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