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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 2007 HSC 8a

One model for the number of mobile phones in use worldwide is the exponential growth model,

`N = Ae^(kt)`,

where `N` is the estimate for the number of mobile phones in use (in millions), and `t` is the time in years after 1 January 2008.

  1. It is estimated that at the start of 2009, when  `t = 1`, there will be 1600 million mobile phones in use, while at the start of 2010, when  `t = 2`, there will be 2600 million. Find `A` and `k`.  (3 marks)

  2. According to the model, during which month and year will the number of mobile phones in use first exceed 4000 million?  (2 marks)

Show Answers Only
  1. `A = (12\ 800)/13, k = log_e\ 13/8`
  2. `text(November 2010)`
Show Worked Solution

i.  `N = Ae^(kt)`

`text(When)\ t = 1,\ N = 1600`

`:. 1600` `= Ae^k`
`A` `= 1600/e^k`

 
`text(When)\ t = 2, N = 2600`

`:. 2600` `= A e^(2k)`
  `= 1600/e^k xx e^(2k)`
  `= 1600 e^k`
`e^k` `= 2600/1600 = 13/8`
`log_e e^k` `= log_e\ 13/8`
`k` `= log_e\ 13/8`
  `=0.4855…`
  `=0.49\ \ \ text{(to 2 d.p.)}`

 

`:. A` `= 1600/(e^(log_e\ 13/8)`
  `= 1600/(13/8)`
  `= (12\ 800)/13`

 

ii.  `text(Find)\ \ t\ \ text(such that)\ N > 4000`

`(12\ 800)/13 xx e^(kt)` `> 4000`
`e^(kt)` `> (13 xx 4000)/(12\800)`
`log_e e^(kt)` `> log_e\ 65/16`
`kt` `> log_e\ 65/16\ \ \ \ (k = log_e\ 13/8)`
`t` `> (log_e\ 65/16)/(log_e\ 13/8`
`t` `> 2.887…\ \ text(years)`
  `>\ text{2 years and 10.6 months (approx)}`

 

`:.\ text(The number of mobile phones in use will exceed)`

`text(4000 million in November 2010.)`

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 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 2011 HSC 10a

The intensity, `I`,  measured in  watt/m2,  of a sound is given by

`I=10^-12xxe^(0.1L)`,

where  `L`  is the loudness of the sound in decibels.

  1. If the loudness of a sound at a concert is 110 decibels, find the intensity of the sound. Give your answer in scientific notation.   (1 mark)

  2. Ear damage occurs if the intensity of a sound is greater than `8.1xx10^-9`  watt/m2.

     

    What is the maximum loudness of a sound so that no ear damage occurs?    (2 marks)

  3. By how much will the loudness of a sound have increased if its intensity has doubled?     (2 marks)

Show Answers Only
  1. `6xx10^-8\ \ text{(1 sig. figure)}`
  2. `90\ text(decibels)`
  3. `7\ text(decibels)`
Show Worked Solution

i.    `text(Find)\ I\ text(when)\ \ L=110`

MARKER’S COMMENT: Note that `e^11 xx 10^-12` is not correct scientific notation.
`I` `=10^-12xxe^(0.1xx110)`
  `=10^-12xxe^11`
  `=5.9874…\ \ xx10^-8`
  `=6xx10^-8\ \ \ text{(to 1 sig. fig.)}`

 

ii.   `text(Find)\ \ L\ \ text(such that)\ \ \ I=8.1xx10^-9`

`text(i.e.)\ \ \  8.1xx10^-9` `=10^-12xxe^(0.1L)`
`e^(0.1L)` `=8.1xx10^3`
`lne^(0.1L)` `=ln8100`
`0.1L xx ln e` `=ln8100`
`L` `=ln8100/0.1`
  `=89.996…`
  `=90\ \ text{(nearest whole)}`

 
`:.\ 90\ text(decibels is the maximum loudness.)`

 

iii.  `text(Let)\ \ I=I_0 e^(0.1L)`

♦♦ Mean mark 32%
MARKER’S COMMENT: Actual values can help here. Calculate the intensity at `L=0` (which equals `1xx10^-12`) and then find `L` when this intensity level is doubled (to `2xx10^-12`).

`text(Find)\ \ L\ \ text(when)\ \ I=2I_0`

`2I_0` `=I_0e^(0.1L)`
`e^(0.1L)` `=2`
`lne^(0.1L)` `=ln2`
`0.1L` `=ln2`
`L` `=ln2/0.1`
  `=6.93147…`
  `=7\ \ text{(nearest whole)}`

 
`:.\ text(The loudness of a sound must increase 7)`

`text(decibels for the intensity to double.)`