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

Answer 1

Scarlett Allison

#(dy)/(dx)=-e^(2t)/t^2#

The derivative of a parametric equation #f(t)=(x(t),y(t))# is given by #(dy)/(dx)=((dy)/(dt))/((dx)/(dt))#

As #x(t)=te^(-t)#, #(dx)/(dt)=e^(-t)+txx(-e^(-t))=e^(-t)(1-t)#

and as #y=e^t/t#, #(dy)/(dt)=(te^t-e^txx1)/t^2#

= #e^t/t^2(t-1)#

and #(dy)/(dx)=e^t/t^2(t-1)xx1/(e^(-t)(1-t))#

= #-e^(2t)/t^2#

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Answer 2

Elijah Abram

To differentiate the parametric equations ( x(t) = te^{-t} ) and ( y(t) = \frac{e^t}{t} ) with respect to ( t ), you apply the chain rule and quotient rule, respectively.

( \frac{dx}{dt} = e^{-t} - te^{-t} )

( \frac{dy}{dt} = \frac{te^t - e^t}{t^2} )

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.