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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)

Show Answers Only
  1. `61.6\ \ (text(to 1 d.p.))`
  2. `−5.9^@text(C/min)`
  3. `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.))`

L&E, 2ADV E1 SM-Bank 14

The spread of a highly contagious virus can be modelled by the function

`f(x) = 8000/(1 + 1000e^(−0.12x))`

Where `x` is the number of days after the first case of sickness due to the virus is diagnosed and `f(x)` is the total number of people who are infected by the virus in the first `x` days.

  1. Calculate `f(0)`.   (1 mark)

  2. Find the value of `f(365)` and interpret it result.   (2 marks)

Show Answers Only
  1. `7.99…`
  2. `text(After 1 year, the model predicts the total number)`
    `text(of people infected by the virus is 8000.)`
Show Worked Solution
i.   `f(0)` `= 8000/(1 + 1000e^0)`
    `= 8000/1001`
    `= 7.99…`

 

ii.    `f(365)` `= 8000/(1 + 1000e^(−0.12 xx 365))`
    `= 8000/(1 + 1000e^(−43.8))`
    `~~ 8000`

 
`text(After 1 year, the model predicts the total number)`

`text(of people infected by the virus is 8000.)`

Algebra, STD2 A4 2016 HSC 29b

The mass `M` kg of a baby pig at age `x` days is given by  `M = A(1.1)^x`  where `A` is a constant. The graph of this equation is shown.
 

2ug-2016-hsc-q29_1

  1. What is the value of `A`?   (1 mark)

  2. What is the daily growth rate of the pig’s mass? Write your answer as a percentage.   (1 mark)

Show Answers Only
  1. `1.5\ text(kg)`
  2. `10text(%)`
Show Worked Solution

i.   `text(When)\ x = 0,`

♦ Mean mark (i) 48%.
♦♦♦ Mean mark part (ii) 6%!
`1.5` `= A(1.1)^0`
`:. A` `= 1.5\ text(kg)`

 
ii.   `text(Daily growth rate)`

`= 0.1`

`= 10text(%)`

Algebra, STD2 A4 2004 HSC 26a

  1. The number of bacteria in a culture grows from 100 to 114 in one hour.

     

    What is the percentage increase in the number of bacteria?   (1 mark)

  2. The bacteria continue to grow according to the formula  `n = 100(1.14)^t`, where `n` is the number of bacteria after `t` hours.

     

    What is the number of bacteria after 15 hours?   (1 mark)

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \text{Time in hours $(t)$} \rule[-1ex]{0pt}{0pt} & \;\; 0 \;\;  &  \;\; 5 \;\;  & \;\; 10 \;\;  & \;\; 15 \;\; \\
\hline
\rule{0pt}{2.5ex} \text{Number of bacteria ( $n$ )} \rule[-1ex]{0pt}{0pt} & \;\; 100 \;\;  &  \;\; 193 \;\;  & \;\; 371 \;\;  & \;\; ? \;\; \\
\hline
\end{array}

  1. Use the values of `n` from  `t = 0`  to  `t = 15`  to draw a graph of  `n = 100(1.14)^t`.

     

    Use about half a page for your graph and mark a scale on each axis.   (4 marks)

  2. Using your graph or otherwise, estimate the time in hours for the number of bacteria to reach 300.   (1 mark)

Show Answers Only
  1. `text(14%)`
  2. `714`
  3. `text(Proof)\ \ text{(See Worked Solutions)}`
  4. `text(8.4 hours)`
Show Worked Solution

i.   `text(Percentage increase)`

COMMENT: Common ADV/STD2 content in new syllabus.

`= (114 -100)/100 xx 100`

`= 14text(%)`

 

ii.  `n = 100(1.14)^t`

`text(When)\ \ t = 15,`

`n` `= 100(1.14)^15`
  `= 713.793\ …`
  `= 714\ \ \ text{(nearest whole)}`

 

iii. 

 

iv.  `text(Using the graph)`

`text(The number of bacteria reaches 300 after)`

`text(approximately 8.4 hours.)`