smc-746-60-Composite function

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.

Express the derivative \(f^{\prime}(t)\) as a hybrid function. (2 marks)

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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)

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

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ii. Find the value of \(t\) for which \(g^{\prime}(t)=10\).
Give your answer correct to three decimal places. (1 mark)

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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)

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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)

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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.

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

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ii. 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)

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

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

\(=22-10e^{-6t}\)

\(\text{Using CAS:}\)

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

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. \(\text{Using CAS:}\)

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

\(\dfrac{d}{dx}(f(x)\times g(x))\)	\(=f(x)g^{′}(x)+g(x)f^{′}(x)\)
	\(=f(-2)g^{′}(-2)+g(-2)f^{′}(-2)\)
	\(=-7\times 2+8\times 3\)
	\(=10\)

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