How do you differentiate the following parametric equation: $x (t) = - 2 t e^{t} + 4 t, y (t) = - 2 t^{2} - 3 e^{t}$ $x(t)=-2te^t+4t, y(t)= -2t^2-3e^(t)$ ?

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How do you differentiate the following parametric equation: $x (t) = - 2 t e^{t} + 4 t, y (t) = - 2 t^{2} - 3 e^{t}$ $x(t)=-2te^t+4t, y(t)= -2t^2-3e^(t)$ ?

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Answer 1 · 2017-07-12T15:56:59

the first derivative, $\frac{d y}{d x} = \frac{d y}{d t} / \frac{d x}{d t}$ $dy/dx = dy/dt/dx/dt$ for parametric equations
the reason this holds true could be seen by treating $d y, d x, and d t$ $dy, dx, and dt$ as differentials (which they are) and upon dividing, $d t$ $dt$ cancels out and you are left with $\frac{d y}{d x}$ $dy/dx$

as a refresher, $\frac{d}{d t} (x^{t}) = t x^{t - 1}$ $d/dt(x^t) = tx^(t-1)$ and $d/dt(e^t) = e^t * t'$

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$dy/dt = d/dt (-2t^2 - 3e^t) = -4t - 3e^t$

$dx/dt = d/dt (-2te^t + 4t) = -2(e^t + te^t) + 4$

$= -2e^t - 2te^t + 4$

$dy/dx = dy/dt/dx/dt = (-4t - 3e^t)/(-2e^t - 2te^t + 4)$

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How do you differentiate the following parametric equation: $x (t) = - 2 t e^{t} + 4 t, y (t) = - 2 t^{2} - 3 e^{t}$ $x(t)=-2te^t+4t, y(t)= -2t^2-3e^(t)$ ?

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How do you differentiate the following parametric equation: $x (t) = - 2 t e^{t} + 4 t, y (t) = - 2 t^{2} - 3 e^{t}$ $x(t)=-2te^t+4t, y(t)= -2t^2-3e^(t)$ ?