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 #?

Question

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  #?

Charles Arocha · Accepted Answer

#(10, 6 - e^3) # #f(t)= (te^(t^2-3t),t^2-e^t)# #f'(t)= (e^(t^2-3t) + t (2t-3) e^(t^2-3t),2t-e^t)# #= (e^(t^2-3t) (1 + t (2t-3)),2t-e^t)# #f'(3) = ( (1)(1 + 3(3)), 6 - e^3) = (10, 6 - e^3) #

Aurora Alvarado · Answer

undefined 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) $.