Consider F=<-y, x> Let C be the unit circle oriented CCW. Let S be the unit square oriented CCW (.e. segments from (1,0) to (1,1) to (0,1) to (0,0) to (1,0)) a) Show that all vectors of F on curve C have a magnitude of 1.

b) Show that all vectors of F on curve C are tangent to the circle C. (Hint: the radius to any point P on a circle is perpendicular to the tangent line at point P.)

c) Show that the line integral of F over C = the length of C.

d) Find the line integral of F over S.

e) Is F path independent? How do you know? Justify your answer in two ways.

a) Paremeterize C by

with 0 ≤ t ≤ 2π. Then any vector of F coiniciding with C takes the form

and any such vector has magnitude

b) If a vector of F on C is tangent to C, then

and this is indeed the case, since

c) Observe that

Then in the line integral of F over C, we have

Recall that for any vector x, we have ||x||² = <x, x>. (The dot product of a vector with itself is the square of its norm.) The two vectors in the integral are the same, and their norm is 1, so the line integral reduces to

which is indeed the circumference of a circle with radius 1.

d) Parameterize each side of S = S₁ U S₂ U S₃ U S₄,

• S₁, from (1, 0) to (1, 1) : x(t) = 1, y(t) = t;

• S₂, from (1, 1) to (0, 1) : x(t) = 1 - t, y(t) = 1;

• S₃, from (0, 1) to (0, 0) : x(t) = 0, y(t) = 1 - t;

• S₄, from (0, 0) to (1, 0) : x(t) = t, y(t) = 0;

each with 0 ≤ t ≤ 1.

Then the line integral of F over S is

The integral over each side is

so the over integral along S is

e) F is not path independent. If it were, then the integral along any closed simple curve would be 0. But we've shown that the integrals along C and S are both positive.

Another way to check would be to confirm that F is not a conservative vector field. If F is conservative, then it is the gradient of some scalar function, so that ∇ f(x, y) = F. This would require

∂f/∂x = -y

∂f/∂y = x

From the first partial derivative, if we integrate both sides with respect to x, then

f(x, y) = -xy + g(y)

Differentiating this with respect to y gives

∂f/∂y = x = -x + dg/dy

so that dg/dy = 2x. But we assume g(y) is a function only of y, so there is no solution to this differential equation, and hence no such function f(x, y).

c

step-by-step explanation:

6.5

step-by-step explanation:

you said it was in the question

~courtney