Question #146034

What is the value of the crosstrack error, governed by the ODE e'(t) = -ke(t), at t=2 given that e(0)=4 and k = 1?

Please give your answer with the precision of 2 decimal places.

Please give your answer with the precision of 2 decimal places.

Expert's answer

Let us rewrite the differential equation as "\\frac{d e(t)}{d t} = - k e(t)", or "\\frac{d e(t)}{e(t)} = - k dt".

Integrating from both sides, obtain "\\ln \\frac{e(t)}{C} = - k t", from where "e(t) = C e^{-k t}".

Using initial condition, "e(0) = 4 = C", hence "e(t) = 4 e^{- k t}".

For "k=1", "e(2) = 4 e^{-2} \\approx 0.54".

Learn more about our help with Assignments: Mechanical Engineering

## Comments

## Leave a comment