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

Question

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

William Abdalla · Accepted Answer

#(-3e^-2,0)# velocity is #vec v# #vec v = ((dx)/(dt),(dy)/(dt)) =(f'(t),g'(t))# # # # # so, we need to know #f'(2)# (1) #(e^(t-t^2))'=e^{t-t^2}(1-2t)# (2) #(e^t/t^2)' = ((e^t*t^2)-(e^t*2t))/t^4# # # - TIP) #(f/g)'=(f'*g-f*g')/g^2# # # - TIP) #(e^t)'=e^t# # # # # each (1), (2) input # t=2# # # 1-1) #e^(2-4)xx(1-4)=##-3e^-2#, 2-1) #((e^2*4)-(e^2*4))/16=0# # # # # #:. vec v = (-3e^-2,0)#

Isabella Ackerman · Answer

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

The first component of the function is $ e^{t - t^2} $, so its derivative with respect to $ t $ is $ \frac{d}{dt} e^{t - t^2} = e^{t - t^2} (1 - 2t) $.

The second component of the function is $ \frac{e^t}{t^2} $, so its derivative with respect to $ t $ is $ \frac{d}{dt} \left( \frac{e^t}{t^2} \right) = \frac{e^t (t^2 \cdot 1 - 2t \cdot e^t)}{t^4} = \frac{e^t (t - 2t^2)}{t^4} $.

Now, we can evaluate these derivatives at $ t = 2 $:

For the first component: $ e^{2 - 2^2} (1 - 2 \cdot 2) = e^{-2} (-3) $.

For the second component: $ \frac{e^2 (2 - 2 \cdot 2^2)}{2^4} = \frac{e^2 (-6)}{16} $.

So, the instantaneous velocity of the object at $ t = 2 $ is $ \left( e^{-2} (-3), \frac{e^2 (-6)}{16} \right) $.