How do you differentiate the following parametric equation: $x (t) = \frac{1}{t}, y (t) = \frac{1}{1 - t^{2}}$ $x(t)=1/t, y(t)=1/(1-t^2)$ ?

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Answer 1 · 2018-03-09T18:05:22

$\frac{d x}{d t} = - \frac{1}{t^{2}}$ $dx/dt = -1/t^2$

$\frac{d y}{d t} = \frac{2 t}{{(1 - t^{2})}^{2}}$ $dy/dt = (2t)/(1-t^2)^2$

Which leads to:

$\frac{d y}{d x} = \frac{- 2 t^{3}}{{(1 - t^{2})}^{2}}$ $dy/dx = (-2t^3)/(1-t^2)^2$

We have:

$x(t) = 1/t \ \$ and $\ \ y(t) = 1/(1-t^2)$

Differentiating each parametric term wrt $t$ we get:

$dx/dt = -1/t^2$

And:

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

This is is technically the answer to the question, however more likely we would be asked to find $dy/dx$ , which we obtain using the chain rule:

$dy/dx = (dy/dt) / (dx/dt)$

$\ \ \ \ \ = ((2t)/(1-t^2)^2) / (-1/t^2)$

$\ \ \ \ \ = (-2t^3)/(1-t^2)^2$

How do you differentiate the following parametric equation: x(t)=1/t, y(t)=1/(1-t^2) x(t)=1t,y(t)=11−t2 x(t)=1/t, y(t)=1/(1-t^2) ?