How do you differentiate the following parametric equation: # x(t)=(1-t)/e^t , y(t)=t/e^(3t) #?
Given that
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To differentiate the given parametric equations ( x(t) = \frac{1-t}{e^t} ) and ( y(t) = \frac{t}{e^{3t}} ) with respect to ( t ), we will use the chain rule and product rule where necessary.
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Differentiate ( x(t) = \frac{1-t}{e^t} ) with respect to ( t ): [ \frac{dx}{dt} = \frac{d}{dt} \left( \frac{1-t}{e^t} \right) ] [ = \frac{d}{dt} \left( (1-t)e^{-t} \right) ] [ = (1-t) \frac{d}{dt} (e^{-t}) + e^{-t} \frac{d}{dt} (1-t) ] [ = (1-t)(-e^{-t}) + e^{-t}(-1) ] [ = -e^{-t} + te^{-t} ]
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Differentiate ( y(t) = \frac{t}{e^{3t}} ) with respect to ( t ): [ \frac{dy}{dt} = \frac{d}{dt} \left( \frac{t}{e^{3t}} \right) ] [ = \frac{d}{dt} \left( te^{-3t} \right) ] [ = t \frac{d}{dt} (e^{-3t}) + e^{-3t} \frac{d}{dt} (t) ] [ = te^{-3t}(-3) + e^{-3t}(1) ] [ = -3te^{-3t} + e^{-3t} ]
So, the derivatives of the given parametric equations with respect to ( t ) are: [ \frac{dx}{dt} = -e^{-t} + te^{-t} ] [ \frac{dy}{dt} = -3te^{-3t} + e^{-3t} ]
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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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