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Trigonometry, 2ADV T3 2020 HSC 31

The population of mice on an isolated island can be modelled by the function.

`m(t) = a sin (pi/26 t) + b`,

where  `t`  is the time in weeks and  `0 <= t <= 52`. The population of mice reaches a maximum of 35 000 when  `t=13`  and a minimum of 5000 when  `t = 39`. The graph of  `m(t)`  is shown.
 

  1. What are the values of `a` and `b`?  (2 marks)

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  2. On the same island, the population of cats can be modelled by the function
     
    `\ \ \ \ \ c(t) = −80cos(pi/26 (t - 10)) + 120`
     
    Consider the graph of  `m(t)`  and the graph of  `c(t)`.

     

    Find the values of  `t, \ 0 <= t <= 52`, for which both populations are increasing.  (3 marks)

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  3. Find the rate of change of the mice population when the cat population reaches a maximum.  (2 marks)

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Show Answers Only
  1. `a = 15\ 000, b = 20\ 000`
  2. `text(Both populations are increasing when)\ 10 < t < 13`
  3. `\text(– 643 mice per week)`
Show Worked Solution
a.    `b` `= (35\ 000 + 5000)/2`
    `= 20\ 000`

 

`a` `=\ text(amplitude of sin graph)`
  `= 35\ 000 – 20\ 000`
  `= 15\ 000`

 

b.   `text(By inspection of the)\ \ m(t)\ \ text(graph)`

♦♦ Mean mark part (b) 30%.

`m^{′}(t) > 0\ \ text(when)\ \ 0 <= t < 13\ \ text(and)\ \ 39 < t <= 52`

`text(Sketch)\ \ c(t):`

`text(Minimum)\ \ (cos0)\ \ text(when)\ \ t = 10`

`text(Maximum)\ \ (cospi)\ \ text(when)\ \ t = 36`

`:. c^{′}(t) > 0\ \ text(when)\ \ 10 < t < 36`

`:. text(Both populations are increasing when)\ \ 10 < t < 13`

 

c.   `c(t)\ text(maximum when)\ \ t = 36`

♦♦♦ Mean mark part (c) 27%.
`m(t)` `= 15\ 000 sin(pi/26 t) + 20\ 000`
`m^{′}(t)` `= (15\ 000pi)/26 cos(pi/26 t)`
`m^{′}(36)` `= (15\ 000pi)/26 · cos((36pi)/26)`
  `= -642.7`

 
`:.\ text(Mice population is decreasing at 643 mice per week.)`

Filed Under: Rates of Change (Y12), Trig Applications (Y12) Tagged With: Band 4, Band 5, Band 6, smc-1091-25-Other Themes, smc-1091-40-Trig Function, smc-1188-10-Population

Calculus, 2ADV C3 2004 HSC 5b

A particle moves along a straight line so that its displacement, `x` metres, from a fixed point `O` is given by  `x = 1 + 3 cos 2t`, where  `t`  is measured in seconds.

  1. What is the initial displacement of the particle?  (1 mark)

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  2. Sketch the graph of  `x`  as a function of  `t`  for  `0 ≤ t ≤ pi`.  (2 marks)

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  3. Hence, or otherwise, find when AND where the particle first comes to rest after  `t = 0`.  (2 marks)

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  4. Find a time when the particle reaches its greatest magnitude of velocity. What is this velocity?  (2 marks)

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Show Answers Only
  1. `text(4 m to the right of)\ O.`
  2. `text(See Worked Solutions)`
  3. `t = pi/2\ text(seconds, 2 m to the left of)\ O.`
  4. `text(6 m s)^(−1)`
Show Worked Solution

i.   `x = 1 + 3 cos 2t`

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

`x` `= 1 + 3 cos 0`
  `= 1 + 3`
  `= 4`

 

`:.\ text(Initial displacement is 4 m to the right of)\ O.`

 

 ii.  `text(Period)\ = (2pi)/n = (2pi)/2 = pi`

`text(Considering the range)`

`-1` `<=cos 2t<=1`
`-3` `<=3cos 2t<=3`
`-2` `<=1 + 3 cos 2t<=4`

 

 Calculus in the Physical World, 2UA 2004 HSC 5b

 

iii.   `x` `= 1 + 3 cos 2t`
  `:.v` `= −6 sin 2t`

 

`text(The particle comes to rest when)\ \ v=0`

`-6 sin 2t` `= 0`
`sin 2t` `= 0`
`2t` `= 0, pi, 2pi…`
`t` `= 0, pi/2, pi…`

 

`:.\ text(After)\ \ t=0, text(particle first comes to rest when)`

`t = pi/2\ text(seconds.)`

`text(When)\ t = pi/2,`

`x` `= 1 + 3 cos 2(pi/2)`
  `= 1 + 3 cos pi`
  `= 1 + 3(−1)`
  `= −2`

 

`:.\ text(Particle first comes to rest at 2 m to the left of)\ O.`

 

iv.  `x = 1 + 3 cos 2t`

`v` `= -6 sin 2t`
`a` `= -12 cos 2t`
   

`text(MAX occurs when)\ \ a=0`

`−12 cos 2t` `= 0`
`cos 2t` `= 0`
`2t` `= pi/2, (3pi)/2, …`
`t` `= pi/4, (3pi)/4, …`

 
`:.\ text(Maximum at)\ \ t=pi/4,\ \ (3pi)/4, …\ text(seconds,)`

 

`text(When)\ \ t = pi/4\ text(seconds,)`

`v` `= -6 sin 2(pi/4)`
  `= -6 sin(pi/2)`
  `= −6`

 
`:.\ text(Maximum is 6 m s)^(−1).`

Filed Under: Motion, Rates of Change (Y12) Tagged With: Band 3, Band 4, Band 5, smc-1091-10-Motion, smc-1091-40-Trig Function

Calculus, 2ADV C4 2015 HSC 15c

Water is flowing in and out of a rock pool. The volume of water in the pool at time `t` hours is `V` litres. The rate of change of the volume is given by

`(dV)/(dt) = 80 sin(0.5t)`

At time  `t = 0`, the volume of water in the pool is 1200 litres and is increasing.

  1. After what time does the volume of water first start to decrease?  (2 marks)

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  2. Find the volume of water in the pool when  `t = 3`.  (2 marks)

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  3. What is the greatest volume of water in the pool?  (1 mark)

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Show Answers Only
  1. `2 pi\ text(hours)`
  2. `1349\ text(litres)`
  3. `1520\ text(litres)`
Show Worked Solution

i.   `(dV)/(dt) = 80 sin (0.5t)`

♦ Mean mark (i) 37%.

`text(Volume decreases when)\ \ (dV)/(dt) < 0`

`(dV)/(dt) < 0\ \ text(when)\ \ t > 2 pi\ text(hours)`

`:.\ text(After 2)\pi\ \text(hours, volume starts to decrease.)`

 

ii.   `V` `= int (dV)/(dt)\ dt`
  `= int 80 sin (0.5t)\ dt`
  `= -160 cos (0.5t) + c`

 

`text(When)\ \ t = 0,\ V = 1200`

`1200` `= -160 cos 0 + c`
`c` `= 1360`
`:.\ V` `= -160 cos (0.5t) + 1360`

 

`text(When)\ \ t = 3`

`V` `= -160 cos (0.5 xx 3) + 1360`
  `= 1348.68…\ \ text(litres)`
  `= 1349\ text{litres  (nearest litre)}`

 

iii.  `V = -160 cos (0.5t) + 1360`

♦♦ Mean mark (iii) 27%.

`=>\ text(Greatest volume occurs when)`

`cos (0.5t) = -1`

`:.\ text(Maximum volume)`

`= -160 (-1) + 1360`

`= 1520\ text(litres)`

Filed Under: Other Integration Applications (Y12), Rates of Change Tagged With: Band 4, Band 5, smc-1091-20-Flow, smc-1091-40-Trig Function, smc-1213-15-Flow

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