How do you differentiate the following parametric equation: $x(t)=(t+1)e^t, y(t)= (t-2)^2+t$ ?

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Answer 1 · 2017-07-26T02:23:47

$(2t-3)/(e^t t + 2e^t)$

$dy/dx = dy/dt / dx/dt$

since $dy/dt / dx/dt = dy/dt * dt/dx = dy/dx$ (cancelling $dt$ in the numerator and denominator)

$x(t) = (t+1)e^t$
$dx/dt = d/dt((t+1)e^t) = (t+1)e^t + e^t$

$y(t) = (t-2)^2 + t$
$dy/dt = d/dt((t-2)^2 + t) = 2(t-2) + 1$

Therefore,
$dy/dx = (2(t-2) + 1)/((t+1)e^t + e^t) = (2t-3)/(e^t t + 2e^t)$

How do you differentiate the following parametric equation: x(t)=(t+1)e^t, y(t)= (t-2)^2+t x(t)=(t+1)e^t, y(t)= (t-2)^2+t?