A particle moves along the x-axis with position function s(t) = e^(cos(x)). How many times in the interval [0, 2π] is the velocity equal to 0?
1
2
3
More than 3< my nswer
I think you must mean
s(t) = e^(cos(t))
velocity is zero when
v(t) = ds/dt = e^cos(t) (-sin(t)) = 0
sin(t)=0 when t is a multiple of pi.
So, it looks like you are in error.
To determine the number of times the velocity is equal to 0 in the interval [0, 2π], we need to find the critical points of the velocity function.
The velocity, v(t), is the derivative of the position function, s(t), with respect to time, t. So, to find the velocity function, we take the derivative of s(t).
The position function is given as s(t) = e^(cos(t)). Taking the derivative of s(t) with respect to t will give us the velocity function.
Let's calculate the derivative of s(t):
s'(t) = d/dt(e^(cos(t)))
To find the derivative of e^(cos(t)), we use the chain rule:
s'(t) = -e^(cos(t)) * sin(t)
Now that we have the velocity function, v(t) = -e^(cos(t)) * sin(t), we can proceed to find the critical points.
To find when the velocity equals 0, we need to solve the equation:
v(t) = -e^(cos(t)) * sin(t) = 0
To do this, we set each factor to 0 and solve for t:
1. -e^(cos(t)) = 0
This equation has no solution since e^(cos(t)) is always positive.
2. sin(t) = 0
This equation has solutions when t = 0, π, 2π, ...
In the interval [0, 2π], there are two solutions for sin(t) = 0: t = 0 and t = π.
Therefore, the velocity is equal to 0 twice in the interval [0, 2π].
So, the correct answer is 2.