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

Question

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

Christopher Agresta · Accepted Answer

Instantaneous velocity of the object is #8.729# Instantaneous velocity of an object moving in accordance to #f(t)=(tlnt,e^(2t))# at #t=2#, will be given by the value #(dy)/(dx)# at #t=2#. As the object moves according to parametric form of equation given by #f(t)#, #(dy)/(dx)=((dy)/(dt))/((dx)/(dt))# = #(2e^(2t))/(txx1/t+1xxlnt)# = #(2e^(2t))/(1+lnt)# and at #t=2#, #(dy)/(dx)=(2e^2)/(1+ln2)# = #(2xx7.389)/(1+0.693)# = #14.778/1.693# = #8.729# Hence, instantaneous velocity of the object is #8.729#

Luke Alvarez · Answer

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

The derivative of $ f(t) = (t \ln t, e^{2t}) $ with respect to time $ t $ is $ f'(t) = (\ln t + 1, 2e^{2t}) $.

Now, we can substitute $ t = 2 $ into $ f'(t) $ to find the instantaneous velocity at $ t = 2 $:

$ f'(2) = (\ln 2 + 1, 2e^{4}) $

So, the instantaneous velocity of the object at $ t = 2 $ is $ (\ln 2 + 1, 2e^{4}) $.