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

Answer 1

Christopher Agresta

#d y=((2(t-2)+1) /(2(t+1)+e^t*l n e))*d x#

#(d x)/(d t)=2*(t+1)*1+e^t*l n e=2(t+1)+e^t*l n e#

#(d y)/( d t)=2*(t-2)*1+1=#2(t-2)+1

#(d y)/(d x)=(d y)/(d t)*(d t)/(d x)=[2(t-2)+1] /(2(t+1)+e^t*l n e)#

#d y=((2(t-2)+1) /(2(t+1)+e^t*l n e))*d x#

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Answer 2

Emilia Aguilar

To differentiate the parametric equations (x(t) = (t + 1)^2 + e^t) and (y(t) = (t - 2)^2 + t) with respect to (t), you simply differentiate each equation separately. Here are the steps:

Differentiate (x(t)) with respect to (t): [ \frac{dx}{dt} = 2(t + 1) + e^t ]
Differentiate (y(t)) with respect to (t): [ \frac{dy}{dt} = 2(t - 2) + 1 ]

So, the derivatives of the parametric equations are: [ \frac{dx}{dt} = 2(t + 1) + e^t ] [ \frac{dy}{dt} = 2(t - 2) + 1 ]

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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.