anonymous
  • anonymous
Evaluate: Please look at the integral
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
\[\int\limits_{C}^{} F.T ds \] for a vector field F=x ^{2}i-yj
anonymous
  • anonymous
from (2,4) to (1.1)
anonymous
  • anonymous
|dw:1352896610110:dw|

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experimentX
  • experimentX
along the straight line? what is T ?
anonymous
  • anonymous
you have to find flow
experimentX
  • experimentX
|dw:1352897206497:dw|
anonymous
  • anonymous
how did u get r(t) i know you used the points but why did u pick t for the second set of points
amistre64
  • amistre64
the t used is just a variable scalar to stretch the vector to all point along the line from the point used to anchor it to the line
anonymous
  • anonymous
so how would i find F. dr/dt where T=dr/dt?
amistre64
  • amistre64
i dont recall the flow stuff to clearly, but does this look familiar? \[\frac{dr}{dt}=\frac{dr}{dx}\frac{dx}{dt}+\frac{dr}{dy}\frac{dy}{dt}\]
anonymous
  • anonymous
looks like chain rule
amistre64
  • amistre64
this is a line integral right?
amistre64
  • amistre64
http://tutorial.math.lamar.edu/Classes/CalcIII/LineIntegralsVectorFields.aspx ive always found this to be a rather good read. so im going over it
anonymous
  • anonymous
r(t)= <2-t, 4-3t>
anonymous
  • anonymous
F=(2-t)^2i-(4-3t)j
amistre64
  • amistre64
i believe we also need r'
anonymous
  • anonymous
how would i get it i know its a derivative would it be r(t)= -i-3j
anonymous
  • anonymous
r'(t)=-i-3j
amistre64
  • amistre64
then thats its; dot F and r' together to get a scalar equation to integrate right?
anonymous
  • anonymous
yes its F. dr/dt
anonymous
  • anonymous
i am not sure abt the dr/dt part
amistre64
  • amistre64
r= <2-t, 4-3t> r'= <(2-t)', (4-3t)'> r'= <-1, -3>
amistre64
  • amistre64
F=<(2-t)^2 ,-4+3t> dot r'=< -1 , -3 > ----------------------- -(2-t^2)+12-9t
amistre64
  • amistre64
got me ^2 in the wrong side ... :/
amistre64
  • amistre64
the line is from t=0 to t=1 giving us\[\int_{0}^{1}-4-t^2+4t-9t+12~dt\]\[\int_{0}^{1}-t^2-5t+8~dt\]
amistre64
  • amistre64
do you have an answer key to check with by chance?
anonymous
  • anonymous
nope
amistre64
  • amistre64
well, weve followed the simple directions from pauls site; so it should be good ;)
anonymous
  • anonymous
these vector fields so confusing
amistre64
  • amistre64
indeed they are
anonymous
  • anonymous
i have another vector question
amistre64
  • amistre64
the Force equation gives us the force at each for each x,y point along the line the lines vector equation gives us the x and y values along the path F(r) is just defining the forces along the path; dotting with r' tho has me a little baffled at an explanation at the moment tho ;)
anonymous
  • anonymous
how do i draw a vector field
amistre64
  • amistre64
for each lattice point on a graph you draw a little arrow indicating the direction and magnitude of the vector associated with the values of x and y (or whatever reference frame your using) at that point

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