Question
I understand how to solve this problem ,however cannot figure out why the .5 degrees is being converted to radians:
PROBLEM:
How would one determine the percent uncertainty in theta and in sin of theta for the following:
1) theta = 15.0 degrees +/- 0.5 degrees
SOLUTION:
f(θ) = sin(θ)
f'(θ) = cos(θ)
For θ = 15°
d(cos(θ))
= ±cos(θ) dθ
= ±cos(15°) (0.5°)
= ±0.9659*(0.5*π/180) <---- Why is .5 degrees converted to radians?
= ±0.00843
PROBLEM:
How would one determine the percent uncertainty in theta and in sin of theta for the following:
1) theta = 15.0 degrees +/- 0.5 degrees
SOLUTION:
f(θ) = sin(θ)
f'(θ) = cos(θ)
For θ = 15°
d(cos(θ))
= ±cos(θ) dθ
= ±cos(15°) (0.5°)
= ±0.9659*(0.5*π/180) <---- Why is .5 degrees converted to radians?
= ±0.00843
Answers
Related Questions
Solve each problem.
Find the quotient and remainder when
x^2 – 5x + 9 is divided by x – 3....
How many kilojoules of heat are absorbed when 1.00L of water is heated from 18 degrees C to 85 degre...
The KC for the reaction
I2 --> 2I
is 3.8 x 10 ^ -5 at 727 degrees C.
Calculate Kc and Kp for t...
The conventional problem-solving strategies usually taught have been:
A. act out the problem, mak...