C = 35.74 + 0.6215T - 35.75v^0.16 + 0.4275Tv^0.16 where v is...

C = 35.74 + 0.6215T - 35.75v^{0.16} + 0.4275Tv^{0.16} where v is the wind speed in miles per hour and T is the temperature in degrees Fahrenheit. The wind speed is 23 ± 3 miles per hour and the temperature is 8° ± 3°. Use do to estimate the maximum possible propagated error (round your answer to four decimal places) and relative error (as a percent) in calculating the wind chill (round your answer to two decimal places). dC = ± \frac{dC}{C} = ± %

Transcript

00:01 Hi there, so for this problem, we are given this expression in here, which is the formula for the wind chill.

00:12 And this, in terms of the wind speed and the temperature in degrees fahrenheit.

00:20 Now, the information that we are given for this speed is that the speed is equal to 21 plus or minus 3 miles per hour.

00:32 And that the temperature is equal to 8 plus or minus 1 fahrenheit degrees.

00:42 So we need to estimate the differential in c and also the differential in c divided by c.

00:50 Now, the differential in c is what we know as the error or the propagated error.

00:57 And then we can recognize in here that this value right here is the speedless.

01:03 Called this speed b -0 and this plus or minus the differential in the speed and the same for the temperature.

01:14 Okay, so to obtain the error from the c equation, we know that this is the partial derivative of the c function with respect to the speed, this times the differential in the speed.

01:30 This plus the partial derivative of the c function would respect to the speed, this plus the partial derivative of the c function the temperature and this times the differential in the temperature.

01:40 So let me calculate this separately.

01:43 So let's start with the partial derivative of c with respect to the speed.

01:47 Now we know that the first and the second terms are zero because they do not depend on the speed.

01:56 Now the third term, its derivative, is going to be minus 35 .75.

02:07 0 .16, this times the speed elevated to 0 .16 minus 1, so that will give us minus 0 .84.

02:21 And the last term in here, as you can see, is just plus 0 .42, 75.

02:32 And this time 0 .16 times the temperateur and the speed elevated to minus 0 .84.

02:45 Okay.

02:46 So now for the partial derivative of c with respect to the temperate, as you can see, the first term and the third term are zero for this partial derivative because they do not depend on the temperature.

03:04 Peritor.

03:06 So we will have, for the second term, we will have just simply 0 .62 .15.

03:18 And for the last term, that will be plus 0 .42 .75 times the speed elevated to 0 .16.

03:35 Now, we just need to simply substitute all of these values into this expression right here.

03:42 But of course, this is a long term.

03:45 So what i'm going to do is to try to just simply evaluate this at the precise values.

03:52 So, for example, the partial derivative of c with respect to be evaluated when b is given.

04:00 Remember that b, b0 is 21, and the temperature is 8.

04:13 So we just need to simply substitute the values for that.

04:16 So let me simplify that.

04:18 That will be minus 35 .75, that times 0 .16.

04:23 That will give us minus 5 .72, this times the speed that is 21, and that elevated to minus 0 .84.

04:38 And the second term is 0 .42, 75.

04:44 Times 0 .16, so that will give us 0 .064.

04:53 And this times the temperature, that is 8, this times the speed that is 21, and that elevated to minus 0 .84.

05:06 So from this, we obtain that the partial derivative evaluated at these values is equal to, that that gave us a value of minus 0 .401...

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