# What is the instantaneous velocity of an object moving in accordance to # f(t)= (e^(t-t^2),2t-te^t) # at # t=-1 #?

and it's direction

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To find the instantaneous velocity of an object moving according to the function ( f(t) = (e^{t - t^2}, 2t - te^t) ) at ( t = -1 ), we need to calculate the derivative of the function with respect to time and then evaluate it at ( t = -1 ).

The derivative of ( f(t) ) with respect to time is given by:

[ f'(t) = \left( \frac{d}{dt} e^{t - t^2}, \frac{d}{dt} (2t - te^t) \right) ]

[ f'(t) = \left( e^{t - t^2}(1 - 2t), 2 - e^t - te^t \right) ]

Now, we evaluate ( f'(t) ) at ( t = -1 ):

[ f'(-1) = \left( e^{-1 - (-1)^2}(1 - 2(-1)), 2 - e^{-1} - (-1)e^{-1} \right) ]

[ f'(-1) = \left( e^{-2}(1 + 2), 2 - \frac{1}{e} + \frac{1}{e} \right) ]

[ f'(-1) = \left( 3e^{-2}, 2 \right) ]

Therefore, the instantaneous velocity of the object at ( t = -1 ) is ( (3e^{-2}, 2) ).

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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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