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Calculus, MET2 2024 VCAA 2

A model for the temperature in a room, in degrees Celsius, is given by

\(f(t)=\left\{
\begin{array}{cc}12+30 t & \quad \quad 0 \leq t \leq \dfrac{1}{3} \\
22 & t>\dfrac{1}{3}
\end{array}\right.\)

where \(t\) represents time in hours after a heater is switched on.

  1. Express the derivative \(f^{\prime}(t)\) as a hybrid function.   (2 marks)
  1. Find the average rate of change in temperature predicted by the model between \(t=0\) and \(t=\dfrac{1}{2}\).
  2. Give your answer in degrees Celsius per hour.   (1 mark)
  1. Another model for the temperature in the room is given by \(g(t)=22-10 e^{-6 t}, t \geq 0\).
  2.  i. Find the derivative \(g^{\prime}(t)\).   (1 mark)
  3. ii. Find the value of \(t\) for which \(g^{\prime}(t)=10\).
  4.     Give your answer correct to three decimal places.   (1 mark)
  1. Find the time \(t \in(0,1)\) when the temperatures predicted by the models \(f\) and \(g\) are equal.
  2. Give your answer correct to two decimal places.   (1 mark)
  1. Find the time \(t \in(0,1)\) when the difference between the temperatures predicted by the two models is the greatest.
  2. Give your answer correct to two decimal places.   (1 mark)
  1. The amount of power, in kilowatts, used by the heater \(t\) hours after it is switched on, can be modelled by the continuous function \(p\), whose graph is shown below.

\(p(t)=\left\{
\begin{array}{cl}1.5 & 0 \leq t \leq 0.4 \\
0.3+A e^{-10 t} & t>0.4
\end{array}\right.\)

The amount of energy used by the heater, in kilowatt hours, can be estimated by evaluating the area between the graph of \(y=p(t)\) and the \(t\)-axis.
 

  1.   i. Given that \(p(t)\) is continuous for \(t \geq 0\), show that \(A=1.2 e^4\).   (1 mark)
  2.  ii. Find how long it takes, after the heater is switched on, until the heater has used 0.5 kilowatt hours of energy.
  3.     Give your answer in hours.   (1 mark)
  4. iii. Find how long it takes, after the heater is switched on, until the heater has used 1 kilowatt hour of energy.
  5.     Give your answer in hours, correct to two decimal places.   (2 marks)

Show Answers Only

a.    \(f^{\prime}(t)=\left\{
\begin{array}{cc}30  & \quad \quad 0 \leq t <\dfrac{1}{3} \\
0 & t>\dfrac{1}{3}
\end{array}\right.\)

b.    \(20^{\circ}\text{C/h}\)

ci.    \(g^{\prime}(t)=60e^{-6t}\)

cii.   \(0.299\ \text{(3 d.p.)}\)

d.    \(0.27\ \text{(2 d.p.)}\)

e.    \(0.12\ \text{(2 d.p.)}\)

fi.   \(\text{Because function is continuous}\)

\(0.3+Ae^{-10t}\) \(=1.5\)
\(Ae^{-10\times 0.4}\) \(=1.2\)
\(A\) \(=\dfrac{1.2}{e^{-10\times 0.4}}\)
  \(=1.2e^4\)

fii.  \(\dfrac{1}{3}\ \text{hours}\)

fiii. \(1.33\ \text{(2 d.p.)}\)

Show Worked Solution

a.    \(f^{\prime}(t)=\left\{
\begin{array}{cc}30  & \quad \quad 0 \leq t < \dfrac{1}{3} \\
0 & t>\dfrac{1}{3}
\end{array}\right.\)

b.    \(\text{When }\ t=0, f(t)=12\ \ \text{and when }\ t=\dfrac{1}{2}, f(t)=22\)

\(\therefore\ \text{Average rate of change}\) \(=\dfrac{22-12}{\frac{1}{2}}\)
  \(=20^{\circ}\text{C/h}\)

 

ci.    \(g(t)\) \(=22-10 e^{-6 t}\)
  \(g^{\prime}(t)\) \(=60e^{-6t},\ \ t\geq 0\)

 

cii.   \(60e^{-6t}\) \(=10\)
  \(-6t\,\ln{e}\) \(=\ln{\dfrac{1}{6}}\)
  \(t\) \(=\dfrac{\ln{\frac{1}{6}}}{-6}\)
      \(=0.2986…\approx 0.299\ \text{(3 d.p.)}\)

 

d.     \(\left\{
\begin{array}{cc}12+30 t & \ \  0 \leq t \leq \dfrac{1}{3} \\
22 & t>\dfrac{1}{3}
\end{array}\right.\)		 \(=22-10e^{-6t}\)

\(\text{Using CAS:}\)

\(\text{Temps equal when}\ t\approx 0.27\ \text{(2 d.p.)}\)

 

e.     \(\text{Difference }(D)\) \(=|g(t)-f(t)|\)
    \(=\left(22-10e^{-6t}\right)-(12+30t)\)
  \(\dfrac{dD}{dt}\) \(=60e^{-6t}-30\)

\(\text{Max time diff when}\ \dfrac{dD}{dt}=0\)

\(\therefore\ 60e^{-6t}-30\) \(=0\)
\(e^{-6t}\) \(=0.5\)
\(-6t\) \(=\ln{0.5}\)
\(t\) \(=0.1155\dots\)
  \(\approx 0.12\ \text{(2 d.p.)}\)

 

fi.   \(\text{Because function is continuous}\)

\(0.3+Ae^{-10t}\) \(=1.5\)
\(Ae^{-10\times 0.4}\) \(=1.2\)
\(A\) \(=\dfrac{1.2}{e^{-10\times 0.4}}\)
  \(=1.2e^4\)

  
fii.  \(\text{Using CAS:}\)

\(\text{Or, considering the graph, the area from 0 to 0.4 }=0.6\ \rightarrow t<0.4\)

\(\therefore\ \text{Solving}\ 1.5t=0.5\ \rightarrow\ t=\dfrac{1}{3}\)

\(\therefore\ \text{It takes }\dfrac{1}{3}\ \text{hours for heater to use 0.5  kilowatts.}\)
  

fiii. \(\text{Using CAS:}\)

\(1.5\times 4+\displaystyle\int_{0.4}^{t}0.3+1.2e^4.e^{-10t}dt\) \(=1\)
\(\Bigg[0.3t+0.12e^4.e^{-10t}\Bigg]_{0.4}^{t}\) \(=0.4\)
\(t\) \(=1.3333\dots\)
  \(\approx 1.33\ \text{(2 d.p.)}\)

 

Calculus, MET1 2023 VCAA SM-Bank 5

Let  \(f: R \rightarrow R\),  where  \(f(x)=2-x^2\).

  1. Calculate the average rate of change of \(f\) between \(x=-1\) and \(x=1\).  (1 mark)

  2. Calculate the average value of \(f\) between \(x=-1\) and \(x=1\).  (2 marks)

  3. Four trapeziums of equal width are used to approximate the area between the functions  \(f(x)=2-x^2\)  and the \(x\)-axis from \(x=-1\) to \(x=1\).
  4. The heights of the left and right edges of each trapezium are the values of \(y=f(x)\), as shown in the graph below.

  1. Find the total area of the four trapeziums.  (2 marks)

Show Answers Only

a.    \(0\)

b.    \(\dfrac{5}{3}\)

c.    \(\dfrac{13}{4}\)

Show Worked Solution
a.     \(\text{Average rate of change}\) \(=\dfrac{f(1)-f(-1)}{1-(-1)}\)
    \(\dfrac{1-1}{2}\)
    \(=0\)

 

b.    \(\text{Avg value}\) \(=\dfrac{1}{1-(-1)}\displaystyle\int_{-1}^{1} \Big(2-x^2\Big)\,dx\)
    \(=\dfrac{1}{2}\displaystyle\Bigg[2x-\dfrac{x^3}{3}\displaystyle\Bigg]_{-1}^{1}\)
    \(=\dfrac{1}{2}\displaystyle\Bigg[\Bigg(2.(1)-\dfrac{(1)^3}{3}\Bigg)-\Bigg(2.(-1)-\dfrac{(-1)^3}{3}\Bigg)\Bigg]\)
    \(=\dfrac{1}{2}\times\dfrac{10}{3}=\dfrac{5}{3}\)

 

c.     \(\text{Total Area}\) \(=2\times\ \text{Area from 0 to 1}\)
    \(=2\times \dfrac{h}{2}\Big(f(0)+2.f(0.5)+f(1)\Big)\quad \text{where }h=\dfrac{1}{2}\)
    \(=\dfrac{1}{2}\Big(2+2\times\dfrac{7}{4}+1\Big)=\dfrac{13}{4}\)

Calculus, MET2 2022 VCAA 17 MC

A function `g` is continuous on the domain `x \in[a, b]` and has the following properties:

  • The average rate of change of `g` between `x=a` and `x=b` is positive.
  • The instantaneous rate of change of `g` at `x=\frac{a+b}{2}` is negative.

Therefore, on the interval `x \in[a, b]`, the function must be

  1. many-to-one.
  2. one-to-many.
  3. one-to-one.
  4. strictly decreasing.
  5. strictly increasing.
Show Answers Only

`A`

Show Worked Solution

The first property shows that the point (`a` , __ ) must be below (`b` , __ ) to give a positive rate of change.

The second property gives the midpoint of `a` and `b`. But this point has a negative instantaneous rate of change.

Therefore, this shows that the function `g(x)` has turning points and is in the shape of a cubic, so many-to-one is the correct description.

 
`=>A`


♦♦ Mean mark 39%.

Calculus, MET2 2021 VCAA 13 MC

The value of an investment, in dollars, after `n` months can be modelled by the function

`f(n) =  2500 xx (1.004)^n`

where  `n ∈ {0, 1, 2, ...}`.

The average rate of change of the value of the investment over the first 12 months is closest to

  1. $10.00 per month.
  2. $10.20 per month.
  3. $10.50 per month.
  4. $125.00 per month.
  5. $127.00 per month.
Show Answers Only

`B`

Show Worked Solution

`text(By CAS):`

`text{Average ROC}` `={f(12) – f(0)}/{12 – 0}`
  `= 10.2229 …`

`=> B`

Calculus, MET2 2019 VCAA 3 MC

Let  `f: R\ text(\){4} -> R, \ f(x) = a/(x - 4),\ \ text(where)\ \ a > 0`.

The average rate of change of  `f`  from  `x = 6`  to  `x = 8`  is

A.   `a log_e(2)`

B.   `a/2 log_e(2)`

C.   `2a`

D.   `-a/4`

E.   `-a/8`

Show Answers Only

`E`

Show Worked Solution

`f(6) = a/2`

`f(8) = a/4`

`:.\ text(Average ROC)` `= (a/4 – a/2)/(8 – 6)`
  `= -a/8`

`=>   E`

Calculus, MET2 2017 VCAA 9 MC

The average rate of change of the function with the rule  `f(x) = x^2 - 2x`  over the interval  `[1, a]`, where  `a > 1`, is `8`.

The value of `a` is

  1.  `9`
  2.  `8`
  3.  `7`
  4.  `4`
  5.  `1+ sqrt2`
Show Answers Only

`A`

Show Worked Solution
`(f(a) – f(1))/(a – 1)` `= ((a^2-2a)-(1-2))/(a-1)`
`8` `=(a-1)^2/(a-1),\ \ \ (a>0)`
`8` `=a-1`
`a` `=9`

`=> A`

Calculus, MET2 2007 VCAA 4 MC

The average rate of change of the function with rule  `f(x) = x^3 - sqrt (x + 1)`  between  `x = 0`  and  `x = 3`  is

A.   `0`

B.   `12`

C.   `26/3`

D.   `25/3`

E.   `8`

Show Answers Only

`C`

Show Worked Solution

`text(Define)\ \ f(x) = x^3 – sqrt (x + 1)\ \ text(on CAS)`

`text(Average ROC)` `= (f(3) – f(0))/(3 – 0)`
  `= 26/3`

`=>   C`

Calculus, MET2 2010 VCAA 2 MC

For  `f(x) = x^3 + 2x`, the average  rate of change with respect to `x` for the interval  `[1, 5]`  is

A.   `18`

B.   `20.5`

C.   `24`

D.   `32.5`

E.   `33`

Show Answers Only

`E`

Show Worked Solution
`text(Average ROC)` `= (f(5) – f(1))/(5 – 1)`
  `=(135-3)/4`
  `= 33`

`=>   E`

Calculus, MET2 2016 VCAA 4 MC

The average rate of change of the function  `f` with rule  `f(x) = 3x^2 - 2 sqrt(x + 1)`, between  `x = 0 and x = 3`, is

A.   `8`

B.   `25`

C.   `53/9`

D.   `25/3`

E.   `13/9`

Show Answers Only

`D`

Show Worked Solution
`text(Average ROC)` `= {f(3) – f(0)}/(3 – 0)`
  `=[(27-2xxsqrt4) – (-2)]/(3-0)`
  `= 25/3`

`=>   D`

Algebra, MET2 2012 VCAA 2 MC

For the function with rule  `f(x) = x^3 - 4x`, the average rate of change of  `f(x)`  with respect to `x` on the interval `[1,3]` is

A.   `1`

B.   `3`

C.   `5`

D.   `6`

E.   `9`

Show Answers Only

`=> E`

Show Worked Solution
`text(Average ROC)` `= (f(3) – f(1))/(3 – 1)`
  `=(15-(-3))/2`
  `= 9`

`=> E`

Calculus, MET2 2013 VCAA 6 MC

For the function  `f(x) = sin (2 pi x) + 2x,`  the average rate of change for  `f(x)`  with respect to  `x`  over the interval  `[1/4, 5]`  is

A.   `0`

B.   `34/19`

C.   `7/2`

D.   `(2 pi + 10)/4`

E.   `23/4`

Show Answers Only

`B`

Show Worked Solution

`text(Average rate of change for)\ \f(x)\ \ text(over)\ \  [1/4, 5]`

`= (f(5) – f(1/4))/(5 – 1/4)`
`= (sin (10 pi) + 10 – (sin\ pi/2 +1/2))/(19/4)`
`= 34/19`

`=>   B`