Let r(t) be the position vector of a curve. Then the normal vector field of the
ID: 2863191 • Letter: L
Question
Let r(t) be the position vector of a curve. Then the normal vector field of the curve is equal to T'(s)/k(s) where T is the unit tangent vector and k is the curvature of r(t). (b) Let z = f(x, y) = x^2 + y^2 be the equation of a surface, then the normal vector of the tangent plane at the point P is grad (f(P)) = 2xi|p + 2yj|_p, that is, the gradient of f evaluated at P. (c) The function = x^2y^2 + xyz^2 is a potential function for the vector field F = (2xy^2 + yz)i + (2x^2y + z^2)j + 2xyzk. (d) Suppose a particle is traveling along a curve r(t). Then the tangential acceleration is a_T = r'(t) middot r"middot(t)/||r'(t)|| and he normal acceleration is a_N = r'(t) times r"(t)/||r'(t)||^2 (e) The curvature of a circle having radius 4 centered at the origin is 1/2. 2. Short Answer Let F be a vector tield defined on a curve C. What condition needed for the path integral of F over C to be independent of path?Explanation / Answer
a. false
b.true
c.false
d.true
2)
curl of F should be zero.
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