# Work done by Force Field

1. May 30, 2011

### 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

2. May 30, 2011

### hunt_mat

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

3. May 30, 2011

### Punkyc7

yes it is

4. May 30, 2011

### hunt_mat

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

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

5. May 30, 2011

### Punkyc7

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

6. May 30, 2011

### hunt_mat

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

7. May 30, 2011

### Punkyc7

would i have to parameterize the x?

8. May 30, 2011

### hunt_mat

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

9. May 30, 2011

### Punkyc7

I still dont see how that is going to work

10. May 30, 2011

### HallsofIvy

Staff Emeritus
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
$$\int_{t= 1}^2 (3t)(2t)dt- (t^4)dt= \int_{t=1}^2 (6t^2- t^4)dt$$

11. May 30, 2011

### 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

12. May 30, 2011

### HallsofIvy

Staff Emeritus
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.