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

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Answer 1 · 2017-02-22T13:43:51

Instantaneous velocity is given by the vector:

# vec(v) = << 1-e^2, 2e^4 >> #, or, #(1-e^2) hat(i)+2e^4hat(j)#, or, #( (1-e^2),(2e^4) )#

We have:

# f(t) = ( x(t), y(t) )# where #x(t)=t-e^t#, #y(t)=e^(2t)#

Where #f(t)# represents the position at tme #t#; Then;

# dx/dt = 1-e^t = 1/(2sqrt(t+2))#
# dy/dt = 2e^(2t) #

So, when #t=2 #:

# dx/dt = 1-e^2 ~~ -6.389#
# dy/dt = 2e^4 \ \ \ \ \ ~~ 109.196#

And so the instantaneous velocity is given by the vector:

# vec(v) = << 1-e^2, 2e^4 >> #, or, #(1-e^2) hat(i)+2e^4hat(j)#, or, #( (1-e^2),(2e^4) )#

If we want the instantaneous speed , it is given by:

# v = || vec(v) || #
# \ \ \= sqrt( (1-e^2)^2 + (2e^4)^2 ) #
# \ \ \= sqrt( 11964.651 ... ) #
# \ \ \~~ 109.383 ... #