One-dimensional motion: v(t) = -(t+1) * sin(t^2 / 2)
how often does a particle following that function change direction in the open interval (0,3).
It's part of an FRQ. I've been working at it for a while, but I have not gotten a single definitive answer.
I'm allowed to graph it using a graphing calculator, but I want to understand why it's once.
the particle changes direction when it stops and reverses. That is, when v(t) = 0
sin(t^2/2) = 0 when t^2/2 = kπ
That is, when t = √2π, √4π, ...
Only √2π is in the interval (0,3)
To find how often the particle changes direction in the open interval (0,3), we need to locate the points where the velocity function changes sign.
The velocity function tells us the rate at which the particle is moving at any given time. When the velocity is positive, the particle is moving in the positive direction, and when the velocity is negative, the particle is moving in the negative direction. Thus, a change in direction occurs when the particle's velocity changes from positive to negative or from negative to positive.
To determine the sign changes of the velocity function, we need to identify the critical points of the function where the velocity is zero or undefined. In this case, the velocity function is v(t) = -(t+1) * sin(t^2 / 2).
First, let's find the points where the velocity is zero by setting v(t) = 0 and solving for t:
-(t+1) * sin(t^2 / 2) = 0
Either -(t+1) = 0 or sin(t^2 / 2) = 0.
Solving -(t+1) = 0 gives t = -1.
Solving sin(t^2 / 2) = 0 gives t^2 / 2 = kπ, where k is an integer other than zero. Since we are interested in the interval (0,3), we can disregard t = 0 as a solution. Thus, we solve t^2 / 2 = π, which gives t = √(2π).
Therefore, the critical points of the velocity function are t = -1 and t = √(2π).
Next, we need to determine the sign of the velocity function for various values of t:
- For t < -1, v(t) is negative because the sin(t^2 / 2) term is negative, and -(t+1) is positive.
- For -1 < t < √(2π), v(t) is positive because both -(t+1) and sin(t^2 / 2) are negative.
- For √(2π) < t < 3, v(t) is negative because -(t+1) is positive, and sin(t^2 / 2) is negative.
Based on this analysis, we can see that the velocity changes sign twice in the interval (0,3), indicating two instances when the particle changes its direction.
Therefore, the particle following the given function changes direction twice in the open interval (0,3).