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

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Answer 1 · 2017-07-02T06:53:10

$\frac{d y}{d x} = \frac{1 - e^{t - 1}}{(t + 2) e^{t}}$ $(dy)/(dx)=(1-e^(t-1))/((t+2)e^t)$

In parametric form of equations, $\frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}}$ $(dy)/(dx)=((dy)/(dt))/((dx)/(dt))$

Here $x (t) = (t + 1) e^{t}$ $x(t)=(t+1)e^t$ and therefore

$\frac{d x}{d t} = e^{t} + (t + 1) e^{t} = (t + 2) e^{t}$ $(dx)/(dt)=e^t+(t+1)e^t=(t+2)e^t$

and as $y (t) = t = e^{t - 1}$ $y(t)=t=e^(t-1)$ , $\frac{d y}{d t} = 1 - e^{t - 1}$ $(dy)/(dt)=1-e^(t-1)$

Hence $\frac{d y}{d x} = \frac{1 - e^{t - 1}}{(t + 2) e^{t}}$ $(dy)/(dx)=(1-e^(t-1))/((t+2)e^t)$

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