Let C be the straight line curve between the points (0,5) and (1,0). Let N be the unit normal vector field on C, oriented so that it points away from the origin. Let F:R2→R2 be the vector field defined by F(x,y)=(x,11). Find the flux of F across the curve C in the direction pointing away from the origin.
∫c F⋅N ds=
To find the flux of the vector field F across the curve C in the direction pointing away from the origin, we can use the formula:
∫c F⋅N ds
where F is the vector field and N is the unit normal vector field on C.
Step 1: Find the tangent vector T to the curve C
The tangent vector T to the curve C can be found by taking the derivative of the parametric equations of the curve, with respect to the parameter t, which defines the curve. In this case, the curve C is a straight line with endpoints (0,5) and (1,0).
Let t vary from 0 to 1:
x = t
y = 5 - 5t
Differentiating with respect to t, we get:
dx/dt = 1
dy/dt = -5
Therefore, the tangent vector T is given by T = (dx/dt, dy/dt) = (1, -5).
Step 2: Calculate the unit normal vector N
The unit normal vector N can be obtained by normalizing the tangent vector T. To normalize a vector, we divide each component by the magnitude of the vector.
Magnitude of T = |T| = sqrt(1^2 + (-5)^2) = sqrt(26)
Therefore, the unit normal vector N is given by N = (1/sqrt(26), -5/sqrt(26)).
Step 3: Calculate F⋅N
The dot product of the vector field F and the unit normal vector N is given by:
F⋅N = (x, 11)⋅(1/sqrt(26), -5/sqrt(26))
Expanding the dot product, we get:
F⋅N = x/sqrt(26) - 5(11)/sqrt(26)
= x/sqrt(26) - 55/sqrt(26)
Step 4: Integrate along the curve C
Finally, we integrate F⋅N ds along the curve C to find the flux:
∫c F⋅N ds = ∫c (x/sqrt(26) - 55/sqrt(26)) ds
The integral can be evaluated by parameterizing the curve C and expressing ds in terms of the parameter t. In this case, since C is a straight line, we can express ds as the magnitude of the derivative of the parametric equations of C:
ds = |(dx/dt, dy/dt)| dt
= sqrt((dx/dt)^2 + (dy/dt)^2) dt
= sqrt(1 + (-5)^2) dt
= sqrt(26) dt
Since x = t, we have dx = dt. Therefore, our integral becomes:
∫c (x/sqrt(26) - 55/sqrt(26)) ds
= ∫c (t/sqrt(26) - 55/sqrt(26)) sqrt(26) dt
= ∫c (t - 55) dt
Evaluating the integral from t = 0 to t = 1, we get:
∫c F⋅N ds = ∫0^1 (t - 55) dt
= [(t^2/2 - 55t)]0^1
= (1/2 - 55) - (0 - 0)
= -109/2
Therefore, the flux of the vector field F across the curve C in the direction pointing away from the origin is -109/2.