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Calculus, 2ADV C3 2022 HSC 20

A scientist is studying the growth of bacteria. The scientist models the number of bacteria, `N`, by the equation

`N(t)=200e^(0.013 t)`,

where `t` is the number of hours after starting the experiment.

  1. What is the initial number of bacteria in the experiment?   (1 mark)

  2. What is the number of bacteria 24 hours after starting the experiment?   (1 mark)

  3. What is the rate of increase in the number of bacteria 24 hours after starting the experiment?   (2 marks)

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a.    `200`

b.    `273`

c.    `3.55\ text{bacteria per hour}`

Show Worked Solution
a.     `N(0)` `=200e^0`
    `=200\ text{bacteria}`

  
b.   `text{Find}\ N\ text{when}\ \ t=24:`

`N(24)` `=200e^(0.013xx24)`
  `=273.23…`
  `=273\ text{bacteria (nearest whole)}`

 

c.     `N` `=200e^(0.013 t)`
  `(dN)/dt` `=0.013xx200e^(0.013t)`
    `=2.6e^(0.013t)`

 
`text{Find}\ \ (dN)/dt\ \ text{when}\ \ t=24:`

`(dN)/dt` `=2.6e^(0.013xx24)`
  `=3.550…`
  `=3.55\ text{bacteria/hr (to 2 d.p.)}`

Calculus, 2ADV C3 2021 HSC 23

A population,  \(P\), which is initially 5000, varies according to the formula

\(P = 5000b^\tfrac{-t}{10}\),

where  \(b\)  is a positive constant and  \(t\)  is time in years,  \(t \geq 0\).

The population is 1250 after 20 years.

Find the value of  \(t\), correct to one decimal place, for which the instantaneous rate of decrease is 30 people per year.   (4 marks)

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\(35.3 \ \text{years}\)

Show Worked Solution

\(P=1250 \text { when } t=20\)

♦ Mean mark 42%.
\(1250\) \(= 5000 \cdot b^\tfrac{-t}{10}\)
\(b^{-2}\) \(= \dfrac{1}{4}\)
\(b\) \(= 2\ \ (b>0)\)

 

\(P\) \(=5000 \cdot 2^{\tfrac{-t}{10}}\)
\(\dfrac{d P}{d t}\) \(=\ln 2 \cdot-\dfrac{1}{10} \cdot 5000 \cdot 2^{-\tfrac{t}{10}}\)
  \(=-500 \ln 2 \cdot 2^{\tfrac{-t}{10}}\)

  
\(\text{Find} \ t \ \text{when} \ \dfrac{d P}{d t}=-30\):

\(-30\) \(=-500 \ln 2 \cdot 2^{\tfrac{-t}{10}}\)
\(2^{\tfrac{-t}{10}}=\) \(=\dfrac{3}{50 \ln 2}\)
\(\ln 2^{\tfrac{-t}{10}}\) \(=\ln \left(\dfrac{3}{50 \ln 2}\right)\)
\(\dfrac{-t}{10}\) \(=\frac{\ln \left(\dfrac{3}{50 \ln 2}\right)}{\ln 2}\)
\(t\) \(=\dfrac{-10 \ln \left(\frac{3}{50 \ln 2}\right)}{\ln 2}\)
  \(=35.301 \ldots\)
  \(=35.3 \ \text{years (1 d.p.)}\)

Calculus, 2ADV C3 2020 HSC 21

Hot tea is poured into a cup. The temperature of tea can be modelled by  `T = 25 + 70(1.5)^(−0.4t)`, where `T` is the temperature of the tea, in degrees Celsius, `t` minutes after it is poured.

  1. What is the temperature of the tea 4 minutes after it has been poured?   (1 mark)

  2. At what rate is the tea cooling 4 minutes after it has been poured?   (2 marks)

  3. How long after the tea is poured will it take for its temperature to reach 55°C?   (3 marks)

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a.    `61.6\ \ (text(to 1 d.p.))`

b.    `-5.9^@text(C/min)`

c.    `5.2\ text(minutes  (to 1 d.p.))`

Show Worked Solution
a.     `T` `= 25 + 70(1.5)^(-0.4 xx 4)`
    `= 61.58…`
    `= 61.6\ \ (text(to 1 d.p.))`

 

b.    `(dT)/(dt)` `= 70 log_e(1.5) xx-0.4(1.5)^(-0.4t)`
    `= -28log_e(1.5)(1.5)^(-0.4t)`

  
`text(When)\ \ t = 4,`

`(dT)/(dt)` `=-28log_e(1.5)(1.5)^(-1.6)`
  `=-5.934…`
  `=-5.9^@text(C/min  (to 1 d.p.))`

 

c.   `text(Find)\ \ t\ \ text(when)\ \ T = 55:`

♦ Mean mark part (c) 44%.
`55` `= 25 + 70(1.5)^(-0.4t)`
`30` `= 70(1.5)^(0.4t)`
`(1.5)^(-0.4t)` `= 30/70`
`-0.4t log_e(1.5)` `= log_e\ 3/7`
`-0.4t` `= (log_e\ 3/7)/(log_e (1.5))`
`:. t` `= (-2.08969)/(-0.4)`
  `= 5.224…`
  `= 5.2\ text(minutes  (to 1 d.p.))`