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

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Answer 1 · 2018-03-09T18:05:22

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

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

Which leads to:

# 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 #