Converting between Celsius and FahrenheitDate: 06/23/2009 at 15:47:27 From: Jeff Subject: Why does the formula °F = 9/5(°C + 40) - 40 work The formulas °F = 9/5(°C + 40) - 40 and °C = 5/9(°F + 40) - 40 both work and are easy to remember because all you do is change 5/9 to 9/5 depending on which way you want to convert and 9/5 > 5/9 and °F > °C. With the regular equations I can never remember do I add then multiply or multiply then add or is it subtract? So the above equations remove the confusion. But when I tried to derive these formulas with algebra using F = mC + b or C = mF + b, I just keep getting the standard forms of °F = 9/5°C + 32 and °C = 5/9(°F - 32). Is there some other way I should be approaching this? Date: 06/23/2009 at 21:39:24 From: Doctor Rick Subject: Re: Why does the formula °F = 9/5(°C + 40) - 40 work Hi, Jeff. The formula F = (9/5)C + 32 is in slope-intercept form, so that's what you get when you work with the slope-intercept form y = mx + b. You can use another form for the equation of a straight line--the point-slope form: y = m(x - h) + k where m is the slope and (h, k) is a point on the line. One point that we know is on the line is (-40, -40)--that is, -40 F and -40 C are the same temperature. Knowing that 9/5 degree F = 1 degree C, the slope is 9/5, so plugging these values into the form, we get: F = (9/5)(C - -40) + -40 F = (9/5)(C + 40) - 40 Likewise, if C is the independent variable, the slope is 5/9. The known point is the same, (-40, -40), since switching coordinates doesn't change it. Thus we get C = (5/9)(F - -40) + -40 C = (5/9)(F + 40) - 40 - Doctor Rick, The Math Forum http://mathforum.org/dr.math/ Date: 06/24/2009 at 16:17:37 From: Jeff Subject: Thank you (Why does the formula °F = 9/5(°C + 40) - 40 work) Thank you very much for your solution to my problem. I hope that you can see why I prefer this form of the equation (less to remember = easier to remember). |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/