The discussion revolves around the evaluation of indefinite integrals in the context of the Net Change Theorem, specifically focusing on the relationship between velocity, displacement, and total distance traveled by a particle over a specified interval.
Participants are actively engaging with the problem, offering insights and corrections to each other's reasoning. There is an ongoing exploration of how to correctly calculate the total distance traveled by considering both forward and backward motion, with some participants suggesting methods to integrate over specific intervals.
There is a noted confusion regarding the definitions of displacement and total distance, as well as the implications of integrating a velocity function that changes sign. Participants are working within the constraints of a homework assignment, which may limit the information available for discussion.
Dick said:Number one is right. The problem with two is that at e.g. v(pi)=(-1/2). So the particle is traveling backwards part of the time.
MitsuShai said:But even if it's traveling backwards, it does not effect the total distance right? Because that will effect the displacement.
Dick said:You've got that backwards. Displacement is the total distance between the starting point and the ending point. It doesn't matter how you got from the start to the end. How you did does affect the total distance travelled.
MitsuShai said:ok then, but I still can't think of a way to solve it.
Dick said:Find out at what time the particle switches from going forward to backwards by solving v(t)=0. Then add the distance it travels forward to the distance it travels backward.
MitsuShai said:v(t)=cos(t)+(1/2)=0
cos (t)= -1/2
t=2pi/3
[sin(0)+(1/2)(0)] + [sin(2pi/3)+(1/2)(2pi/3)]
0 + squareroot(3)/2 + pi/6...
Dick said:(1/2)(2pi/3)=pi/3. So if you mean sqrt(3)/2+pi/3, then, yes, that's the distance it travels forward. Now how far does it travel backwards?
MitsuShai said:that was only the forward one; i thought I was solving for both. ok let's see now:
[sin(2pi/3)+(1/2)(2pi/3)] + [sin(pi)+(1/2)(pi)]
sqrt(3)/2+pi/3 + 0
=sqrt(3)/2+pi/3
forward + backward=
(sqrt(3)/2+pi/3) + (sqrt(3)/2+pi/3)
2(sqrt(3)/2+pi/3)
Dick said:To get the backwards distance you are integrating v(t) from 2pi/3 to pi and then reversing the sign since you know the result will be negative, and you want the positive distance, right? So shouldn't [sin(2pi/3)+(1/2)(2pi/3)] + [sin(pi)+(1/2)(pi)] be [sin(2pi/3)+(1/2)(2pi/3)] - [sin(pi)+(1/2)(pi)]??
MitsuShai said:oh yeah that's right, but won't you still end up with the same answer?
sqrt(3)/2+pi/3
then add the forward distance with the backward distance? But then the answer turns out weird...
Dick said:No, I don't get sqrt(3)/2+pi/3 for the backwards distance. The difference between the forward distance and the backwards distance should be the displacement, right? It's not zero. Try the backwards distance again.
MitsuShai said:[sin(2pi/3)+(1/2)(2pi/3)] - [sin(pi)+(1/2)(pi)]
[squareroot(3)/2 + pi/3] - [0+ pi/2] ---sorry about that I assumed it was zero because of the forward direction one, that was a stupid mistake
squareroot(3)/2 + pi/3 - pi/2
squareroot(3)/2 -pi/6
forward + backwards=
sqrt(3)/2+pi/3 + squareroot(3)/2 -pi/6
2(sqrt(3)/2)+ 3pi/18
= sqrt(3) + 3pi/18
Dick said:That's better. You can simplify 3pi/18 to pi/6. But now I think it's right.