# What is the instantaneous velocity of an object with position at time t equal to # f(t)= (te^(-3t),t^2-te^t) # at # t=2 #?

By definition, velocity is the rate of change in position and may hence be found by differentiating the position function with respect to time. This will require application of the product rule and the result is

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To find the instantaneous velocity at ( t = 2 ), first differentiate the position function with respect to time to get the velocity function. Then evaluate the velocity function at ( t = 2 ) to find the instantaneous velocity.

The position function ( f(t) = (te^{-3t}, t^2 - te^t) ).

Differentiating with respect to time: [ \frac{{df}}{{dt}} = \left( e^{-3t} - 3te^{-3t}, 2t - e^t - te^t \right) ]

Evaluate at ( t = 2 ): [ \frac{{df}}{{dt}}\Bigr|_{t=2} = \left( e^{-6} - 6e^{-6}, 4 - e^2 - 2e^2 \right) ]

[ = \left( e^{-6} - 6e^{-6}, 4 - 3e^2 \right) ]

So, the instantaneous velocity at ( t = 2 ) is ( \left( e^{-6} - 6e^{-6}, 4 - 3e^2 \right) ).

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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.

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