Logarithms and Exponentials: Applications 2

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Christian Lawson-Perfect 2 years, 7 months ago

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Ben McGovern 3 years, 10 months ago

Created this as a copy of Logarithms and Exponentials: Applications 1.

Name	Status	Author	Last Modified
Logarithms and Exponentials: Applications 1	Ready to use	Ben McGovern	10/10/2022 12:59
Logarithms and Exponentials: Applications 2	Ready to use	Ben McGovern	10/10/2022 12:59

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Question statement

The temperature $T$ , in degrees $C$ , of a chemical reaction is given by

$T = \var{temp0}e^{\var{k}t}, \quad t\geq0$ where

$t$ is time, measured in seconds.

Give any introductory information the student needs.

			Name	Type	Generated Value

temp0	integer	50
k	rational	3/50
time1	integer	12
temp1	number	102.72
temp2	integer	170
time2	number	20.40
timedub	number	11.55

Automatically regenerate variables?

Name

Data type

Value

xxxxxxxxxx
 
random(50..80 #10)

JME function reference

Description

Describe what this variable represents, and list any assumptions made about its value.

Can an exam override the value of this variable?

Generated value: `integer`

→ Used by:

temp1
time2

This variable doesn't seem to be used anywhere.

Parts

Part a) Gap-fill
1. Gap Mathematical expression
  Add
Part b) Gap-fill
1. Gap Mathematical expression
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Part c) Gap-fill
1. Gap Mathematical expression
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Gap-fill

Ask the student a question, and give any hints about how they should answer this part.

What is the temperature at $\var{time1}$ seconds?

Unnamed gap $^\circ \,C$

(Give your answer to 2 decimal places where necessary)

Advice

Part a:

To calculate the temperature at $\var{time1}$ seconds, we use the formula given with $t=\var{time1}$ :

$\begin{split} T &\,= \var{temp0}\,e^{\var{k}\times\var{time1}} \\ &\,= \var{temp1}\, ^\circ C \text{ (2 d.p.)} \end{split}$

Part b:

To find how long it takes for the temperature to reach $\var{temp2}\, ^\circ C$ , we use the formula given with $T=\var{temp2}$ , and solve for $t$ :

$\begin{split} \var{temp0} \, e^{\var{k}t} &\,= \var{temp2} \\ e^{\var{k}t} &\,= \simplify{{temp2/temp0}} \\ \var{k} t &\,= \ln \left(\simplify[fractionNumbers]{{temp2/temp0}} \right) \\ t &\,=\frac{\ln \left(\simplify[fractionNumbers]{{temp2/temp0}}\right)}{\var{k}} \\ t &\,= \var{time2} \text{ seconds (2 d.p.)}. \end{split}$

Part c:

To calculate the time taken for the temperature to double from the initial temperature, we want to first check the initial temperature by setting $t=0$ , which gives $T_0 = \var{temp0} \,^\circ C$ .

Therefore, we want to find the time taken for the temperature to reach $\var{2*temp0} \,^\circ C$ :

$\begin{split} \var{temp0} \, e^{\var{k}t} &\,= \var{2*temp0} \\ e^{\var{k}t} &\,= 2 \\ \var{k} t &\,= \ln (2) \\ t &\,=\frac{\ln (2)}{\var{k}} \\ t &\,= \var{timedub} \text{ seconds (2 d.p.)}. \end{split}$

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This question is used in the following exams:

Logarithms and Exponentials - Applications by Ben McGovern in MASH Bath: Moodle quizzes and TS.
All the good stuff by Chris Templet in Chris's workspace.
EGD125_Quiz1 by Matt Falk in Matt's workspace.
Core Maths Test 2025 by Angus Rosenburgh in Angus's workspace.

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Christian Lawson-Perfect 2 years, 7 months ago

Ben McGovern 3 years, 8 months ago

Ben McGovern 3 years, 10 months ago

Error in variable testing condition

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Generated value: `integer`

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Logarithms and Exponentials: Applications 2

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

Christian Lawson-Perfect 2 years, 7 months ago

Ben McGovern 3 years, 8 months ago

Ben McGovern 3 years, 10 months ago

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Gap-fill

Generated value: `integer`