A particle moves along the x-axis with position function s(t) = esin(x). How many times in the interval [0, 2π] is the velocity equal to 0?
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?
To find when the velocity is equal to 0, we need to find the critical points of the velocity function. The velocity is the derivative of the position function, so we'll start by finding the derivative of the position function.
Given the position function, s(t) = esin(x), we can find the velocity function by taking the derivative with respect to time:
v(t) = d/dt (esin(x))
To find the derivative of esin(x), we'll need to use the chain rule. Let's break down the calculation step-by-step:
Step 1: Apply the chain rule by differentiating the outer and inner functions separately.
v(t) = d/dt (e^sin(x))
= d/dx (e^sin(x)) * dx/dt
Step 2: Find the derivative of e^sin(x) with respect to x.
To find the derivative of e^sin(x), we'll need to use the chain rule again.
d/dx (e^sin(x)) = e^sin(x) * cos(x)
Step 3: Multiply the derivative of e^sin(x) with dx/dt
Since x is a function of t, we can rewrite dx/dt as the derivative of x with respect to t.
dx/dt = x'(t)
Now we can put everything together:
v(t) = e^sin(x) * cos(x) * x'(t)
To find when the velocity is equal to 0, we need to solve the equation v(t) = 0. In other words, we need to find where the function e^sin(x) * cos(x) * x'(t) equals zero.
However, it's important to note that the position function s(t) = esin(x) depends on both t and x, so we have two variables to consider.
Without additional information about x(t) or x'(t), it is not possible to determine specifically how many times the velocity is equal to 0 on the interval [0, 2π].
To determine the number of times the velocity is equal to zero during the interval [0, 2π], we first need to find the expression for velocity using the given position function.
The velocity of a particle is the derivative of its position with respect to time. Therefore, we need to differentiate the position function s(t) = esin(t) with respect to t to find the velocity function v(t).
Let's compute the derivative of s(t) using the chain rule:
s'(t) = d/dt (e * sin(t))
= e * d/dt (sin(t)) (Applying the chain rule: d/dx(f(g(x))) = f'(g(x)) * g'(x))
= e * cos(t)
Now we have the expression for velocity, which is v(t) = e * cos(t).
To find the number of times the velocity is equal to zero in the interval [0, 2π], we need to determine the values of t that satisfy v(t) = e * cos(t) = 0.
However, we want to find the values of t within the given interval [0, 2π], so we can rewrite the equation as cos(t) = 0 and solve it.
The solutions to the equation cos(t) = 0 are given by t = (n + 0.5)π, where n is an integer.
In the interval [0, 2π], we have:
When n = 0, t = (0 + 0.5)π = 0.5π
When n = 1, t = (1 + 0.5)π = 1.5π
Therefore, there are two times in the interval [0, 2π] when the velocity is equal to zero.