Finding unit vector tangential to a curve in cylindrical coordinates?

Question

Hi, I'm a math student. I'm having trouble with this problem. If someone could help explain the problem, I'd really appreciate it. The answers are given below.

Find the expression for the unit vector tangential to the curve given in cylindrical coordinates by (r^2)sin [2(phi)] = 1, z = 0. Then obtain the unit vectors tangential to the curve at the points: a) (1, pi/4, pi/2) b) (1, pi/2, pi)

Answer: +-(cos 2(phi) ar - sin 2(phi) ap); a) +-ap; b) +-((3^(1/2))/2 ar - (1/2) ap))
Thank You

Note: ar = unit vector of in radial direction. ap = unit vector in phi direction. +- means plus or minus

Answer 1

If z is everywhere 0, then you have a curve in the x,y plane (or r,Φ plane) expressed in polar coordinates. Take dr/dΦ as the slope of the vector, and normalize it to a unit length with Pythagorean Theorem.


Doug

Answer 2

42.