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

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

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Answer 1 · 2016-03-19T17:47:22

$\frac{d y}{d x} = \frac{t (t - 8)}{{(4 - t)}^{2}}$ $(dy)/(dx)=(t(t-8))/(4-t)^2$

Explanation:

Differential of a parametric of the type $y = y (t)$ $y=y(t)$ and $x = x (t)$ $x=x(t)$ is given by

$\frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}}$ $(dy)/(dx)=((dy)/(dt))/((dx)/(dt))$

Here $\frac{d y}{d t} = \frac{(- t + 4) \times 2 t - t^{2} (- 1)}{{(- t + 4)}^{2}}$ $(dy)/(dt)=((-t+4)xx2t-t^2(-1))/(-t+4)^2$

= $\frac{(8 t - 2 t^{2}) + t^{2}}{{(- t + 4)}^{2}} = \frac{t (8 - t)}{{(4 - t)}^{2}}$ $((8t-2t^2)+t^2)/(-t+4)^2=(t(8-t))/(4-t)^2$

As $x = - t - 2$ $x=-t-2$ , $\frac{d x}{d t} = - 1$ $(dx)/(dt)=-1$

Hence $\frac{d y}{d x} = \frac{t (8 - t)}{{(4 - t)}^{2}} \times (- 1) = \frac{t (t - 8)}{{(4 - t)}^{2}}$ $(dy)/(dx)=(t(8-t))/(4-t)^2xx(-1)=(t(t-8))/(4-t)^2$

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

1 Answer

Explanation:

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How do you differentiate the following parametric equation: x(t)=−t−2,y(t)=t2−t+4 x(t)=-t-2, y(t)= t^2/(-t+4?

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Explanation:

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