How Is Work Calculated in a Force Field Along a Parametrized Curve?

[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

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

 

[hunt_mat]

The work done is given by:
<br /> W=\int_{\gamma}\mathbf{F}\cdot d\mathbf{r}<br />
where d\mathbf{r}=dx\mathbf{i}+dy\mathbf{j}

 

[Punkyc7]

yes it is

 

[hunt_mat]

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

Also on the curve you know that y=x^1/2, and so...

 

[Punkyc7]

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

 

[hunt_mat]

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

 

[Punkyc7]

would i have to parameterize the x?

 

[hunt_mat]

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

 

[Punkyc7]

I still don't see how that is going to work

 

[HallsofIvy]

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

 

[Punkyc7]

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]

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.