Question 545·Medium·Math

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A meteorologist observed that the air temperature fell at a constant rate of $1.8$ degrees Celsius per hour for $10$ consecutive hours. At the end of this period, the temperature was $-2$ degrees Celsius.

What was the air temperature, in degrees Celsius, at the beginning of the 10-hour period?

Your answer

A constant rate of cooling means the temperature falls by the same number of degrees each hour. Multiply the hourly rate by the elapsed time to find the total drop, then write starting temperature minus drop equals ending temperature. A one-unknown Desmos regression finds the start. Subtracting the drop from the ending temperature moves in the wrong direction.

Hints

Hint 1
A constant rate means the same temperature change happens each hour. Multiply the hourly drop by the number of hours to find how much the temperature fell over the whole period.
Hint 2
The negative temperature is the reading at the end, not the start. If you call the starting value $x$ , what equation says that $x$ minus the total drop equals the ending reading?

Step-by-step

Work backward from the ending temperature
Step 1
Find the total temperature drop
A constant rate means the $1.8$ -degree decrease repeats each hour, so multiply it by all $10$ hours. Type $1.8(10)$ in Desmos; it shows $18$ , the total drop in degrees Celsius.
Step 2
Connect the start to the end
The temperature ended at $-2$ degrees Celsius. Starting temperature minus the total drop equals ending temperature. Let $x$ be the starting temperature, so
$x-18=-2.$
Step 3
Solve for the starting temperature
Replace $x$ with $x_1$ and $=$ with $\sim$ : type $x_1-18\sim-2$ . Desmos treats $x_1$ as the unknown starting temperature and shows $x_1=16$ under PARAMETERS. So the air temperature at the beginning was $16$ degrees Celsius. Grid in 16.

Question 545·Medium·Math

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Hints

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