Work done by Force Field


by Punkyc7
Tags: field, force, work
Punkyc7
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#1
May30-11, 03:42 PM
P: 420
Find the work done by the force field F= 3yi - x^2j
in moving a particle along the curve y=x^(1/2) from (1,1) to (4,2)


so i parameterize and get x=1+3t, y= 1+t

so

3(1+t)-(1+t)^2

then take the derivative of x and y and multiple to ge
9(1+t)-(1+t)^2

and i integrate that from 0-1
and i get 13/2


but the answer should be 39/5

so my question is where am i going wrong
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hunt_mat
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#2
May30-11, 03:51 PM
HW Helper
P: 1,584
The work done is given by:
[tex]
W=\int_{\gamma}\mathbf{F}\cdot d\mathbf{r}
[/tex]
where [itex]d\mathbf{r}=dx\mathbf{i}+dy\mathbf{j}[/itex]
Punkyc7
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#3
May30-11, 03:52 PM
P: 420
yes it is

hunt_mat
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#4
May30-11, 03:54 PM
HW Helper
P: 1,584

Work done by Force Field


if y=x^1/2, then dy=...

Also on the curve you know that y=x^1/2, and so....
Punkyc7
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#5
May30-11, 03:58 PM
P: 420
dy=1/2x^-(1/2)
hunt_mat
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#6
May30-11, 04:26 PM
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P: 1,584
the correct answer is [itex]dy=1/2x^-(1/2)dx[/itex], insert this into F.dr to find that F.dr=...
Punkyc7
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#7
May30-11, 04:32 PM
P: 420
would i have to parameterize the x?
hunt_mat
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#8
May30-11, 04:35 PM
HW Helper
P: 1,584
No, just work out F.dr in terms of x and dx.
Punkyc7
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#9
May30-11, 07:43 PM
P: 420
I still dont see how that is going to work
HallsofIvy
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#10
May30-11, 07:48 PM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,881
Quote Quote by Punkyc7 View Post
Find the work done by the force field F= 3yi - x^2j
in moving a particle along the curve y=x^(1/2) from (1,1) to (4,2)


so i parameterize and get x=1+3t, y= 1+t
This is wrong. If y= 1+ t, then t= y- 1 so x= 1+ 3(y- 1)= 3y- 2. That is linear- it is the straight line through (1, 1) and (4, 2).

Use [itex]x= t^2[/itex], [itex]y= t[/itex], with t from 1 to 2 instead.
Now F= 3ti- t^4 j, dx= 2tdt, dy= dt so
[tex]\int_{t= 1}^2 (3t)(2t)dt- (t^4)dt= \int_{t=1}^2 (6t^2- t^4)dt[/tex]

so

3(1+t)-(1+t)^2

then take the derivative of x and y and multiple to ge
9(1+t)-(1+t)^2

and i integrate that from 0-1
and i get 13/2


but the answer should be 39/5

so my question is where am i going wrong
Punkyc7
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#11
May30-11, 08:28 PM
P: 420
Thanks Halls, that made sense. I have one more question how do you know to to parameterize it the way you did? is it because y=x^1/2 so choose x=t^2 ad y=t
HallsofIvy
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#12
May30-11, 09:16 PM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,881
Yes. Of course you could always use [itex]x= t[/itex], [itex]y= t^{1/2}[/itex] with t from 1 to 4. There are many different ways to parameterize any curve.


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