1. 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.

2. 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?

3. 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

4. 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

5. Originally Posted by qpmathelp