# Work done by Force Field

by Punkyc7
Tags: field, force, work
 P: 418 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
 HW Helper P: 1,584 The work done is given by: $$W=\int_{\gamma}\mathbf{F}\cdot d\mathbf{r}$$ where $d\mathbf{r}=dx\mathbf{i}+dy\mathbf{j}$
 P: 418 yes it is
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....
 P: 418 dy=1/2x^-(1/2)
 HW Helper P: 1,584 the correct answer is $dy=1/2x^-(1/2)dx$, insert this into F.dr to find that F.dr=...
 P: 418 would i have to parameterize the x?
 HW Helper P: 1,584 No, just work out F.dr in terms of x and dx.
 P: 418 I still dont see how that is going to work
PF Patron
Thanks
Emeritus
P: 38,412
 Quote by Punkyc7 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 $x= t^2$, $y= t$, with t from 1 to 2 instead.
Now F= 3ti- t^4 j, dx= 2tdt, dy= dt so
$$\int_{t= 1}^2 (3t)(2t)dt- (t^4)dt= \int_{t=1}^2 (6t^2- t^4)dt$$

 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
 P: 418 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
 PF Patron Sci Advisor Thanks Emeritus P: 38,412 Yes. Of course you could always use $x= t$, $y= t^{1/2}$ with t from 1 to 4. There are many different ways to parameterize any curve.

 Related Discussions Classical Physics 4 Advanced Physics Homework 1 Introductory Physics Homework 2 Calculus & Beyond Homework 4 Introductory Physics Homework 5