Related Errors (Calculus)

Question

For rotationally symmetric objects that have two orientations that are stable when set on a flat surface (for example, a standard model for a chess pawn (Links to an external site.) or (approximately) a Hershey's Kiss (Links to an external site.)), the relationship between the tipping angle θ (the maximum angle you can lean the object over before it falls over) and the probability p that the object will land upright when randomly tossed is given by

Based on tossing a particular rotationally symmetric object many times, it has been determined that that p = 0.16 ± 0.069. If this experimentally determined value for p is used to calculate a value for the tipping angle θ, what is the approximate margin of error for θ? Give your answer correctly rounded to 4 decimal places.

Answer 1

p=sin²(θ/2)=½-½cos(θ) (using trig identity).

dp/dθ=½sin(θ).

For small changes ∆p and ∆θ, ∆p/∆θ=dp/dθ.

Therefore ∆p=0.069=½sin(θ)∆θ, ∆θ=0.138/sin(θ).

When p=0.16, cos(θ)=1-2sin²(θ/2)=1-2p=0.68.

sin(θ)=√(1-cos²(θ))=√(1-0.4624)=0.7332 approx.

∆θ=0.138/0.7332=0.1882.

So the margin of error in θ is 0.1882 radians (10.7838º).

When p=0.16, sin²(θ/2)=0.16, sin(θ/2)=0.04, θ=2arcsin(0.04)=0.0800 (4.5849º)

θ=0.08±0.1882 or 4.5849±10.7838 degrees [-0.1082,0.2682] or [-6.1989º,15.3687º].