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Work done by Force Field 
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#1
May3011, 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 01 and i get 13/2 but the answer should be 39/5 so my question is where am i going wrong 


#2
May3011, 03:51 PM

HW Helper
P: 1,583

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] 


#3
May3011, 03:52 PM

P: 420

yes it is



#4
May3011, 03:54 PM

HW Helper
P: 1,583

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.... 


#5
May3011, 03:58 PM

P: 420

dy=1/2x^(1/2)



#6
May3011, 04:26 PM

HW Helper
P: 1,583

the correct answer is [itex]dy=1/2x^(1/2)dx[/itex], insert this into F.dr to find that F.dr=...



#7
May3011, 04:32 PM

P: 420

would i have to parameterize the x?



#8
May3011, 04:35 PM

HW Helper
P: 1,583

No, just work out F.dr in terms of x and dx.



#9
May3011, 07:43 PM

P: 420

I still dont see how that is going to work



#10
May3011, 07:48 PM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,488

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] 


#11
May3011, 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



#12
May3011, 09:16 PM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,488

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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