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)=55.6-22.1v^.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 O 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.
a) To find W'(20), we need to take the derivative of the wind chill function W(v) with respect to v and then evaluate it at v = 20.
The wind chill function is given as W(v) = 55.6 - 22.1v^0.16.
To find the derivative, we apply the power rule to the term -22.1v^0.16:
W'(v) = 0 - 0.16*22.1*v^(0.16 - 1) = -3.536v^(-0.84).
Substituting v = 20 into this expression gives:
W'(20) = -3.536 * 20^(-0.84) ≈ -0.2855.
The meaning of W'(20) in terms of wind chill is the rate at which the wind chill is changing with respect to wind velocity at v = 20 mph. In other words, it represents how much the wind chill will decrease (since the derivative is negative) for each mph increase in wind velocity when it is at 20 mph.
b) To find the average rate of change of W over the interval 5 ≤ v ≤ 60, we need to find the change in W divided by the change in v over that interval.
W(v) = 55.6 - 22.1v^0.16.
The change in W over the interval is: ΔW = W(60) - W(5).
The change in v over the interval is: Δv = 60 - 5.
Substituting these values into the wind chill function, we can calculate ΔW and Δv.
Then, the average rate of change of W over the interval is given by:
Average rate of change = ΔW / Δv.
For the instantaneous rate of change of W to be equal to the average rate of change over the interval, we need to find the value of v at which W'(v) (the derivative of W) is equal to the average rate of change.
To find this value, we set W'(v) = average rate of change and solve for v.
c) To find the rate of change of the wind chill with respect to time, we need to differentiate W(v) with respect to time (t) instead of v.
We are given that the air temperature is constant at 32 degrees Fahrenheit over the time interval 0 ≤ t ≤ 4 hours. This means the wind chill function W(v) remains constant with respect to time (since air temperature is not affecting it).
At time t = 0, the wind velocity is v = 20 mph. We are given that the wind velocity increases at a constant rate of 5 mph per hour. This means the wind velocity can be expressed as v(t) = 20 + 5t.
To find the rate of change of the wind chill with respect to time at t = 3 hours, we differentiate W(v(t)) with respect to t and substitute t = 3.
Remember that the chain rule would be used to differentiate W(v(t)) with respect to t. Afterwards, plug in the values into the expression and simplify to find the rate of change with respect to time at t = 3 hours.