Posted by Leannna on Sunday, January 23, 2011 at 10:59pm.
The wind chill is the temperature, in degrees Fahrenheit, a human feels based on the air temperature, in degrees Fahrenheit, and the wind velocity (v), in miles per hour (mph). If the air temperature is 32degrees Fahrenheit, then the wind chill is given by W(v)=55.622.1v^.16 and is valid for 5 is < or = v which is < or = 60.
a) Find W'(20). Using the correct units, explain the meaning of W'(20) in terms of wind chill.
I got W'(20)=.2855 but don't know what that means.
b) Find the average rate of change of W over the interval 5 is < or = v which is < or = 60. Find the value of v at which the instantaneous rate of change of W is equal to the average rate of change of W over the interval 5 is < or = v which is < or = 60.
I got the average rate of change as .2538 but i don't know how to do the instantaneous rate of change.
c) Over the time interval 0 is < or = to t which is < or = to 4 hours, the air temperatures is a constant 32degrees Fahrenheit. At time t=0, the wind velocity is v=20mph. if the wind velocity increases at a constant rate of 5mph per hour, what is the rate of change of the wind chill with respect to time at t=3 hours? Indicate units of measure.
All I know is that you have to do the derivative but i don't know about c.

Calc  Reiny, Sunday, January 23, 2011 at 11:40pm
a) The negative means that at 20 mph , the rate at which the windchill factor is "decreasing" is .2855 degrees per 1 mph
b) you obviously had the derivative correct in a) to get the correct answer.
Assuming your average  .2538 is correct , set
3.536 v^(.84) = .2538
v^(84/100) = .071776
v^(84/100) = 1/.071776 = 13.93223
raise both sides to the 100/84
v = 23.01
velocity is appr. 23 mph
c) at t=0 , v=20
at t=1, v = 25
at t = 3 , v = 35
put v = 35 into your derivative
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