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 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.
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
a) To find W'(20), we need to differentiate the wind chill function W(v) with respect to v and then substitute v = 20.
W(v) = 55.6 - 22.1v^0.16
Taking the derivative of W(v) with respect to v, we get:
W'(v) = -22.1 * 0.16 * v^(-0.84)
Substituting v = 20 into the derivative, we get:
W'(20) = -22.1 * 0.16 * 20^(-0.84) ≈ -0.2855
So, your calculation for W'(20) is correct.
In terms of wind chill, W'(20) represents the instantaneous rate of change of the wind chill with respect to wind velocity when the wind velocity is 20 mph. Since the derivative is negative, it means that for every 1 mph increase in wind velocity, the wind chill decreases by approximately 0.2855 degrees Fahrenheit.
b) The average rate of change of W over the interval 5 ≤ v ≤ 60 can be found by calculating the difference in values of W at each endpoint and dividing by the length of the interval:
Average rate of change = (W(60) - W(5)) / (60 - 5)
Substituting the values into the wind chill function, we get:
Average rate of change = (55.6 - 22.1(60)^0.16 - 55.6 + 22.1(5)^0.16) / (60 - 5)
Simplifying the expression gives the average rate of change as approximately -0.2538.
To find the value of v at which the instantaneous rate of change of W is equal to the average rate of change, we need to set W'(v) equal to the average rate of change and solve for v:
-22.1 * 0.16 * v^(-0.84) = -0.2538
Unfortunately, this equation cannot be solved algebraically, and we will need to use numerical methods or a graphing calculator to find the approximate value of v.
c) To find the rate of change of the wind chill with respect to time at t = 3 hours, we need to differentiate the wind chill function W(v) with respect to time. Since the wind velocity increases at a constant rate of 5 mph per hour, we can express v as a function of time: v(t) = 20 + 5t.
Now we substitute v(t) into the wind chill function:
W(t) = 55.6 - 22.1(20 + 5t)^0.16
To find the rate of change of W with respect to t, we differentiate W(t) with respect to t:
W'(t) = -22.1 * 0.16 * (20 + 5t)^(-0.84) * 5
Evaluating W'(t) at t = 3 gives the rate of change of the wind chill with respect to time at t = 3 hours.
a) To find W'(20), we need to take the derivative of W(v) with respect to v and then substitute v = 20 into the derivative.
W(v) = 55.6 - 22.1v^0.16
To differentiate this function, we can use the power rule and chain rule:
W'(v) = -22.1 * 0.16 * v^(-0.84)
Now substitute v = 20:
W'(20) = -22.1 * 0.16 * 20^(-0.84) ≈ -0.2855
The units for W'(20) are (degrees Fahrenheit) / (miles per hour).
In terms of wind chill, W'(20) represents the rate at which the wind chill temperature changes with respect to wind velocity at v = 20 mph. Since the value is negative, it means that as wind velocity increases by 1 mph, the wind chill temperature decreases by approximately 0.2855 degrees Fahrenheit.
b) The average rate of change of W over the interval 5 ≤ v ≤ 60 can be found by calculating (W(60) - W(5)) / (60 - 5).
W(60) = 55.6 - 22.1 * 60^0.16
W(5) = 55.6 - 22.1 * 5^0.16
Average rate of change = (W(60) - W(5)) / (60 - 5)
Substitute the values into the equation and calculate to find the average rate of change.
To find the value of v at which the instantaneous rate of change of W is equal to the average rate of change over the interval 5 ≤ v ≤ 60, we need to set W'(v) equal to the average rate of change and solve for v:
W'(v) = -22.1 * 0.16 * v^(-0.84) = average rate of change
Solve the equation to find the value of v.
c) To find the rate of change of the wind chill with respect to time at t = 3 hours, we need to consider that wind velocity increases at a constant rate of 5 mph per hour.
At t = 0, v = 20 mph. Over 3 hours, the wind velocity increases by 5 mph per hour, so at t = 3, v = 20 + (5 * 3) = 35 mph.
Now we have the wind velocity (v = 35 mph) and the air temperature (32 degrees Fahrenheit). We can use the chain rule to find the rate of change of the wind chill with respect to time (t):
dW/dt = dW/dv * dv/dt
Apply the chain rule and substitute the appropriate values to find the rate of change of the wind chill with respect to time at t = 3 hours. The units of measure will be (degrees Fahrenheit) / (hours).
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. The derivative of v^0.16 is 0.16 * v^(0.16-1) = 0.16 * v^(-0.84).
Now, taking the derivative of W(v) = 55.6 - 22.1v^0.16:
W'(v) = 0 - 22.1 * 0.16 * v^(-0.84) = -3.536 * v^(-0.84).
Substituting v = 20 into the derivative:
W'(20) = -3.536 * 20^(-0.84) ≈ -0.2855.
The units of wind chill are degrees Fahrenheit (°F). Therefore, W'(20) = -0.2855 means that for every 1 mph increase in wind velocity (v) when the air temperature is 32°F, the wind chill decreases by approximately 0.2855°F. It represents the rate of change of wind chill with respect to wind velocity.
b) To find the average rate of change of W over the interval 5 ≤ v ≤ 60, we need to calculate the change in W divided by the change in v over that interval.
Change in W = W(60) - W(5) = (55.6 - 22.1*60^0.16) - (55.6 - 22.1*5^0.16)
Change in v = 60 - 5 = 55.
Using the given wind chill function, we can calculate the change in W and then divide it by the change in v:
Average rate of change = (Change in W) / (Change in v) = (W(60) - W(5)) / (60 - 5)
Now, to find the value of v at which the instantaneous rate of change of W is equal to the average rate of change over the interval 5 ≤ v ≤ 60, we need to find the value of v where W'(v) is equal to the calculated average rate of change. This can be done by solving the equation W'(v) = Average rate of change.
c) To find the rate of change of the wind chill with respect to time at t = 3 hours, we need to consider the change in wind velocity with respect to time.
Given that the wind velocity increases at a constant rate of 5 mph per hour, at t = 3 hours, the wind velocity will be 20 + (5 * 3) = 35 mph.
Now, using the wind chill formula W(v) = 55.6 - 22.1v^0.16, we need to differentiate W(v) with respect to v, and then substitute v = 35.
The derivative of W(v) is W'(v) = -3.536 * v^(-0.84).
Substituting v = 35 into W'(v):
Rate of change of wind chill with respect to time at t = 3 hours = W'(35).
After evaluating the expression W'(35), you will get the rate of change of wind chill with respect to time at t = 3 hours. The units of measure will be degrees Fahrenheit per hour (°F/hr).