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

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To find the instantaneous velocity of an object with position at time ( t ) equal to ( f(t) = (te^{t^2-3t}, t^2 - e^t) ) at ( t = 3 ), we need to find the derivative of the position function with respect to time and then evaluate it at ( t = 3 ).

The position function ( f(t) = (te^{t^2-3t}, t^2 - e^t) ) consists of two components: ( x(t) = te^{t^2-3t} ) for the ( x )-coordinate and ( y(t) = t^2 - e^t ) for the ( y )-coordinate.

To find the instantaneous velocity, we take the derivative of each component with respect to time:

( v_x(t) = \frac{dx}{dt} = \frac{d}{dt}(te^{t^2-3t}) )

( v_y(t) = \frac{dy}{dt} = \frac{d}{dt}(t^2 - e^t) )

Then, we evaluate these derivatives at ( t = 3 ) to find the velocities at that specific time.

Taking the derivatives:

( v_x(t) = e^{t^2-3t} + (t)(2t-3)e^{t^2-3t} )

( v_y(t) = 2t - (-e^t) )

Evaluating at ( t = 3 ):

( v_x(3) = e^{3^2-3(3)} + (3)(2(3)-3)e^{3^2-3(3)} )

( v_y(3) = 2(3) - (-e^3) )

Calculating the values:

( v_x(3) = e^0 + (3)(3)e^0 = 1 + 9 = 10 )

( v_y(3) = 6 + e^3 )

So, the instantaneous velocity at ( t = 3 ) is approximately ( (10, 6 + e^3) ).

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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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