# Line integrals

• Mar 28th 2007, 05:23 PM
pakman
Line integrals
F, i, j, r, q, p are vectors

F(x,y) = (2xy)i - (x^2)j

div F(x,y) = 2y

curl F(x,y) = -4x which is not equal to zero, allowing no potential.

Now this is the confusing part. Work = 25/3

Here is how they got it

Row (or atleast I think that's what it is) is parameterized by r(t) = p + t(q-p)
x(t) = 1+t
y(t) = 3+t
p=<1,3>
q=<2,4>
0 <= t <= 1

INT(F dot dr) = INT(0 to 1) 2(1+t)(3+t)dt - (1+t)^2 dt
u = 1+t
du = dt
then on from there to get 25/3

My question is... what is all of that stuff for Work? I don't understand where the r(t) = p + t(q-p) comes from... same goes for all the values of x(t), y(t) etc...

Someone help please! I had to postpone my finals due to military training and now I've forgotten so much.
• Mar 28th 2007, 06:38 PM
ThePerfectHacker
Quote:

Originally Posted by pakman
F, i, j, r, q, p are vectors

F(x,y) = (2xy)i - (x^2)j

div F(x,y) = 2y

curl F(x,y) = -4x which is not equal to zero, allowing no potential.

Now this is the confusing part. Work = 25/3
.

Since it is not path independent, you need to specify what path are you intergrating?
• Mar 29th 2007, 05:24 AM
topsquark
Quote:

Originally Posted by ThePerfectHacker
Since it is not path independent, you need to specify what path are you intergrating?

I don't really understand the parametrization part, but isn't he specifying a path there? (It's just a guess, it's too confusing for me to follow.)

-Dan
• Mar 29th 2007, 06:55 AM
qpmathelp
Quote:

Originally Posted by pakman
F, i, j, r, q, p are vectors

F(x,y) = (2xy)i - (x^2)j

div F(x,y) = 2y

curl F(x,y) = -4x which is not equal to zero, allowing no potential.

Now this is the confusing part. Work = 25/3

Here is how they got it

Row (or atleast I think that's what it is) is parameterized by r(t) = p + t(q-p)
x(t) = 1+t
y(t) = 3+t
p=<1,3>
q=<2,4>
0 <= t <= 1

INT(F dot dr) = INT(0 to 1) 2(1+t)(3+t)dt - (1+t)^2 dt
u = 1+t
du = dt
then on from there to get 25/3

My question is... what is all of that stuff for Work? I don't understand where the r(t) = p + t(q-p) comes from... same goes for all the values of x(t), y(t) etc...

Someone help please! I had to postpone my finals due to military training and now I've forgotten so much.

INT(0 to 1) 2(1+t)(3+t)dt - (1+t)^2 dt

just expand and integrate term by term you do not need a substitution

if you are asking about r(t) = p + t(q-p), i think it is the equation of a line passing through points with position vector p and q where t is a parameter

you might have been asked to evaluate the work done in moving from p=<1,3> to q=<2,4>

substituting into r(t) = p + t(q-p) and simplifying,

you get r(t) =<1+t , 3+t>

if r(t) = <x(t),y(t)>

<x(t),y(t)>=<1+t , 3+t> will give you x(t) = 1+t y(t) = 3+t

work done is line integral of F.dr

replace the x and y in F(x,y) = (2xy)i - (x^2)j with x(t) = 1+t y(t) = 3+t
and take the dot product with r(t) =<1+t , 3+t>

to get INT(0 to 1) 2(1+t)(3+t)dt - (1+t)^2 dt

when t =0 and 1, you get p and q from r(t) =<1+t , 3+t> hence the limits
• Mar 29th 2007, 07:03 AM
ecMathGeek
Quote:

Originally Posted by qpmathelp