QUESTION 3
Easy
Consider the relationship between temperature measured in degrees Celsius(°C) and degrees Fahrenheit(°F), which is given by °C = a °F + b.
Find a and b, given that ice melts at 0 degrees Celsius and 32 degrees Fahrenheit, and water boils at 100 degrees Celsius and 212 degrees Fahrenheit.
(Hint: When °C = 0, °F = 32 and when °C = 100, °F = 212. Use this information to find a and b, and thus , the linear relationship between °C and °F.
SOLUTION
1
Consider the given relationship
We have: \(°C = a°F + b\)
2
Use the two given points
When \(°C = 0, °F = 32\), the equation becomes
\(0 = a(32) + b\) ---(I)
When \(°C = 100, °F = 212\), the equation becomes
\(100 = a(212) + b\) ---(II)
\(0 = a(32) + b\) ---(I)
When \(°C = 100, °F = 212\), the equation becomes
\(100 = a(212) + b\) ---(II)
3
Solve the equations to find \(a\) and \(b\)
Subtracting equation (I) from (II), we get
\(100 − 0 = a(212 − 32) + (b − b) \)
⇒ \(100 = 180a\)
⇒ \(a = \frac{100}{180} \)
⇒ \(a = \frac{5}{9} \)
From (I), we have
\(0 = \frac{5}{9} × 32 + b\)
⇒ \(b = − \frac{160}{9} \)
\(100 − 0 = a(212 − 32) + (b − b) \)
⇒ \(100 = 180a\)
⇒ \(a = \frac{100}{180} \)
⇒ \(a = \frac{5}{9} \)
From (I), we have
\(0 = \frac{5}{9} × 32 + b\)
⇒ \(b = − \frac{160}{9} \)
4
Write the final relationship
\(°C = \frac{5}{9} °F − \frac{160}{9} \)
Multiply by 9 to get rid of fractions:
\(9°C = 5°F − 160 \)
⇒ \(5°F = 9°C + 160\)
⇒ \(°F = \frac{9}{5}°C + 32\), which is the standard form.
Multiply by 9 to get rid of fractions:
\(9°C = 5°F − 160 \)
⇒ \(5°F = 9°C + 160\)
⇒ \(°F = \frac{9}{5}°C + 32\), which is the standard form.
🏆
Final Answer :
\(a = \frac{5}{9} , b = − \frac{160}{9} \)
Linear relationship: \(°F = \frac{9}{5}°C + 32\)
Concept Note
The two measurement scales for temperature are related linearly because equal changes in one correspond to equal changes in the other. The slope \(\frac{5}{9}\) means a change of 9°F equals a change of 5°C. The intercept adjusts for the different zero points(0°C = 32°F).