help with a calculus problem please!?

Question

the set up is:

a particle moves along the x-axis in such a way that at time t>0 its position coordinate is x= sin (e^t)

so now the question is: at what time does the particle first have zero velocity?

i know you have to take the derivative of the position function given to get velocity and i got: v= cos (e^t)(e^t)dt, is this right? if so, then how would you find at what time the particle has a zero velocity?

thanks!

Answer 1

You are right about taking the derivative to get the velocity.

x(t) = sin(e^t)
v(t) = (e^t)cos(e^t) = 0

e^t will never equal zero so divide by e^t

cos(e^t) = 0
e^t = arccos(0) = π/2 + πn, where n is an integer
ln(e^t) = ln(π/2 + πn)
t = ln(π/2 + πn)

Answer 2

v(t) = d(sin(e^t))/dt = (e^t) cos(e^t) with t>0

zero velocity:

v(t)=0
=> (e^t) cos(e^t) = 0
=> cos (e^t) = 0 because (e^t) is always >0
=> e^t = pi/2 because cos (pi/2) is the first zero of the cos function
=> t = ln(pi/2)

Answer 3

assuming you did the derivative right you would set v=0 and then solve for v, you dont need the dt and the end of the function

so you are gonna solve 0=cos(e^t)*e^t for t.

Answer 4

v= e^t * cos (e^t)=0
As e^t never is zero cos(e^t)=0.
The smallest value which makes cos = 0 is pi/2

so e^t= pi/2 and t = log(pi/2) in base e t=0.4516

Answer 5

v = 0 means cos(e^t) = 0. Therefore,
e^t = pi/2
t = ln pi/2 = 0.4516