Differentiation: Exponential and Logarithmic functions - Max Temperature

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Evi Papadaki

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Needs to be tested

2 years, 7 months ago

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Christian Lawson-Perfect 2 years, 5 months ago

Published this.

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Evi Papadaki 2 years, 7 months ago

Gave some feedback: Needs to be tested

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Evi Papadaki 2 years, 8 months ago

Created this as a copy of Differentiation: Exponential and Logarithmic functions - Gradient of Hill.

Name	Status	Author	Last Modified
Differentiation: Exponential and Logarithmic functions - Gradient of Hill	Needs to be tested	Evi Papadaki	30/01/2025 09:40
Differentiation: Exponential and Logarithmic functions - Max Temperature	Needs to be tested	Evi Papadaki	12/04/2024 19:18

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Question statement

Iron is mixed with an acid in an insulated container. The temperature, $T$ °C, above room temperature, at time $t$ minutes after the metal is added to the acid is given by:

$T=\var{a}te^{-t}$

Give any introductory information the student needs.

			Name	Type	Generated Value

a	integer	57
t1	integer	4
Ansa	number	-3.13
Ansb	number	20.97

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Part b) Gap-fill
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Calculate the rate of change of the temperature $\var{t1}$ minutes after the metal is added.

When $t=\var{t1}$ minutes then $\frac{dT}{dt}=$ Unnamed gap $^\circ \mathrm{C}/\mathrm{min}$ .

Give your answer to two decimal places, if needed.

Advice

a) To find the rate of change at a specific time, we need to first differentiate the function

$T=\var{a}te^{-t}$

with respect to $t$ .

To differentiate the function, we need to notice that $T$ is of the form $T=u(t) \times v(t)$ . In other words $T$ is the product of the functions $u=\var{a}t$ and $v=e^{-t}$ .

Therefore, we can use the product rule:

$\begin{split} \frac{dT}{dt}&= \frac{du}{dt}\times v+u\times \frac{dv}{dt} \end{split}$

First, we need to calculate the derivatives $\frac{du}{dt}$ and $\frac{dv}{dt}$ :

$\begin{split} u(t)&=\var{a}t\qquad \text{and} \qquad v(t)&=e^{-t} \\ \frac{du}{dt}&=\var{a}~~~~~\qquad \qquad \frac{dv}{dt}&=-e^{-t}\end{split}$

Substituting these results into the product rule formula, we can obtain the derivative of $T$ :

$\begin{split} \frac{dT}{dt}&= \frac{du}{dt}v+u\frac{dv}{dt} \\ &=\var{a}e^{-t}+\var{a}t \left(-e^{-t} \right) \\ &=\var{a}e^{-t}-\var{a}te^{-t}\end{split}$

Factorising,

$\begin{split} \frac{dT}{dt}&= \var{a}e^{-t}(1-t)\end{split}$

Now, we can substitute $t=\var{t1}$ and calculate the rate of change of the temperature $\var{t1}$ minutes after the metal is added:

$\begin{split} \frac{dT}{dt}&= \var{a}e^{-\var{t1}}(1-\var{t1}) \\&= \simplify{{a}e^{-{t1}}(1-{t1})} \\ &=\var{ansa} ~~\text{[2.dp]}\end{split}$

Therefore, the rate of change of the temperature when $t=\var{t1}$ is $\frac{dT}{dt}=\var{ansa} ^\circ C/min$ .

b) When the temperature reaches maximum, the rate of change of the temperature will be $0$ . So, to find the time when the temperature reaches its maximum we must solve $\frac{dT}{dt}=0$ . Using our answer from a):

$\begin{split} \var{a}e^{-t}(1-t)&=0 \end{split}$

Since, $\var{a}e^{-t} > 0$ for every value of $t$ , this equation can only be equal to zero when

$\begin{split} 1-t &=0 \\ t&=1 \end{split}$

Thus, the maximum temperature will be reached when $t=1$ .

Finally, to calculate the temperature we need to substitute $t=1$ in the original function $T$ :

$\begin{split} T&=\var{a}te^{-t} \\\\ T_{max}&=\var{a}\times 1\times e^{-1} \\&=\frac{\var{a}}{e} \\ &=\var{ansb} ~~\text{[2.d.p.]} \end{split}$

Thus, the maximum temperature of the mixture is $\var{ansb} ^\circ C$ above room temperature.

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This question is used in the following exams:

Further Differentiation by Evi Papadaki in MASH Bath: Moodle quizzes and TS.
Quiz 3 by Kate Vincent in Kate's workspace.
Natural Sciences by Evi Papadaki in Evi's workspace.
Natural Sciences by Evi Papadaki in Evi's workspace.

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Differentiation: Exponential and Logarithmic functions - Max Temperature

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Christian Lawson-Perfect 2 years, 5 months ago

Evi Papadaki 2 years, 7 months ago

Evi Papadaki 2 years, 8 months ago

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Differentiation: Exponential and Logarithmic functions - Max Temperature

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Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

Christian Lawson-Perfect 2 years, 5 months ago

Evi Papadaki 2 years, 7 months ago

Evi Papadaki 2 years, 8 months ago

Error in variable testing condition

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Warning

Generated value: integer

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Generated value: `integer`