# Vector field along curve

• May 22nd 2010, 04:37 AM
Vector field along curve
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?
• May 22nd 2010, 09:31 AM
HallsofIvy
Quote:

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 22nd 2010, 03:07 PM
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 22nd 2010, 06:30 PM
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.