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L&E, 2ADV E1 2024 HSC 13

The graph shows the populations of two different animals, \(W\) and \(K\), in a conservation park over time. The \(y\)-axis is the size of the population and the \(x\)-axis is the number of years since 1985 .

Population \(W\) is modelled by the equation  \(y=A \times(1.055)^x\).

Population \(K\) is modelled by the equation  \(y=B \times(0.97)^x\).
 

Complete the table using the information provided.   (3 marks)

\begin{array}{|l|c|c|}
\hline
\rule{0pt}{2.5ex}\rule[-1ex]{0pt}{0pt}&\text {Population } W & \text {Population } K \\
\hline
\rule{0pt}{2.5ex}\text { Population in 1985 }\rule[-1ex]{0pt}{0pt} & A=34 & B=\ \ \ \ \  \\
\hline
\rule{0pt}{2.5ex}\text { Percentage yearly change in the population }\rule[-1ex]{0pt}{0pt} & & \\
\hline
\rule{0pt}{2.5ex}\text { Predicted population when } x=50 \rule[-1ex]{0pt}{0pt}& & 61 \\
\hline
\end{array}

--- 5 WORK AREA LINES (style=lined) ---

Show Answers Only

\begin{array}{|l|c|c|}
\hline
\rule{0pt}{2.5ex}\rule[-1ex]{0pt}{0pt}&\text {Population } W & \text {Population } K \\
\hline
\rule{0pt}{2.5ex}\text { Population in 1985 }\rule[-1ex]{0pt}{0pt} & A=34 & B=280 \\
\hline
\rule{0pt}{2.5ex}\text { Percentage yearly change in the population }\rule[-1ex]{0pt}{0pt} &+5.5\% & -3.0\%\\
\hline
\rule{0pt}{2.5ex}\text { Predicted population when } x=50 \rule[-1ex]{0pt}{0pt}&494 & 61 \\
\hline
\end{array}

Show Worked Solution

\begin{array}{|l|c|c|}
\hline
\rule{0pt}{2.5ex}\rule[-1ex]{0pt}{0pt}&\text {Population } W & \text {Population } K \\
\hline
\rule{0pt}{2.5ex}\text { Population in 1985 }\rule[-1ex]{0pt}{0pt} & A=34 & B=280 \\
\hline
\rule{0pt}{2.5ex}\text { Percentage yearly change in the population }\rule[-1ex]{0pt}{0pt} &+5.5\% & -3.0\%\\
\hline
\rule{0pt}{2.5ex}\text { Predicted population when } x=50 \rule[-1ex]{0pt}{0pt}&494 & 61 \\
\hline
\end{array}

\(\text{Calculations:}\)

\(\text{Population } W: \ (1.055)^x \Rightarrow \text { increase of 5.5% each year}\)

\(\text {Population } K: \ (0.97)^x \Rightarrow 1-0.97=0.03 \Rightarrow \text { decrease of 3.0% each year}\)

\(\text {Predicted population }(W)=34 \times(1.055)^{50}=494\)

Filed Under: Graphs and Applications (Adv-2027), Graphs and Applications (Y11) Tagged With: 2adv-std2-common, Band 4, smc-6456-20-Exponential Graphs, smc-966-10-Exponential graphs, smc-966-20-Population

Calculus, 2ADV C4 EQ-Bank 2

The population, `D`, of Tasmanian Devils in a sanctuary is given by  `D(t)`, where  `t`  is the time in years after the sanctuary was established.

The devil population changes at a rate modelled by the function  `(dD)/(dt) = 28 e^(0.35t)`.

Calculate the increase in the number of Tasmanian Devils at the end of the first 8 years. Give your answer correct to three significant figures.  (3 marks)

Show Answers Only

`1240 \ text((to 3 sig. fig.))`

Show Worked Solution
`int_0^8 28e^(0.35t)` `= [28 xx (1)/(0.35) e^(0.35t)]_0^8`
  `= 80(e^(0.35 xx 8) – e°)`
  `= 80(16.44 … – 1)`
  `= 1235.57 …`
  `= 1240 \ text((to 3 sig. fig.))`

Filed Under: Other Integration Applications (Y12) Tagged With: Band 4, smc-1213-20-Population, smc-966-20-Population

L&E, 2ADV E1 SM-Bank 2

The population of Indian Myna birds in a suburb can be described by the exponential function

`N = 35e^(0.07t)`

where `t` is the time in months.

  1.  What will be the population after 2 years?  (1 mark)

    --- 1 WORK AREA LINES (style=lined) ---

  2.  Draw a graph of the population.  (2 marks)

    --- 8 WORK AREA LINES (style=lined) ---

Show Answers Only
  1. `188\ text(birds)`
  2.  

Show Worked Solution

i.   `N = 35e^(0.07t)`

`text(Find)\ N\ text(when)\ \ t = 24:`

`N` `= 35e^(0.07 xx 24)`
  `= 35e^(1.68)`
  `= 187.79…`
  `= 188\ text(birds)`

 

ii.

Filed Under: Graphs and Applications (Adv-2027), Graphs and Applications (Y11) Tagged With: Band 3, smc-6456-20-Exponential Graphs, smc-966-10-Exponential graphs, smc-966-20-Population

Algebra, STD2 A4 2012 HSC 30c

In 2010, the city of Thagoras modelled the predicted population of the city using the equation

`P = A(1.04)^n`.

That year, the city introduced a policy to slow its population growth. The new predicted population was modelled using the equation

`P = A(b)^n`.

In both equations, `P` is the predicted population and `n` is the number of years after 2010.  

The graph shows the two predicted populations.
 

  1. Use the graph to find the predicted population of Thagoras in 2030 if the population policy had NOT been introduced.   (1 mark)

    --- 1 WORK AREA LINES (style=lined) ---

  2. In each of the two equations given, the value of `A` is 3 000 000.

     

    What does `A` represent?   (1 mark)

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  3. The guess-and-check method is to be used to find the value of `b`, in  `P = A(b)^n`.

     

    (1) Explain, with or without calculations, why 1.05 is not a suitable first estimate for `b`.  (1 mark)

     

    (2) With  `n = 20`  and  `P = 4\ 460\ 000`, use the guess-and-check method and the equation  `P = A(b)^n`  to estimate the value of `b` to two decimal places. Show at least TWO estimate values for `b`, including calculations and conclusions.  (2 marks)

    --- 4 WORK AREA LINES (style=lined) ---

  4. The city of Thagoras was aiming to have a population under 7 000 000 in 2050. Does the model indicate that the city will achieve this aim?

     

    Justify your answer with suitable calculations.  (2 marks)

    --- 5 WORK AREA LINES (style=lined) ---

Show Answers Only
  1.  `6\ 600\ 000`
  2.  `text(The population in 2010.)`
  3.  `text{(1)  See Worked Solution}`

     

    `(2)\ \ b = 1.03, 1.02`

  4. `text(See Worked Solution)`
Show Worked Solution

i.    `text(2030 occurs at)\ \ n = 20\ \ text(on the)\ x text(-axis.)`

`text(Expected population (no policy) ) = 6\ 600\ 000`

COMMENT: Common ADV/STD2 content in new syllabus.

 

ii.   `A\ text(represents the population when)\ \ n=0` 

`text(which is the population in 2010.)`
 

♦ Mean mark part (ii) 48%

iii. (1)  `P = A(1.05)^n\ text(would be steeper and lie above)`

    `P = A(1.04)^n\ text(since)\ 1.05 > 1.04`
 

iii. (2)  `text(Let)\ \ b = 1.03`

`P` `= 3\ 000\ 000 xx 1.03^20`
  `= 5\ 418\ 000`

 
`text(Let)\ \ b = 1.02`

`P` `= 3\ 000\ 000 xx 1.02^20`
  `= 4\ 457\ 800`

 
`:. b = 1.02`
 

iv.  `text(In 2050,)\ n = 40`

`P` `= 3\ 000\ 000 xx 1.02^40`
  `= 6\ 624\ 119\ \ (text(nearest whole))`

 
`text(S)text(ince the population is below 7 million,)`

`text(the model will achieve the aim.)`

Filed Under: Exponential/Quadratic (Projectile), Graphs and Applications (Y11), Non-Linear: Exponential/Quadratics (Std 2) Tagged With: Band 4, Band 5, Band 6, common-content, smc-830-30-Exponential, smc-966-10-Exponential graphs, smc-966-20-Population

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