# continuity

• Apr 22nd 2006, 02:31 PM
becky
continuity
Suppose air temp. is T (degrees F) and wind speed is v (mph). Then the equivalent windchill temp. is given by the function

W(v) = T if 0 <=v<=4
91.4 + (91.4-T)(.0203v -.304sqrt v - .474) if 4<v<45
1.6T -55 if v>=45

For T=30 degrees F, what wind speed corresponds to a windchill temp. of 0 degrees F?

I assume in this problem that I would set the second equation (beginning with 91.4 +(91.4-T), to 0 and solve for v, but that approaches isn't working very well so far. Am I wrong in my approach to the solution?
• Apr 22nd 2006, 03:48 PM
topsquark
Quote:

Originally Posted by becky
Suppose air temp. is T (degrees F) and wind speed is v (mph). Then the equivalent windchill temp. is given by the function

W(v) = T if 0 <=v<=4
91.4 + (91.4-T)(.0203v -.304sqrt v - .474) if 4<v<45
1.6T -55 if v>=45

For T=30 degrees F, what wind speed corresponds to a windchill temp. of 0 degrees F?

I assume in this problem that I would set the second equation (beginning with 91.4 +(91.4-T), to 0 and solve for v, but that approaches isn't working very well so far. Am I wrong in my approach to the solution?

Is your problem in how to solve for v? I get

$0=91.4+(91.4-30)(.0203v-.304 \sqrt v-.474)$

$0 = 1.2464v-18.6656 \sqrt v+62.2964$

Let $t = \sqrt v$

Then $0 = 1.2464t^2-18.6656t+62.2964$

Using the quadratic formula I get t = 9.9545 and t = 5.02087, which corresponds to v = 99.092 and v = 25.2092.

The solution, then, is v = 25.2092, since that's inside the correct range for the formula.

-Dan