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< My answer
2
3
More than 3
I don't really get this question will someone please explain it to me?
v(x) = -sin(x) e^cos(x)
since e^cos(x) is never zero, how many times in sin(x) = 0 in the interval?
Your answer is correct on the interval (0,2π), but not on [0,2π]
v(t) = 0 at 0,π,2π
If you exclude the endpoints of the interval, your answer is correct. But, we are working with the closed interval, including the endpoints. So, there are 3 points where v(t) = 0.
Better review interval notation.
what is the difference between the two?
Certainly! In this question, we're given a position function s(t) = e^cos(t) that describes the motion of a particle along the x-axis. Our goal is to determine how many times the velocity of the particle is equal to zero within the interval [0, 2π].
To find the velocity function, we need to take the derivative of the position function with respect to time. In this case, we have s(t) = e^cos(t), so the velocity function v(t) is given by the derivative of s(t):
v(t) = d/dt(e^cos(t))
To find when the velocity is zero, we need to solve the equation v(t) = 0. So, let's find the derivative v(t) and set it equal to zero:
v(t) = -sin(t)e^cos(t)
Setting v(t) = 0, we get:
-sin(t)e^cos(t) = 0
Now, we have two factors: -sin(t) = 0 and e^cos(t) = 0. But e^cos(t) can never be equal to zero for any value of t, so we can ignore that factor.
To solve -sin(t) = 0, we need to find all the values of t within the interval [0, 2π] where sin(t) equals zero. Sin(t) equals zero at several points: t = 0, t = π, and t = 2π, as sin(0) = sin(π) = sin(2π) = 0.
Therefore, within the interval [0, 2π], the velocity is equal to zero at three distinct points: t = 0, t = π, and t = 2π.
Hence, the correct answer to the question is 3.