Question 2
A substance is being heated such that its temperature $T$ in °C after $t$ minutes is given by the function $T=2e^{0.5t}$
The first integer value of $t$ for which the instantaneous rate of change of temperature is greater than 100 °C per minute is
That’s it. Step through the hints below to see why. In 2021, 44.4% of Queensland students chose (A).
Not quite. The rate reaches 100 at $t=2\ln(100)\approx9.21$. Rounding that down gives 9, but at $t=9$ the rate is only about 90.02 °C per minute. The first integer past 9.21 is 10. In 2021, 15.7% of Queensland students chose (B).
Not quite. $t=8$ is the first integer where the temperature itself is over 100 °C: $2e^{0.5t}\gt100$ gives $t\gt2\ln(50)\approx7.82$. The question asks about the rate of change, so differentiate first. In 2021, 34.9% of Queensland students chose (C).
Not quite. 7 comes from solving $2e^{0.5t}=100$, which uses the temperature rather than its rate of change, then rounding $t\approx7.82$ down. Differentiate first, and round up to the next integer. In 2021, 4.6% of Queensland students chose (D).
At $t=2\ln(100)\approx9.21$ the rate is exactly 100 °C per minute. It has to be greater than 100 at an integer value of $t$, so round up to $t=10$, not down to 9.
At $t=10$ the rate is about 148.41 °C per minute, over 100. At $t=9$ it was about 90.02, still under. So 10 is the first integer that works.
At $t=9$ the rate is about 90.02 °C per minute, still under 100. The rate only reaches 100 at $t\approx9.21$, so 9 is too early.
At the rate is °C per minute, under 100. Move to a later time.
The rate is over 100 here, but it was already over 100 at $t=10$, so this isn’t the first integer.
Here the rate is about °C per minute, 100. The question wants an integer value of $t$, so check the whole numbers either side of the crossing.
The temperature passes 100 °C at $t=2\ln(50)\approx7.82$, which leads to (C) $t=8$, or to (D) $t=7$ if you round down. The question asks when the rate of change is greater than 100, so compare $\tfrac{dT}{dt}$ with 100, not $T$.
The instantaneous rate of change of temperature is the derivative $\tfrac{dT}{dt}$. By the chain rule, the 0.5 comes down in front.
$$\frac{dT}{dt}=2\times0.5e^{0.5t}=e^{0.5t}$$
Set the rate greater than 100 and take the natural log of both sides.
$e^{0.5t}\gt100$$\Rightarrow\ 0.5t\gt\ln(100)$$\Rightarrow\ t\gt2\ln(100)\approx9.21$
You can also graph $y=e^{0.5x}$ and $y=100$ on your calculator and find where they cross.
$t$ has to be an integer greater than 9.21, so round up, not to the nearest. Check the integers either side.
$t=9:\ e^{4.5}\approx90.02\lt100$$t=10:\ e^{5}\approx148.41\gt100$
So the first integer is $t=10$.
$$\frac{dT}{dt}=e^{0.5t}$$
$\displaystyle e^{0.5t}\gt100$$\displaystyle \Rightarrow\ t\gt2\ln(100)\approx9.21$
The first integer greater than 9.21 is 10. At $t=9$ the rate is about 90.02 °C per minute, and at $t=10$ it is about 148.41 °C per minute.
How Queensland students answered in 2021
From the QCAA Mathematical Methods subject report for 2021. The percentages can add to less than 100 because some students didn’t answer.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording is from the 2021 QCAA Mathematical Methods external assessment, © State of Queensland (QCAA) 2021, licensed under CC BY 4.0, and has been adapted. Tangent Tuition is not affiliated with the QCAA.
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