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December 20, 2014

December 20, 2014

Posted by **Leannna** on Sunday, January 23, 2011 at 10:59pm.

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 11:40pma) 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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