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What is the distance a particle has moved if I know the work done

by Psinter
Tags: distance, moved, particle, work
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Psinter
#1
Feb26-12, 12:54 PM
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P: 94
1. The problem statement, all variables and given/known data
I was asked to find the distance a particle had traveled and was given its integral solved, except the upper limit was an equation.

[itex](x^2+2x)[/itex] is the force in pounds that acts on the particle


2. Relevant equations
[itex]\int_{0}^{x^2+2}(x^2+2x)dx = 108[/itex]


3. The attempt at a solution
I know I must find what number the top limit equation equals, but I've been inserting random numbers with no success. Also, I don't know if it is possible to find the value of x and plug it in. This is ridiculous, why couldn't they just give me the integral with the limits and I solve it.
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alanlu
#2
Feb26-12, 12:58 PM
P: 65
What is the integral of x2 + 2x?
Psinter
#3
Feb26-12, 01:19 PM
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P: 94
Quote Quote by alanlu View Post
What is the integral of x2 + 2x?
Its [itex]\frac{1}{3}x^3+x^2[/itex]

Ray Vickson
#4
Feb26-12, 02:01 PM
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What is the distance a particle has moved if I know the work done

Quote Quote by Psinter View Post
1. The problem statement, all variables and given/known data
I was asked to find the distance a particle had traveled and was given its integral solved, except the upper limit was an equation.

[itex](x^2+2x)[/itex] is the force in pounds that acts on the particle


2. Relevant equations
[itex]\int_{0}^{x^2+2}(x^2+2x)dx = 108[/itex]


3. The attempt at a solution
I know I must find what number the top limit equation equals, but I've been inserting random numbers with no success. Also, I don't know if it is possible to find the value of x and plug it in. This is ridiculous, why couldn't they just give me the integral with the limits and I solve it.
Maybe the problem is poor notation; this problem's notation is terrible--it double-uses the symbol x. Much better: write [tex] \int_{0}^{x^2+2}(x'^2+2x')dx' = 108,[/tex] which makes it clear what is the integration variable (x') and what is the unknown (x).

RGV
Psinter
#5
Feb26-12, 04:59 PM
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P: 94
Quote Quote by Ray Vickson View Post
Maybe the problem is poor notation; this problem's notation is terrible--it double-uses the symbol x. Much better: write [tex] \int_{0}^{x^2+2}(x'^2+2x')dx' = 108,[/tex] which makes it clear what is the integration variable (x') and what is the unknown (x).

RGV
I see what you did there, but how does that makes it easier to solve? I still don't know what to do.

Rewinding everything I know:

The force exerted on the particle: [tex](x'^2+2x')[/tex]

The initial point of the particle: 0

The distance traveled by the particle (that's my upper limit) given by: [tex](x^2+2)[/tex]

And the total work done by the particle after traveling the distance I'm looking for: 108

Still not helps writing all that. How can I find the upper limit value or the x value to plug it in and get 108 as answer?
alanlu
#6
Feb26-12, 05:38 PM
P: 65
I agree with RGV. It may be easier to think about using a different variable name for the integral than the x you're trying to solve for.

Try plugging in the limits of integration as if you are evaluating the integral.
Psinter
#7
Feb26-12, 05:57 PM
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P: 94
Quote Quote by alanlu View Post
I agree with RGV. It may be easier to think about using a different variable name for the integral than the x you're trying to solve for.

Try plugging in the limits of integration as if you are evaluating the integral.
I kept trying putting in random values and found that the upper value must be 6. In other words:
[tex] \int_{0}^{6}(x'^2+2x')dx' = 108,[/tex]
However, isn't there a process to reach that? I did it by brute force.
Ray Vickson
#8
Feb26-12, 07:54 PM
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Quote Quote by Psinter View Post
I kept trying putting in random values and found that the upper value must be 6. In other words:
[tex] \int_{0}^{6}(x'^2+2x')dx' = 108,[/tex]
However, isn't there a process to reach that? I did it by brute force.
You have already evaluated the indefinite integral [itex] F(x') = \int(x'^2 + 2x')\, dx'.[/itex] Using your formula for F, what would be [itex] \int_0^a (x'^2 + 2x') \,dx'[/itex]? What is preventing you from substituting in [itex] a = x^2 + 2[/itex]? Mind you, the final result will not be pretty, but it gives you a place to start.

RGV
Drood
#9
Feb26-12, 08:10 PM
P: 8
Here's what I did, perhaps you may find the process of some use.

1) I computed the integral and applied the limits.
2) I know that the integral with the limit substitutions = 108
3) Now it's mostly a matter of expanding the terms and factoring to find solutions.
4) The only applicable solution I found was x=2
5) The upper limit with the x=2 substitution is indeed equal to 6.

Sorry for the short reply, I did it by hand and did the expansions and factoring by calculator since they were quite long. If you would like, I could write it out or post a picture of my solution if you did not understand my steps above.

Hope that helps,

Drood
Psinter
#10
Feb26-12, 08:50 PM
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P: 94
Quote Quote by Ray Vickson View Post
You have already evaluated the indefinite integral [itex] F(x') = \int(x'^2 + 2x')\, dx'.[/itex] Using your formula for F, what would be [itex] \int_0^a (x'^2 + 2x') \,dx'[/itex]? What is preventing you from substituting in [itex] a = x^2 + 2[/itex]? Mind you, the final result will not be pretty, but it gives you a place to start.

RGV
It gives me x = 2, but I wonder if there is a rule or something for me to find that the value was 6 instead of doing it by guessing like I did.

Quote Quote by Drood View Post
Here's what I did, perhaps you may find the process of some use.

1) I computed the integral and applied the limits.
2) I know that the integral with the limit substitutions = 108
3) Now it's mostly a matter of expanding the terms and factoring to find solutions.
4) The only applicable solution I found was x=2
5) The upper limit with the x=2 substitution is indeed equal to 6.

Sorry for the short reply, I did it by hand and did the expansions and factoring by calculator since they were quite long. If you would like, I could write it out or post a picture of my solution if you did not understand my steps above.

Hope that helps,

Drood
Yes please, an image would help a lot. Thanks. Its just that its weird to have to solve it by guessing, but you seem to have some process there.
Drood
#11
Feb26-12, 09:00 PM
P: 8
I hope you can read what I wrote. The site kills photo quality apparently and I didn't have enough time to copy it with my scanner.


Click image for larger version

Name:	IMAG0630.jpg
Views:	12
Size:	50.5 KB
ID:	44430

Hope that helps,

Drood
Psinter
#12
Feb26-12, 09:14 PM
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P: 94
Quote Quote by Drood View Post
I hope you can read what I wrote. The site kills photo quality apparently and I didn't have enough time to copy it with my scanner.


Attachment 44430

Hope that helps,

Drood
Oh, I understand what you did there. Thank you very much. I feel dumb now because you make it look so easy and it took me so long.


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