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 #?

Question

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 #?

Scarlett Allison · Accepted Answer

#(-5/e^6, 4-3e^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

#therefore v(t)=dx/dt=f'(t)#

#=(-3te^(-3t)+e^(-3t) , 2t-te^t-e^t)#

#v(2)=(-6/(e^6)+1/(e^6),4-2e^2-e^2) #

#=(-5/e^6, 4-3e^2)#

So if this is a normal 2 dimensional vector space with standard notations and units, then the instantaneous velocity at #t=2# may be represented in terms of the standard basis unit vectors as

#v(2)=-5/e^6 hati + (4-3e^2) hatj# #m//s#

Christopher Allers · Answer

undefined 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) $.