Answer to Question #146034 in Mechanical Engineering for Yashwanth

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.
1
Expert's answer
2020-12-23T04:56:17-0500

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".


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