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Calculus, MET1 2024 VCAA 1

  1. Let  \(y=e^x \cos (3 x)\).
  2. Find  \(\dfrac{d y}{d x}\)   (1 mark)

  3. Let  \(f(x)=\log _e\left(x^3-3 x+2\right)\).
  4. Find  \(f^{\prime}(3)\)   (2 marks)

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a.   \(e^x (\cos(3x)-3\sin(3x))\)

b.   \(\dfrac{6}{5}\)

Show Worked Solution

a.     \(y\) \(=e^x \cos(3x)\)
  \(\dfrac{dy}{dx}\) \(=e^x.(-3\sin(3x))+\cos(3x).e^x\)
    \(=e^x(\cos(3x)-3\sin(3x))\)

 

b.     \(f(x)\) \(=\log_{e}(x^3-3x+2)\)
  \(f^{\prime}(x)\) \(=\dfrac{3x^2-3}{x^3-3x+2}\)
  \(f^{\prime}(3)\) \(=\dfrac{3(3)^2-3}{(3)^3-3(3)+2}\)
    \(=\dfrac{24}{20}=\dfrac{6}{5}\)

 

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 \(t=\dfrac{1}{2}\).

    Give your answer in degrees Celsius per hour.  (1 mark)

  1. Another model for the temperature in the room is given by \(g(t)=22-10 e^{-6 t}, t \geq 0\).
    1. Find the derivative \(g^{\prime}(t)\).  (1 mark)
    1. Find the value of \(t\) for which \(g^{\prime}(t)=10\).

      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.

    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.

    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. Given that \(p(t)\) is continuous for \(t \geq 0\), show that \(A=1.2 e^4\).  (1 mark)

    1. Find how long it takes, after the heater is switched on, until the heater has used 0.5 kilowatt hours of energy.

      Give your answer in hours.  (1 mark)

    1. Find how long it takes, after the heater is switched on, until the heater has used 1 kilowatt hour of energy.

      Give your answer in hours, correct to two decimal places.  (2 marks)

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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, MET2 2023 VCAA 3

Consider the function \(g:R \to R, g(x)=2^x+5\).

  1. State the value of \(\lim\limits_{x\to -\infty} g(x)\).   (1 mark)
  2. The derivative, \(g'(x)\), can be expressed in the form \(g'(x)=k\times 2^x\).
  3. Find the real number \(k\).   (1 mark)
    1. Let \(a\) be a real number. Find, in terms of \(a\), the equation of the tangent to \(g\) at the point \(\big(a, g(a)\big)\).   (1 mark)
    1. Hence, or otherwise, find the equation of the tangent to \(g\) that passes through the origin, correct to three decimal places.   (2 marks)

Let \(h:R\to R, h(x)=2^x-x^2\).

  1. Find the coordinates of the point of inflection for \(h\), correct to two decimal places.   (1 mark)
  2. Find the largest interval of \(x\) values for which \(h\) is strictly decreasing.
  3. Give your answer correct to two decimal places.   (1 mark)
  4. Apply Newton's method, with an initial estimate of \(x_0=0\), to find an approximate \(x\)-intercept of \(h\).
  5. Write the estimates \(x_1, x_2,\) and \(x_3\) in the table below, correct to three decimal places.   (2 marks)
      

    \begin{array} {|c|c|}
    \hline
    \rule{0pt}{2.5ex} \qquad x_0\qquad \ \rule[-1ex]{0pt}{0pt} & \qquad \qquad 0 \qquad\qquad \\
    \hline
    \rule{0pt}{2.5ex} x_1 \rule[-1ex]{0pt}{0pt} &  \\
    \hline
    \rule{0pt}{2.5ex} x_2 \rule[-1ex]{0pt}{0pt} &  \\
    \hline
    \rule{0pt}{2.5ex} x_3 \rule[-1ex]{0pt}{0pt} &  \\
    \hline
    \end{array}
  6. For the function \(h\), explain why a solution to the equation \(\log_e(2)\times (2^x)-2x=0\) should not be used as an initial estimate \(x_0\) in Newton's method.   (1 mark)
  7. There is a positive real number \(n\) for which the function \(f(x)=n^x-x^n\) has a local minimum on the \(x\)-axis.
  8. Find this value of \(n\).   (2 marks)
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a.    \(5\)

b.    \(\log_{e}{2}\ \ \text{or}\ \ \ln\ 2\)

ci.  \(y=2^a\ \log_{e}{(2)x}-(a\ \log_{e}{(2)}-1)\times2^a+5\)

\(\text{or}\ \ y=2^a\ \log_{e}{(2)x}-a\ 2^a\ \log_{e}{(2)}+2^a+5\)

cii. \(y=4.255x\)

d. \((2.06 , -0.07)\)

e. \([0.49, 3.21]\)

f. 

\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex} \qquad x_0\qquad \ \rule[-1ex]{0pt}{0pt} & \qquad \qquad 0 \qquad\qquad \\
\hline
\rule{0pt}{2.5ex} x_1 \rule[-1ex]{0pt}{0pt} &  -1.433 \\
\hline
\rule{0pt}{2.5ex} x_2 \rule[-1ex]{0pt}{0pt} &  -0.897 \\
\hline
\rule{0pt}{2.5ex} x_3 \rule[-1ex]{0pt}{0pt} &  -0.773 \\
\hline
\end{array}

g. \(\text{See worked solution.}\)

h. \(n=e\)

Show Worked Solution

a.    \(\text{As }x\to -\infty,\ \ 2^x\to 0\)

\(\therefore\ 2^x+5\to 5\)
  

b.     \(g(x)\) \(=2^x+5\)
    \(=\Big(e^{\log_{e}{2}}\Big)^x\)
    \(=e\ ^{x\log_{e}{2}}+5\)
  \(g'(x)\) \(=\log_{e}{2}\times e\ ^{x\log_{e}{2}}\)
     \(=\log_{e}{2}\times 2^x\)
  \(\therefore\ k\) \(=\log_{e}{2}\ \ \text{or}\ \ \ln\ 2\)
   
ci.  \(\text{Tangent at}\ (a, g(a)):\)

\(y-(2^a+5)\) \(=\log_{e}{2}\times2^a(x-a)\)
\(\therefore\ y\) \(=2^a\ \log_{e}{(2)x}-a\ 2^a\ \log_{e}{(2)}+2^a+5\)

♦♦ Mean mark (c)(i) 50%.

cii.  \(\text{Substitute }(0, 0)\ \text{into equation from c(i) to find}\ a\)

\( y\) \(=2^a\ \log_{e}{(2)x}-a\ 2^a\ \log_{e}{(2)}+2^a+5\)
\(0\) \(=2^a\ \log_{e}{(2)\times 0}-a\ 2^a\ \log_{e}{(2)}+2^a+5\)
\(0\) \(=-a\ 2^a\ \log_{e}{(2)}+2^a+5\)

  
\(\text{Solve for }a\text{ using CAS }\rightarrow\ a\approx 2.61784\dots\)

\(\text{Equation of tangent when }\ a\approx 2.6178\)

\( y\) \(=2^{2.6178..}\ \log_{e}{(2)x}+0\)
\(\therefore\  y\) \(=4.255x\)

♦♦♦ Mean mark (c)(ii) 30%.
MARKER’S COMMENT: Some students did not substitute (0, 0) into the correct equation or did not find the value of \(a\) and used \(a=0\).
d.     \(h(x)\) \(=2^x-x^2\)
  \(h'(x)\) \(=\log_{e}{(2)}\cdot 2^x-2x\ \ \text{(Using CAS)}\)
  \(h”(x)\) \(=(\log_{e}{(2)})^2\cdot 2^x-2\ \ \text{(Using CAS)}\)

\(\text{Solving }h”(x)=0\ \text{using CAS }\rightarrow\ x\approx 2.05753\dots\)

\(\text{Substituting into }h(x)\ \rightarrow\ h(2.05753\dots)\approx-0.070703\dots\)

\(\therefore\ \text{Point of inflection at }(2.06 , -0.07)\ \text{ correct to 2 decimal places.}\)

e.    \(\text{From graph (CAS), }h(x)\ \text{is strictly decreasing between the 2 turning points.}\)

\(\therefore\ \text{Largest interval includes endpoints and is given by }\rightarrow\ [0.49, 3.21]\)


♦♦ Mean mark (e) 40%.
MARKER’S COMMENT: Round brackets were often used which were incorrect as endpoints were included. Some responses showed the interval where the function was strictly increasing.

f.    \(\text{Newton’s Method }\Rightarrow\  x_a-\dfrac{h(x_a)}{h'(x_a)}\) 

\(\text{for }a=0, 1, 2, 3\ \text{given an initial estimation for }x_0=0\)

\(h(x)=2x-x^2\ \text{and }h'(x)=\ln{2}\times 2^x-2x\)

\begin{array} {|c|l|}
\hline
\rule{0pt}{2.5ex} \qquad x_0\qquad \ \rule[-1ex]{0pt}{0pt} & \qquad \qquad 0 \qquad\qquad \\
\hline
\rule{0pt}{2.5ex} x_1 \rule[-1ex]{0pt}{0pt} &  0-\dfrac{2^0-2\times 0}{\ln2\times 2^0\times 0}=-1.433 \\
\hline
\rule{0pt}{2.5ex} x_2 \rule[-1ex]{0pt}{0pt} &  -1.433-\dfrac{2^{-1.433}-2\times -1.433}{\ln2\times 2^{-1.433}\times -1.433}=-0.897 \\
\hline
\rule{0pt}{2.5ex} x_3 \rule[-1ex]{0pt}{0pt} &  -0.897-\dfrac{2^{-0.897}-2\times -0.897}{\ln2\times 2^{-0.897}\times -0.897}=-0.773 \\
\hline
\end{array}

g.    \(\text{The denominator in Newton’s Method is}\ h'(x)=\log_{e}{(2)}\cdot 2^x-2x\)

\(\text{and the calculation will be undefined if }h'(x)=0\ \text{as the tangent lines are horizontal}.\)

\(\therefore\ \text{The solution to }h'(x)=0\ \text{cannot be used for }x_0.\)

♦♦♦ Mean mark (g) 20%.

h.    \(\text{For a local minimum }f(x)=0\)

\(\rightarrow\ n^x-x^n=0\)

\(\rightarrow\ n^x=x^n\ \ \ (1)\)

\(\text{Also for a local minimum }f'(x)=0\)

\(\rightarrow\ \ln(n)\cdot n^x-nx^{n-1}=0\ \ \ (2)\)

\(\text{Substitute (1) into (2)}\)

\(\ln(n)\cdot x^n-nx^{n-1}=0\) 

\(x^n\Big(\ln(n)-\dfrac{n}{x}\Big)=0\)

\(\therefore\ x^n=0\ \text{or }\ \ln(n)=\dfrac{n}{x}\)

\(x=0\ \text{or }x=\dfrac{n}{\ln(n)}\)

\(\therefore\ n=e\)


♦♦♦ Mean mark (h) 10%.
MARKER’S COMMENT: Many students showed that \(f'(x)=0\) but failed to couple it with \(f(x)=0\). Ensure exact values are given where indicated not approximations.

Calculus, MET1 2023 VCAA 1a

Let  \(y=\dfrac{x^2-x}{e^x}\).

Find and simplify \(\dfrac{dy}{dx}\).   (2 marks)

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\(-\Bigg(\dfrac{x^2-3x+1}{e^x}\Bigg) \)

Show Worked Solution

\(\text{Using the quotient rule:}\)

\(\dfrac{dy}{dx}\) \(=\dfrac{e^x(2x-1)-(x^2-x)e^x}{(e^x)^2}\)
  \(=\dfrac{e^x(-x^2+3x-1)}{e^{2x}}\)
  \(=\dfrac{-x^2+3x-1}{e^x}\)

Calculus, MET1-NHT 2018 VCAA 1a

Let  `f(x) = (e^x)/((x^2 - 3))`.

Find  `f′(x)`.  (2 marks)

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`{e^x(x^2 – 2x – 3)}/{(x^2 – 3)^2}`

Show Worked Solution

`text(Let) \ \ u = e^x \ \ => \ \ u′ = e^x`

 `v = (x^2 – 3) \ \ => \ \ v′ = 2x`

`f′(x)` `= {e^x(x^2 – 3) – 2x e^x}/{(x^2 – 3)^2}`
  `= {e^x(x^2 – 2x – 3)}/{(x^2 – 3)^2}`

Calculus, MET1-NHT 2019 VCAA 1a

Let  `y = (2e^(2x) - 1)/e^x`.

Find  `(dy)/(dx)`.  (2 marks)

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`(dy)/(dx) = 2e^x + e^(-x)`

Show Worked Solution

`text(Method 1)`

`y` `= 2e^x – e^(-x)`
`(dy)/(dx)` `= 2e^x + e^(-x)`

 

`text(Method 2)`

`(dy)/(dx)` `= (4e^(2x) ⋅ e^x – (2e^(2x) – 1) e^x)/(e^x)^2`
  `= (4e^(3x) – 2e^(3x) + e^x)/e^(2x) `
  `= (2e^(2x) + 1)/e^x`

Calculus, MET1 2018 VCAA 1b

Let  `f(x) = (e^x)/(cos(x))`.

Evaluate  `f′(pi)`.  (2 marks)

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`text(See Worked Solutions)`

Show Worked Solution

`f′(x) = (e^x)/(cos(x))`

`u` `= e^x` `v` `= cos(x)`
`u′` `= e^x` `v′` `= −sin(x)`
`f′(x)` `= (u′v – uv′)/(v^2)`
  `= (e^x · cos(x) + e^x sin(x))/(cos^2(x))`

 

`f′(pi)` `= (e^pi · cospi + e^pi sinpi)/(cos^2 pi)`
  `= (e^pi(−1) + e^pi · 0)/((−1)^2)`
  `= −e^pi`

Calculus, MET1 2007 ADV 2ai

Differentiate with respect to `x`:

`(2x)/(e^x + 1)`  (2 marks)

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`(2(e^x + 1 – xe^x))/((e^x + 1)^2)`

Show Worked Solution

`y = (2x)/(e^x + 1)`

`u` `= 2x` `v` `= e^x + 1`
`u′` `= 2` `v′` `= e^x`
`(dy)/(dx)` `= (u′v – uv′)/(v^2)`
  `= (2(e^x + 1) – 2x(e^x))/((e^x + 1)^2)`
  `= (2e^x + 2 – 2x · e^x)/((e^x + 1)^2)`
  `= (2(e^x + 1 – xe^x))/((e^x + 1)^2)`

Calculus, MET2 2009 VCAA 7 MC

For  `y = e^(2x) cos (3x)`  the rate of change of `y` with respect to `x` when  `x = 0`  is

  1. `0`
  2. `2`
  3. `3`
  4. `– 6`
  5. `– 1`
Show Answers Only

`B`

Show Worked Solution
`y` `= e^(2x) cos (3x)`
`dy/dx` `=e^(2x) xx -3sin(3x) + 2e^(2x) xx cos (3x)`
  `=e^(2x)(-3sin(3x) + 2cos(3x))`

 
`text(When)\ \ x = 0,`

`dy/dx= 2`

`=>   B`