Fahrenheit Celsius Equation Derivation William Greco Dec. 19, 2009 2404 6j4w6h

Fahrenheit Celsius Equation Derivation William Greco Dec. 19, 2009 2404 Greensward N. Warrington, Pa. 18976 [email protected]

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Introduction (Executive Summary) One of the most common linear relationships found in everyday situations deals with temperature. Recall that water freezes at 32 degrees Fahrenheit and 0 degrees Celsius, while it boils at 212 degrees Fahrenheit and 100 degrees Celsius. The equation is very familiar to most people, however the derivation of this equation is not well known. The ordered pairs (100,212) and (0,32) are shown in figure-1 on axes showing Fahrenheit as a function of Celsius.

The Derivation (100,212)-(0,32) Slope is equal to the change in y over the change in x, or 'rise over run'. change in y m change in x The change in x is equal to the difference in x-coordinates (also called run), and the change in y is equal to the difference in y-coordinates (also called rise). y2  y1 m x2  x1 Substitute in the values of x and y into the equation to find the slope. 32  212 m 0  100

Fahrenheit Celsius Equation Derivation William Greco Dec. 19, 2009 2404 Greensward N. Warrington, Pa. 18976 [email protected]

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Combine all similar expressions. m

32  212 100

Multiply -1 by the 212  32  212  m     100  Subtract 212 from 32 to get -180.  180  m     100  Reduce the expression

180 100

by removing a factor of 20 from the numerator and

denominator. 9 m    5 Multiply -1 by the (9)/(5) inside the parentheses.  9 m     5 9 Multiply -1 by the  inside the parentheses. 5 9 m 5 After finding the slope between the points, use point-slope form to setup the equation. y2  y1 Point-slope is derived from the equation for slope m  x2  x1

y  212  95 x  100  Remove the parentheses around the expression y  212  y  212  

9 x  100 5

Fahrenheit Celsius Equation Derivation William Greco Dec. 19, 2009 2404 Greensward N. Warrington, Pa. 18976 [email protected]

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Multiply -1 by the 100 inside the parentheses. 9 x  100  y  212   5 Multiply -1 by the 212 inside the parentheses. 9 x  100  y  212  5 Since -212 does not contain the variable to solve for, move it to the right-hand side of the equation by adding 212 to both sides. 9 x  100  y  212  5 Simplify the right-hand side of the equation. 9 x  100  212  5 Multiply each term by a factor of 1 that will equate all the denominators. In this case, all need a denominator of 5.  5  9 x  100  212    5 5 Multiply the expression by a factor of 1 to create the least common denominator (LCD) of 5. 212 5  9 x  100   5 5 Multiply 212 by 5 to get 1060. 1060 9 x  100   5 5 The numerators of expressions that have equal denominators can be combined. In this 9 x  100  1060 case, and have the same denominator of 5, so the numerators can be 5 5 combined. 1060 9 x  100  5 Simplify the numerator of the expression. 1060  9x  900 5

Fahrenheit Celsius Equation Derivation William Greco Dec. 19, 2009 2404 Greensward N. Warrington, Pa. 18976 [email protected]

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Subtract 900 from 1060 to get 160. 160  9x 5 Reorder the polynomial 160+9x alphabetically from left to right, starting with the highest order term. 9x  160 5 After finding the slope between the points, use point-slope form to setup the equation. y2  y1 Point-slope is derived from the equation for slope m  x2  x1

y  32  95 x  0  Remove the parentheses around the expression y-(32). 9 y  32   x  0  5 Multiply -1 by the 0 inside the parentheses. 9 x  0  y  32   5 Remove the 0 from the polynomial; adding or subtracting 0 does not change the value of the expression. 9 x  y  32   5 Multiply 9 by x in the numerator. 9x y  32   5 Multiply -1 by the 32 inside the parentheses. 9x y  32  5 Since -32 does not contain the variable to solve for, move it to the right-hand side of the equation by adding 32 to both sides. 9x y  32  5 Move all not containing y to the right-hand side of the equation. 9x y  32 5

Fahrenheit Celsius Equation Derivation William Greco Dec. 19, 2009 2404 Greensward N. Warrington, Pa. 18976 [email protected]

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The two possible equations in point slope form are 9x  160 and y  9x  32 y 5 5

Fahrenheit 

9 Celsius   160  and Fahrenheit  9 Celsius   32 5

5

Re-arrangement to solve for Celsius : y

9x  160 

5 Since x is on the right-hand side of the equation, switch the sides so it is on the left-hand side of the equation. 9x  160   y 5 Multiply each term in the equation by 5. 9x  160  5  y 5  5 Simplify the left-hand side of the equation by canceling the common factors. 9x  160   y 5 Multiply y by 5 to get 5y. 9x  160  5y Since 160 does not contain the variable to solve for, move it to the right-hand side of the equation by subtracting 160 from both sides. 9x  160  5y Move all not containing x to the right-hand side of the equation. 9x  5y  160 Divide each term in the equation by 9. 9x 5y 160   9 9 9 Simplify the left-hand side of the equation by canceling the common factors. 9 x 5y 160 5y 160   x  9 9 9 9 9

Fahrenheit Celsius Equation Derivation William Greco Dec. 22, 2009 2404 Greensward N. Warrington, Pa. 18976 [email protected]

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Simplify the right-hand side of the equation by simplifying each term. 5y 160  9 9 Combine the numerators of all expressions that have common denominators. 5y  160 9 Factor out the GCF of 5 from each term in the polynomial. 5y  5 32  9 Factor out the GCF of 5 from 5y-160. 5 y  32  9 5 y  32  x 9 x

Celsius 

5 Fahrenheit  32 9