Thread: 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.

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?

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

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.

if you are asking about the integral

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

Originally Posted by qpmathelp

if you are asking about the integral

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

He doesn't understand the setup for the integration. And I don't blame him. Look at the steps he used to arrive at that integration. They make almost no sense.