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

Question

2247 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-16T18:50:50

# vec v = << -3e^(-2), 2e^2-4 >> #
# \ \ \ \ ~~ << -0.406, 10.778 >> #

We have the following displacement function:

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

Then differentiating wrt #t# we get:

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

# dy/dt = 2e^t-2t #

And so when #t=2# we have:

# dx/dt = (1-4)e^(2-4) = -3e^(-2) \ \ (~~-0.406) #

# dy/dt = 2e^2-4 \ \ (~~10.778) #

Hence, assuming a standard Cartesian coordinate system, the instantaneous velocity when #t=2# is the vector:

# vec v = << -3e^(-2), 2e^2-4 >> #

Note:
If we wanted the instantaneous speed , then this would be given by:

# v = || vec v || = sqrt( (-3e^(-2))^2 + (2e^2-4)^2 ) #

Answer 2 · 2017-05-16T19:06:44

The instantaneous velocity is #=(-0.41,10.78)#

The velocity is the derivative of the position.

#f(t)=(e^(t-t^2), 2e^t-t^2))#

#v(t)=f'(t)=((1-2t)e^(t-t^2),2e^t-2t))#

Therefore,

#v(2)=(-3e^-2, 2e^2-4)#

#=(-0.41,10.78)#