Indefinite Integrals & The Net Change Theorem

[MitsuShai]

Homework Statement

2. The attempt at a solution

I integrated the velocity equation and got this: s(t)= sin(t) + (1/2)x

1. [sin(pi)+(1/2)(pi)] - [sin(0)+(1/2)(0)]
= pi/2

2. [sin(pi)+(1/2)(pi)] + [sin(0)+(1/2)(0)]
= pi/2

The first answer is right but the second one isn't

 

[Dick]

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]

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.

But even if it's traveling backwards, it does not effect the total distance right? Because that will effect the displacement.

 

[Dick]

But even if it's traveling backwards, it does not effect the total distance right? Because that will effect the displacement.

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]

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.

ok then, but I still can't think of a way to solve it.

 

[Dick]

ok then, but I still can't think of a way to solve it.

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]

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.

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]

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...

(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]

(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?

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]

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)

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]

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)]??

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 wierd...

 

[Dick]

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 wierd...

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]

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.

[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]

[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

That's better. You can simplify 3pi/18 to pi/6. But now I think it's right.

 

[MitsuShai]

That's better. You can simplify 3pi/18 to pi/6. But now I think it's right.

Thank you so much; it is right :)