Vector field along curve

May 2010
74
1
c1 = x(1-x) c2 = 4x(1-x) The Vector field F is given by F(x,y) =((x+y), (X^2 + y)

to find Integral of F in respect to c1 do i just put x= x+y into c1 then integrate or am i making it too simple?

and C is the curve going along c1 then c2, how do i find the integral of F in respect to C do i just add them together?
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
c1 = x(1-x) c2 = 4x(1-x) The Vector field F is given by F(x,y) =((x+y), (X^2 + y)

to find Integral of F in respect to c1 do i just put x= x+y into c1 then integrate or am i making it too simple?

and C is the curve going along c1 then c2, how do i find the integral of F in respect to C do i just add them together?
I see that you have \(\displaystyle \vec{F}= (x+y)\vec{i}+ (x^2+ y)\vec{j}\) but what do you mean by "c1= x(1- x)", "c2= 4x(1-x)"? Do you mean that the curve is given by the parametric equations (x, y)= (t(1-t), 4t(1-t))? I have changed the parameter in the definition of the curve to t so as not to confuse it with the "x" in the vector function. The integral of a vector function \(\displaystyle \vec{F}\) on a curve is \(\displaystyle \int \vec{F}\cdot d\vec{s}\) where \(\displaystyle d\vec{s}\) is the tangent differential to the curve.

Here, if the curve really is \(\displaystyle (x, y)= (t- t^2, 4t- 4t^2)\), then \(\displaystyle d\vec{s}= ((1- 2t)\vec{i}+ (4- 8t)\vec{j}) dt\).

\(\displaystyle \vec{F}\cdot d\vec{s}= (x+y)(1-2t)dt+ (x^2+ y)(4-8t)dt\).

You will need to put \(\displaystyle x= t- t^2\), \(\displaystyle y= 4t- 4t^2\) into that so that you have an integral in t only.
 
May 2010
74
1
Oh sorry i didnt mention what c1 and c2 are c1 is the equation of one curve and c2 is the equation of another curve. The whole curve moves along C1 then to C2, the graph shows two upside down parabolas intersecting at 0 and 1
 
May 2010
74
1
Sorry i mean C1 is the name given to the function y=x(1-x) and c2 is the name given to the function y=4x(1-x)
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
(Headbang) Parabolas don't intersect at numbers, they intersect at points. These intersect at (0, 0) and (1, 1)

So you mean the y= x(1- x) for c2. Well, in that case, you can take "x" itself as parameter- write, for c1, x= t, \(\displaystyle y= t(1- t)= t- t^2\) so that dx= dt, dy= (1- t)dt. Now \(\displaystyle \vec{F}(t)\cdot d\vec{s}= (t+ t(1-t))dt+ (t^2+ t(1-t))(1- t)dt\). Integrate that from t= 0 to t= 1.

On c2, y= 4x(1- x) so take x= t, \(\displaystyle y= 4t(1- t)= 4t- 4t^2\), dx= dt, dy= (4- 8t)dt. Integrate \(\displaystyle \vec{F}(t)\cdot d\vec{s}= (t+ t(1-t)dt+ (t^2+ t(1-t))(4-8t)dt\) from t= 1 to 0 (note the order) so that you complete the loop.
 
May 2010
74
1
sorry i didnt explain it properly haha But THANKS A LOT! I know how to do it now!