Calculus Question

Question

Calculate the slope of the line tangent to the graph of r=2+4sin∅ at the point corresponding to ∅= pi/6

Answer 1

Let a = theta (angle) The x = rcos(a) and y = rsin(a) but r = 2+4sin(a)

Then

x = cos(a){2+4sin(a)} and y = sin(a){2+4sin(a)}

or x = 2cos(a) + 4cos(a)sin(a) and y = 2sin(a) + 4sin2(a)}

then dx/da = -2sin(a) - 4sin2(a) + 4cos2(a) and dy/da = 2cos(a) + 8sin(a)cos(a)

then

dy/dx = slope of the tangent line = dy/da/dx/da = (2cos(a)+8sin(a)cos(a))/(-2sin(a)-4sin²(a)+4cos²(a))

@ pi/6 = 30 degrees dy/dx = { 2(.866)+8(.5)(.866) }/{ -2(.5)-4(.5)²+4(.866)² } = 5.196/.999 = 5.197