The Velocity function is:
v(t) = 3e^{-t} - 2 and the initial position is s(0) = 0.
Here are my calculations:
s(t) = 3t(e^{-t}) - 2t + C
Since s(0) = 0 we have c=0, Thus,
s(t) = 3t(e^{-t}) - 2t + 0
Did I do this one correct?
The Velocity function is:
v(t) = 3e^{-t} - 2 and the initial position is s(0) = 0.
Here are my calculations:
s(t) = 3t(e^{-t}) - 2t + C
Since s(0) = 0 we have c=0, Thus,
s(t) = 3t(e^{-t}) - 2t + 0
Did I do this one correct?
I do not understand.... in the question it says s(0) = 0 so would c not equal 0? and I also do not understand how you got -3e^-t.....It looks to me like the derivative of that would not be 3e^-t.......perhaps I need more coffee
I suggest going back to basics and just looking up what the derivative is of a function of the type [e^{ax}] and then reasoning backwards to figure out what the integral, knowing that integration is the opposite of differentiation.
thanks, to both of you, and I did just that rousseau, sometimes all this calculus just jiggles around for me lol, but I do see, now, how the derivative of e^-x would in fact equal -e^(-x)..thanks! so that now makes since.....
-3e^-t - 2t + C However I still am unsure why you got a 3 in place of the C(Constant) when the constant was defined as 0 (at least for the starting point)
Sorry if I sound like an idiot, teaching calculus to oneself is quite a task!