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

Answer 1

Evelyn Agard

#dx/dt= (e^t 4t^2 -8te^t)/(16t^4)#

#dy/dt= 1-e^t#

Both the equations would need to be differentiated seperatly as follows:

#x(t)= e^t /(4t^2)-1#

Therefore #dx/dt= (e^t 4t^2 -8te^t)/(16t^4)#

#y(t)=t-e^t#

therefore #dy/dt= 1-e^t#

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Answer 2

William Abdalla

To differentiate the parametric equations (x(t) = \frac{e^t}{(t + t)^2 - 1}) and (y(t) = t - e^t), we will use the chain rule and product rule where necessary. The derivative of (x(t)) with respect to (t) is:

[ \frac{dx}{dt} = \frac{e^t((t + t)^2 - 1) - e^t(2(t + t))}{((t + t)^2 - 1)^2} ]

And the derivative of (y(t)) with respect to (t) is simply:

[ \frac{dy}{dt} = 1 - e^t ]

So, the derivatives of the parametric equations are:

[ \frac{dx}{dt} = \frac{e^t(2t^2 - 2) - 2e^t(2t)}{(2t^2 - 1)^2} ] [ \frac{dy}{dt} = 1 - e^t ]

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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.