Question

Let C be the straight line curve between the points (0,3) 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,19).

Find the flux of F across the curve C in the direction pointing away from the origin.

Answer 1

In this case, the flux of F across C in the direction pointing away from the origin is 19. The flux of the vector field F across the curve C can be calculated using the dot product between F and the unit normal vector field on C.

Step-by-step explanation:

The flux of a vector field across a curve can be calculated using the line integral formula. In this case, we have the vector field F(x, y) = (x, 19) and the curve C between the points (0, 3) and (1, 0). To calculate the flux of F across C in the direction pointing away from the origin, we need to find the dot product between F and the unit normal vector field on C.

The unit normal vector field on C can be found by finding the tangent vector to C and then rotating it 90 degrees counterclockwise. The tangent vector to C is given by T(t) = (1-t, -3t), where t is the parameterization of C. The rotated vector N(t) = (3t, 1-t) is the unit normal vector field on C.

The dot product between F and N can be calculated as follows: (F ⋅ N) = (x, 19) ⋅ (3t, 1-t) = 3xt + 19(1-t). We need to find the parameter t that corresponds to the points on C. The equation of the line passing through the points (0, 3) and (1, 0) is given by y = -3x + 3. Substituting x = t and y = -3t + 3, we get -3t + 3 = -3t + 3, which is true for all t. Therefore, the dot product (F ⋅ N) is constant along C.

To calculate the flux of F across C, we need to evaluate the dot product at any point on C. Let's choose t = 0. Then (F ⋅ N) = 19(1-0) = 19. The flux of F across C in the direction pointing away from the origin is 19.