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Calculus, 2ADV C3 2024 HSC 17

In a particular electrical circuit, the voltage  \(V\)  (volts) across a capacitor is given by

\(V(t)=6.5\left(1-e^{-k t}\right)\),

where  \(k\)  is a positive constant and  \(t\)  is the number of seconds after the circuit is switched on.

  1. Draw a sketch of the graph of  \(V(t)\), showing its behaviour as  \(t\)  increases.   (2 marks)
     

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  1. When  \(t=1\), the voltage across the capacitor is 2.6 volts.
  2. Find the value of  \(k\), correct to 3 decimal places.   (2 marks)

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  1. Find the rate at which the voltage is increasing when  \(t=2\), correct to 3 decimal places.   (2 marks)

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Show Answers Only

a.   
       

b.    \(k=0.511\)

c.    \(1.195\ \text{V/s}\)

Show Worked Solution

a.   
       
♦ Mean mark (a) 49%.

b.    \(V=6.5(1-e^{-kt})\)

\(\text{When}\ \ t=1, V=2.6:\)

\(2.6\) \(=6.5(1-e^{-k})\)
\(1-e^{-k}\) \(=0.4\)
\(e^{-k}\) \(=0.6\)
\(-k\) \(=\ln(0.6)\)
\(k\) \(=0.5108…\)
  \(=0.511\ \text{(3 d.p.)}\)

   
c.
    \(V=6.5-6.5e^{-kt}\)

\(\dfrac{dV}{dt}=6.5ke^{-kt}\)

\(\text{Find}\ \dfrac{dV}{dt}\ \text{when}\ \ t=2:\)

\(\dfrac{dV}{dt}\) \(=6.5 \times 0.511 \times e^{-2 \times 0.511}\)  
  \(=1.1953…\)  
  \(=1.195\ \text{V/s (3 d.p.)}\)  

Filed Under: Rates of Change, Rates of Change Tagged With: Band 3, Band 4, Band 5, smc-1091-25-Other Themes, smc-1091-30-Log/Exp Function, smc-7135-40-Other Themes

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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a.    `a = 15\ 000, b = 20\ 000`

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

c.    `\text(Decreasing by 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 (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 (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: Modelling with Functions, Rates of Change, Rates of Change, Trig Applications Tagged With: Band 4, Band 5, Band 6, smc-1091-25-Other Themes, smc-1091-40-Trig Function, smc-1188-10-Population, smc-7125-10-Trig Applications, smc-7125-30-Population Models, smc-7135-40-Other Themes

Calculus, 2ADV C3 2016 HSC 16b

Some yabbies are introduced into a small dam. The size of the population,  `y`, of yabbies can be modelled by the function

`y = 200/(1 + 19e^(-0.5t)),`

where  `t`  is the time in months after the yabbies are introduced into the dam.

  1. Show that the rate of growth of the size of the population is
  2. `qquad qquad (1900 e^(-0.5t))/(1 + 19 e^(-0.5t))^2`.   (2 marks)

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  3. Find the range of the function `y`, justifying your answer.   (2 marks)

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  4. Show that the rate of growth of the size of the population can be written as
  5. `qquad qquad y/400 (200-y)`.   (1 mark)

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  6. Hence, find the size of the population when it is growing at its fastest rate.   (2 marks)

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a.    `text(Proof)\ \ text{(See Worked Solutions)}`

b.    `10 <= y < 200`

c.    `text(Proof)\ \ text{(See Worked Solutions)}`

d.    `100`

Show Worked Solution
a.     `y` `= 200/(1 + 19 e^(-0.5t))`
  `(dy)/(dt)` `= 200/(1 + 19 e^(-0.5t))^2 xx d/(dt) (1 + 19 e^(-0.5t))`
    `= (-200)/(1 + 19 e^(-0.5t))^2 xx -0.5 xx 19 e^(-0.5t)`
    `= (1900 e^(-0.5t))/(1 + 19 e^(-0.5t))^2\ \ text(… as required)`

 

b.    `text(When)\ \ t = 0,`

♦♦♦ Mean mark (b) 21%.

`y = 200/(1 + 19) = 10`

`text(As)\ \ t -> oo,\ \ (1 + 19^(-0.5t)) -> 1`

`:. y -> 200`

`:.\ text(Range)\ \ \ 10 <= y < 200`

 

c.    `(dy)/(dt) = (1900 e^(-0.5t))/(1 + 19 e^(-0.5t))^2`

♦♦♦ Mean mark (c) 18%.

`text(S) text(ince)\ \ y = 200/(1 + 19 e^(-0.5t))`

`-> (1 + 19 e^(-0.5t)) = 200/y`

`-> 19 e^(-0.5t) = 200/y-1 = (200-y)/y`

`text(Substituting into)\ \ (dy)/(dt):`

`(dy)/(dt)` `= (100 ((200-y)/y))/(200/y)^2`
  `= 100 ((200-y)/y) xx y^2/200^2`
  `= y/400 (200-y)\ \ text(… as required)`

 

d.    `(dy)/(dt) = -y^2/400 + y/2`

♦♦♦ Mean mark (d) 14%.

`text(Sketching the parabola:)`

`(-y^2)/400 + y/2` `= 0`
`-y^2 + 200y` `= 0`
`y (200-y)` `= 0`

 

hsc-2016-16bi

`:.\ text(Maximum)\ \ (dy)/(dt)\ \ text(occurs when)\ \ y = 100.`

Filed Under: Rates of Change, Rates of Change, Rates of Change Tagged With: Band 4, Band 6, smc-1091-25-Other Themes, smc-1091-30-Log/Exp Function, smc-7135-40-Other Themes

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