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Calculus Level 2

A curve has parametric equations x = 8 e − 2 t − 4 x = 8e^{-2t} - 4 and y = 2 e 2 t + 4 y = 2e^{2t} + 4 .

Find d x d t \dfrac{dx}{dt} .

− 4 e − 2 t -4e^{-2t} − 16 e − 2 t -16e^{-2t} It doesn't exist − 8 e − 2 t -8e^{-2t}

Rishabh Sood by Rishabh Sood , 20, India

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1 solution

Tom Engelsman
Oct 9, 2016

We only need to find dx/dt. Since we're given x(t) = 8 exp(-2t), dx/dt = 8 * (-2 exp(-2t)) = -16 * exp(-2t).

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