Respuesta :
a) Paremeterize C by
[tex]\vec r(t) = \langle \cos(t), \sin(t) \rangle[/tex]
with 0 ≤ t ≤ 2π. Then any vector of F coiniciding with C takes the form
[tex]\vec F(\vec r(t)) = \langle -\sin(t), \cos(t) \rangle[/tex]
and any such vector has magnitude
[tex]\left\| \vec F(\vec r(t)) \right\| = \sqrt{(-\sin(t))^2 + (\cos(t))^2} = \sqrt{\sin^2(t) + \cos^2(t)} = \sqrt{1} = 1[/tex]
b) If a vector of F on C is tangent to C, then
[tex]\vec F(\vec r(t)) \cdot \vec r(t) = 0[/tex]
and this is indeed the case, since
[tex]\vec F(\vec r(t)) \cdot \vec r(t) = \langle -\sin(t), \cos(t) \rangle \cdot \langle \cos(t), \sin(t) \rangle \\\\ \vec F(\vec r(t)) \cdot \vec r(t) = -\sin(t)\cos(t) + \cos(t)\sin(t) \\\\ \vec F(\vec r(t)) \cdot \vec r(t) = 0[/tex]
c) Observe that
[tex]\dfrac{d\vec r(t)}{dt} = \langle -\sin(t), \cos(t) \rangle = \vec F(\vec r(t))[/tex]
Then in the line integral of F over C, we have
[tex]\displaystyle \int_C \vec F \cdot d\vec r = \int_0^{2\pi} \vec F(\vec r(t)) \cdot \frac{d\vec r}{dt} \, dt[/tex]
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
[tex]\displaystyle \int_C \vec F \cdot d\vec r = \int_0^{2\pi} dt = 2\pi[/tex]
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
[tex]\displaystyle \int_S \vec F \cdot d\vec r = \sum_{i=1}^4 \int_{S_i} \vec F \cdot d\vec r[/tex]
The integral over each side is
[tex]\displaystyle \int_0^1 \vec F(1,t) \cdot \frac{d\langle1,t\rangle}{dt} \, dt = \int_0^1 \langle -t, 1 \rangle \cdot \langle 0, 1 \rangle \, dt = \int_0^1 dt = 1[/tex]
[tex]\displaystyle \int_0^1 \vec F(1-t,1) \cdot \frac{d\langle1-t,1\rangle}{dt} \, dt = \int_0^1 \langle -1, 1-t \rangle \cdot \langle -1, 0 \rangle \, dt = \int_0^1 dt = 1[/tex]
[tex]\displaystyle \int_0^1 \vec F(0,1-t) \cdot \frac{d\langle0,1-t\rangle}{dt} \, dt = \int_0^1 \langle -1+t, 0 \rangle \cdot \langle 0, -1 \rangle \, dt = \int_0^1 0 \, dt = 0[/tex]
[tex]\displaystyle \int_0^1 \vec F(t,0) \cdot \frac{d\langle t,0\rangle}{dt} \, dt = \int_0^1 \langle 0, t \rangle \cdot \langle 1, 0 \rangle \, dt = \int_0^1 0 \, dt = 0[/tex]
so the over integral along S is
[tex]\displaystyle \int_S \vec F \cdot d\vec r = 1 + 1 + 0 + 0 = 2[/tex]
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).
