How do you differentiate the following parametric equation: # x(t)=t-(t+1)e^t, y(t)= t^2-e^(t-1) #?
Solution:
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To differentiate the parametric equations ( x(t) = t - (t+1)e^t ) and ( y(t) = t^2 - e^{t-1} ) with respect to ( t ), use the chain rule for differentiation.
( \frac{dx}{dt} = \frac{d}{dt}(t) - \frac{d}{dt}[(t+1)e^t] )
( \frac{dx}{dt} = 1 - [(1 + t)e^t + e^t] )
( \frac{dx}{dt} = 1 - (1 + t + 1)e^t )
( \frac{dx}{dt} = 1 - (t + 2)e^t )
Similarly,
( \frac{dy}{dt} = \frac{d}{dt}(t^2) - \frac{d}{dt}[e^{t-1}] )
( \frac{dy}{dt} = 2t - e^{t-1} )
Therefore, the derivatives of ( x(t) ) and ( y(t) ) are:
( \frac{dx}{dt} = 1 - (t + 2)e^t )
( \frac{dy}{dt} = 2t - e^{t-1} )
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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.
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.
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.
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.
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