Exponential growth and decay practice question 7
An object’s temperature is degrees Celsius after minutes. Which differential equation does satisfy, and what is the room temperature?
Check two things in each choice: does really satisfy its equation, and is its room temperature where ends up in the long run?
Differentiate and compare with .
As , what happens to , and so to ? The object settles at the temperature of the room.
Theory UBC Math 100 §11.4.1CLP-1 §3.3Key results
Worked solution
Strategy
Newton’s law says the rate of change is proportional to the difference from the surroundings. Differentiating the given formula shows exactly that, with the surroundings equal to the constant term.
Working
Differentiate and rewrite in terms of T.
The object approaches the room temperature, 24 °C, and started 60 degrees above it at 84 °C.
Answer
The answer is T prime equals negative 0.1 times the quantity T minus 24, and the room is at 24 degrees.
Check
At t = 0 the formula gives T = 84 and T′ = −6, and −0.1(84 − 24) = −6 as well, while 0.1(T − 24) would give +6.
Where this goes wrong
Reading 84 °C, the initial temperature, as the room temperature. The room is where the object ends up, which is the constant term 24.
The idea behind it
The exponential growth and decay lesson, a figure to move, in a few steps
Leans on
The derivative of ekt, from The chain rule and logarithms
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