Question

Fing the angle between the radius vector and the tangent to the curve r=a(1+sin@) atπ/6

Answer 1

To find the direction angle of the tangent, we need dy/dx.

Using the relationships x = rcos(theta) and y = rsin(theta), and the fact that r is a function of theta, dy/dx= (dy/dtheta)/(dx/dtheta), which is then:

[dr/dtheta * sin(theta) + r * cos(theta)] / [dr/dtheta * cos(theta) - r * sin(theta)].

dr/dtheta = acos(theta). At theta = pi/6, r = (3a)/2 and dr/dtheta = (a*sqrt(3))/2.

Putting these values into the expression for slope, dy/dx, and simplifying, we get a zero denominator, so the tangent is vertical at theta = pi/6.

Since the direction angle of the radius vector is pi/6 and the tangent is vertical, the angle between them is pi/3.