Linear approximation of error

Question

The distance, in feet, a stone drops in t seconds is given by d(t) =16t^2. The depth of a hole is to be approximated by dropping a rock and listening for it to hit the bottom. Assume that time measurement is accurate to 1 / 10^th of a second. Use a linear approximation to estimate an upper bound for the propagated error if the measured time is:

a) 3 seconds

b) 6 seconds

I have found a tangent line that models this line well, y=96x-144. What is the question asking for when it says "estimate the upper bound for propagated error"? How would I solve for this?

Answer 1

Using the tangent line as (y - y₀) = f'(t₀)(t-t₀) (or equivalently Δy = dy/dt Δt), I will call the left side the propagated error. I also exchanged d for y for aesthetic purposes.

a) Δy = 32 t Δt = 32 × 3 × 0.1 = 0.96 ft

b) Δy = 32 t Δt = 32 × 6 × 0.1 = 1.92 ft

This is an upper bound because the largest Δt was used.