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

Answer 1

#(dy)/(dx)=(2t^3-e^(t-t^2)t^2(1-2t))/(t^2-te^(t^2-t+1)(2t-1)+e^(t^2-t+1))#

When a functiion is given in parametric form such as #f(t)=(x(t),y(t))#, its dervative is given by #(dy)/(dx)=((dy)/(dt))/((dx)/(dt))#

Here we have #y(t)=t^2-e^(t-t^2)# hence #(dy)/(dt)=2t-e^(t-t^2)(1-2t)#

and #x(t)=t-(e^(t^2-t+1))/t# hence #(dx)/(dt)=1-(te^(t^2-t+1)(2t-1)-e^(t^2-t+1))/t^2#

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

Hence #(dy)/(dx)=(t^2(2t-e^(t-t^2)(1-2t)))/(t^2-te^(t^2-t+1)(2t-1)+e^(t^2-t+1))#

= #(2t^3-e^(t-t^2)t^2(1-2t))/(t^2-te^(t^2-t+1)(2t-1)+e^(t^2-t+1))#

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Answer 2

Christopher Allers

To differentiate the parametric equations (x(t)) and (y(t)):

Differentiate each equation with respect to (t).
Apply the chain rule whenever necessary.

Here are the derivatives:

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

[ \frac{dy}{dt} = 2t - e^{t - t^2} - 2te^{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.