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

Question

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

Aurora Alvarado · Accepted Answer

#f'(-1)=(-2e,2)#.

The law #f(t)# is the position of the object at the time #t#. To find the instantaleous velocity we have to find the derivative of the previous law. #f'(t)=(2te^(t^2),2-e^t-te^t)# and so: #f'(-1)=(-2e,2-1/e+1/e)=(-2e,2)#.

Aurora Alvarado · Answer

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

The velocity vector $ v(t) $ is given by the derivative of the function $ f(t) $ with respect to $ t $:

$$ v(t) = \frac{d}{dt} (e^{t^2}, 2t - te^t) $$

Differentiating each component separately, we get:

$$ v(t) = (\frac{d}{dt} e^{t^2}, \frac{d}{dt} (2t - te^t)) $$

$$ v(t) = (2te^{t^2}, 2 - (t + 1)e^t) $$

Now, we can evaluate $ v(t) $ at $ t = -1 $:

$$ v(-1) = (2(-1)e^{(-1)^2}, 2 - ((-1) + 1)e^{-1}) $$

$$ v(-1) = (-2e^{-1}, 2 - (0)e^{-1}) $$

$$ v(-1) = (-2e^{-1}, 2) $$

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