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Calculus, 2ADV C4 2024 HSC 10 MC

The diagram shows the graph  \(y = f(x)\).
 

The point \(Q\) is a horizontal point of inflection.

Let  \(A(x)= \displaystyle \int_0^x f(t)\,dt\).

How many points of inflection does the graph  \(y=A(x)\)  have?

  1. \(2\)
  2. \(3\)
  3. \(4\)
  4. \(5\)
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\(B\)

Show Worked Solution

\(A(x) = \displaystyle \int_0^x f(t)\,dt \ \text{(note this definite integral produces a}\ F(x)\text{)}\)

\(A^{′}(x) = f(x) \)

\(A^{″}(x) = f^{′}(x) \)

♦♦ Mean mark 41%.

\(\text{POI requirements:}\ A^{″}(x) = 0\ \text{and sign (concavity) changes either side.}\)

\(\text{Inspect graph of}\ f(x)\ \text{to find where}\ \ f^{′}(x)=0\ \ \text{and gradient changes}\)

\(\text{either side of possible points.}\)

\(\text{→ the three turning points all qualify}\)

\(\text{→ point}\ Q\ \text{does not qualify (gradient is positive both sides)}\)

\(\Rightarrow B\)

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)
     

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

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.   \(V=6.5-6.5e^{-kt}\)

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

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

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

Calculus, 2ADV C4 2021 HSC 27

Kenzo has a solar powered phone charger. Its power, `P`, can be modelled by the function

`P(t) = 400 sin(pi/12 t),\ \ 0 <= t <= 12`,

where  `t`  is the number of hours after sunrise.

  1. Sketch the graph of  `P` for  `0 ≤ t ≤ 12`.  (2 marks)

Power is the rate of change of energy. Hence the amount of energy, `E` units, generated by the solar powered phone charger from  `t = a`  to  `t = b`,  where  `0 ≤ a ≤ b ≤ 12` is given by

`E = int_a^b P(t)\ dt`.

  1. Show that  `E = 4800/pi (cos\ (api)/12 - cos\ (bpi)/12)`.  (2 marks)

  2. To make a phone call, a phone battery needs at least 300 units of energy. Kenzo woke up 3 hours after sunrise and found that his phone battery had no units of energy. He immediately began to use his solar powered charger to charge his phone battery.
  3. Find the least amount of time he needed to wait before he could make a phone call. Give your answer correct to the nearest minute.  (3 marks)

  4. The next day, Kenzo woke up 6 hours after sunrise and again found that his phone battery had no units of energy. He immediately began to use his solar powered charger to charge his phone battery.
  5. Would it take more time or less time or the same amount of time, compared to the answer in part (c), to charge his phone battery in order to make a phone call? Explain your answer by referring to the graph drawn in part (a).  (1 mark)

Show Answers Only
  1.  
  2. `text(See Worked Solution)`
  3. `text(57 minutes)`
  4. `text(The power produced is at its peak when)\ t = 6`
  5. `:.\ text(It will charge in less time.)`
Show Worked Solution

a.

b.    `E` `= int_a^b 400 sin(pi/12 t)\ dt`
    `= [-400 · 12/pi cos(pi/12 t)]_a^b`
    `= -4800/pi cos(pi/12 b) – (-4800/pi cos(pi/12 a))`
    `= 4800/pi(cos\ (api)/12 – cos\ (bpi)/12)`

 

♦♦ Mean mark part (c) 29%
COMMENT: It is arguable that the nearest minute may also be 3h 58 m as the phone is not adequately charged at 3h 57 m.

c.   `text(Find)\ \ b\ \ text(given)\ \ E = 300\ \ text(and)\ \ a = 3:`

`300` `= 4800/pi (cos\ pi/4 – cos\ (bpi)/12)`
`(300pi)/4800` `= 1/sqrt2 – cos\ (bpi)/12`
`cos\ (bpi)/12` `= 1/sqrt2 – pi/16`
`(bpi)/12` `= cos^(-1) (1/sqrt2 – pi/16)`
`b` `= 12/pi cos^(-1)(1/sqrt2 – pi/16)`
  `= 3.952…`
  `= 3\ text{h 57 m  (nearest minute)}`

 
`:.\ text(Least time before phone is charged = 57 minutes)`

 

♦♦ Mean mark part (d) 22%

d.   `text(The power produced is at its peak when)\ \ t = 6.`

`:.\ text(It will charge in less time.)`

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)

  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)

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

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.)`

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)

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

  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)

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

Show Answers Only

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 (ii) 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 (iii) 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 (iv) 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.`