error and relative error

Question

For a car with standard tires stopping on dry asphalt, its speed S (in mph) can be found from the length of its skid marks according to

S = 9.4L^0.37

where L is the length (in feet) of the skid marks. Find the error and the relative error in the speed for the skid-mark length if the error in measuring the length is ±10 feet.† (Round your answers to two decimal places.)

150 feet

Answer 1

The relative error: dS/S = 0.37 dL/L

You can derive that formula in many ways but the fastest is to take a log and then a differential:

S = 9.4 L^0.37

ln(S) = ln(9.4) + 0.37ln(L)

Take a differential on both sides: dS/S = 0.37 dL/L.

The error will be dS = 0.37 S dL/L

Note: in general when you have a formula like this F = constant * Aⁿ * B^m, using the same method you can show that the relative error is dF/F = n*dA/A + m*dB/B. Then most texts take the absolute values of the two added terms so they always add regardless of the signs of dA and dB (thus giving the maximal possible error when the errors add up in the same direction), or even square root of the sum of their squares, if these are random errors: dF/F = root( (n*dA/A)² + (m*dB/B)² )