`
 

ii.  `(dP)/dt` `= 300 xx (-0.05) xx e^(-0.05t)`
  `= -15 e^(-0.05t)`

 
`text(When)\ \ t = 10`

`(dP)/dt` `= -15 e^(-0.05 xx 10)`
  `= -15 e^(-0.5)`
  `= -9.097…`
  `= -9.1\ text{(to 1 d.p.)}`

 

`:.\ text(After 10 years, the bird population will be)`

`text(decreasing at a rate of 9.1 birds per year.)`

 

iii.  `text(As)\ t rarr oo`

`300 e^(-0.05t) rarr 0`

`:. P = 150 + 300 e^(-0.05t) rarr 150`

`:.\ text(The model predicts a limiting population)`

`text(of 150 birds.)`

 

iv.  `text(Find)\ t\ text(when)\ P < 200`

`150 + 300 e^(-0.05t)` `< 200`
`300 e^(-0.05t)` `< 50`
`e^(-0.05t)` `< 50/300`
`ln e^(-0.05t)` `< ln­ 1/6`
`-0.05t` `< ln­ 1/6`
`t` `> (ln­ 1/6)/(-0.05t)`
`t` `> 35.83…`

 

`:.\ text(The model predicts the population)`

`text(will fall below 200 after 35.83… years.)`

Calculus, EXT1* C1 2014 HSC 13b

A quantity of radioactive material decays according to the equation 

`(dM)/(dt) = -kM`,

where  `M`  is the mass of the material in kg,  `t`  is the time in years and  `k`  is a constant.

  1. Show that  `M = Ae^(–kt)`  is a solution to the equation, where  `A`  is a constant.   (1 mark)

  2. The time for half of the material to decay is 300 years. If the initial amount of material is 20 kg, find the amount remaining after 1000 years.   (3 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `1.98\ text(kg)\ text{(2 d.p.)}`
Show Worked Solution
i. `M` `= Ae^(-kt)`
  `(dM)/(dt)` `= -k * Ae^(-kt)`
    `= -kM\ \ text(… as required)`

 

ii.    `text(At)\ \ t = 0,\ M = 20`
`=> 20` `= Ae^0`
`A` `= 20`
`:.\ M` `= 20 e^(-kt)`

`text(At)\ \ t = 300,\ M = 10`

TIP: Many students find it efficient to save the exact value of `k` in the memory function of their calculator for these questions.
`=> 10` `= 20 e^(-300 xx k)`
`e^(-300k)` `= 10/20`
`ln e^(-300k)` `= ln 0.5`
`-300 k` `= ln 0.5`
`k` `= – ln 0.5/300`
  `= 0.00231049…`

 
`text(Find)\ M\ text(when)\ t = 1000`

`M` `= 20 e^(-1000k)`
  `= 20 e^(-1000 xx 0.00231049…)`
  `= 20 xx 0.099212…`
  `= 1.9842…`
  `= 1.98\ text(kg)\ text{(2 d.p.)}`

Calculus, EXT1* C1 2008 HSC 5c

Light intensity is measured in lux. The light intensity at the surface of a lake is 6000 lux. The light intensity,  `I`  lux, a distance  `s`  metres below the surface of the lake is given by

`I=Ae^(-ks)`

where  `A`,  and  `k`  are constants.

  1. Write down the value of  `A`.   (2 marks)

  2. The light intensity 6 metres below the surface of the lake is 1000 lux. Find the value of  `k`.    (2 marks)

  3. At what rate, in lux per metre, is the light intensity decreasing 6 metres below the surface of the lake?    (2 marks)

Show Answers Only
  1. `6000`
  2. `- 1/6 ln(1/6)\ text(or)\ 0.29863`
  3. `299`
Show Worked Solution

i.   `I=Ae^(-ks)`

`text(Find)\ A,\ text(given)\   I=6000\ text(at)\  s=0`

`6000` `=Ae^0`
`:.\ A` `=6000`

 

ii.   `text(Find)\ k\ text(given)\  I=1000\ text(at)  s=6`

MARKER’S COMMENT: Many students used the “log” function on their calculator rather than the `log_e` function. BE CAREFUL!
`1000` `=6000e^(-6xxk)`
`e^(-6k)` `=1/6`
`lne^(-6k)` `=ln(1/6)`
`-6k` `=ln(1/6)`
`k` `=- 1/6 ln(1/6)`
  `=0.2986…`
  `=0.30\ \ \ text{(2 d.p.)}`

 

iii.  `text(Find)\ \ (dI)/(ds)\ \ text(at)\ \ s=6`

ALGEBRA TIP: Tidy your working by calculating `(dI)/(ds)` using `k` and then only substituting for `k` in part (ii) at the final stage.
`I` `=6000e^(-ks)`
`:.(dI)/(ds)` `=-6000ke^(-ks)`

 
`text(At)\ s=6,`

`(dI)/(ds)` `=-6000ke^(-6k),\ \ \ text(where)\ k=- 1/6 ln(1/6)`
  `=-298.623…`
  `=-299\ \ text{(nearest whole number)}`

 
`:. text(At)\ s=6,\ text(the light intensity is decreasing)`

`text(at 299 lux per metre.)`

Calculus, EXT1* C1 2009 HSC 6b

Radium decays at a rate proportional to the amount of radium present. That is, if  `Q(t)`  is the amount of radium present at time  `t`,  then  `Q=Ae^(-kt)`,  where  `k`  is a positive constant and  `A`  is the amount present at  `t=0`. It takes 1600 years for an amount of radium to reduce by half.

  1. Find the value of  `k`.   (2 marks)

  2. A factory site is contaminated with radium. The amount of radium on site is currently three times the safe level.

     

    How many years will it be before the amount of radium reaches the safe level.    (2 marks)

Show Answers Only
  1. `(-ln(1/2))/1600\ \ text(or 0.000433)`
  2. `2536\ text(years)`
Show Worked Solution

i.   `Q=Ae^(-kt)`

MARKER’S COMMENT: Students must be familiar with “half-life” and the algebra required. i.e. using `Q=1/2 A` within their calculations.

`text(When)\ \ t=0,\ \ Q=A`

`text(When)\ \ t=1600,\ \ Q=1/2 A`

`:.1/2 A` `=A e^(-1600xxk)`
`e^(-1600xxk)` `=1/2`
`lne^(-1600xxk)` `=ln(1/2)`
`-1600k` `=ln(1/2)`
`k` `=(-ln(1/2))/1600`
  `=0.0004332\ \ text{(to 4 sig. figures)}`

 

ii.   `text(Find)\ \ t\ \ text(when)\ \ Q=1/3 A :`

IMPORTANT: Know how to use the memory function on your calculator to store the exact value of `k` found in part (i) to save time.
`1/3 A` `=A e^(-kt)`
`e^(-kt)` `=1/3`
`lne^(-kt)` `=ln(1/3)`
`-kt` `=ln(1/3)`
`:.t` `=(-ln(1/3))/k,\ \ \ text(where)\ \ \ k=(-ln(1/2))/1600`
  `=(ln(1/3) xx1600)/ln(1/2)`
  `=2535.940…`

 
`:.\ text(It will take  2536  years.)`

Calculus, EXT1* C1 2013 HSC 16b

Trout and carp are types of fish. A lake contains a number of trout. At a certain time, 10 carp are introduced into the lake and start eating the trout. As a consequence, the number of trout,  `N`,  decreases according to

`N=375-e^(0.04t)`,

where `t` is the time in months after the carp are introduced.

The population of carp,  `P`,  increases according to  `(dP)/(dt)=0.02P`.

  1. How many trout were in the lake when the carp were introduced?    (1 mark)

  2. When will the population of trout be zero?    (1 mark)

  3. Sketch the number of trout as a function of time.     (1 marks)

  4. When is the rate of increase of carp equal to the rate of decrease of trout?    (3 marks)

  5. When is the number of carp equal to the number of trout?    (2 marks)

Show Answers Only
  1. `text(374 trout)`
  2. `text{148 months (nearest month)}`
  3.  
     
    2UA 2013 HSC 16b Answer
  4. `text{After 80 months (nearest month)}`
  5. `text{After 135 months (nearest month)}`
Show Worked Solution

i.    `text(Carp introduced at)\ \ t=0`

MARKER’S COMMENT: A number of students did not equate `e^0=1` in this part.

`N=375-e^0=374`

`:.\ text(There was 374 trout when carp were introduced.)`

 

ii.    `text(Trout population will be zero when)`

NOTE: The last line of the solution isn’t necessary but is included as good practice as a check that the answer matches the exact question asked.
`N` `=375-e^(0.04t)=0`
`e^(0.04t)` `=375`
`0.04t` `=ln375`
`t` `=ln375/0.04`
  `=148.173 …`
  `=148\ text{months (nearest month)}`

 
`:.\ text(After 148 months, the trout population will be zero.)`
 

iii.   2UA 2013 HSC 16b Answer

♦♦ Mean mark 33% for part (iii)

 

iv.   `text(We need) \ |(dN)/(dt)|=(dP)/(dt)`

`text(Given)\ N=375-e^(0.04t)`

`(dN)/(dt)=-0.04e^(0.04t)`
  

`text(Find)\ P\ text(in terms of)\  t`

`text(Given)\ (dP)/(dt)=0.02P`

`=> P=Ae^(0.02t)`

♦♦♦ Mean mark 20%
COMMENT: Students who progressed to `0.2e^(0.02t)“=0.04e^(0.04t)` received 2 full marks in this part. Show your working!

 `text(Find)\ A\ \ =>text(when)\  t=0,\ P=10`

`10` `=Ae^0`
`:.A` `=10`
`=>(dP)(dt)` `=10xx0.02e^(0.02t)`
  `=0.2e^(0.02t)`

 
`text(Given that)\ \ (dP)/(dt)=|(dN)/(dt)|`

`0.2e^(0.02t)` `=0.04e^(0.04t)`
`5e^(0.02t)` `=e^(0.04t)`
`e^(0.04t)/e^(0.02t)` `=5`
`e^(0.04t-0.02t)` `=5`
`lne^(0.02t)` `=ln5`
`0.02t` `=ln5`
`t` `=ln5/0.02`
  `=80.4719…`
  `=80\ text{months (nearest month)}`

 

v.   `text(Find)\ t\ text(when)\ N=P`

`text(i.e.)\ \ 375-e^(0.04t)` `=10e^(0.02t)`
`e^(0.04t)+10e^(0.02t)-375` `=0`

`text(Let)\ X=e^(0.02t),\ text(noting)\ \ X^2=(e^(0.02t))^2=e^(0.04t)`

♦♦♦ Mean mark 4%!
MARKER’S COMMENT: Correctly applying substitution to exponentials to form a solvable quadratic proved very difficult for almost all students.
`:.\ X^2+10X-375` `=0`
`(X-15)(X+25)` `=0`

`X=15\ \ text(or)\ \ –25`

 
`text(S)text(ince)\  X=e^(0.02t)`

`e^(0.02t)` `=15\ \ \ \ (e^(0.02t)>0)`
`lne^(0.02t)` `=ln15`
`0.02t` `=ln15`
`t` `=ln15/0.02`
  `=135.4025…`
  `=135\ text(months)`