help with polar equation: r= -4cos3θ?

Question

a) find the range for theta so that the graph is traced exactly one time
b) find the maximum value of r and where it occurs in the interval you wrote in part a
c) where are the zeros of the graph?
d)what type of symmetry does this graph have?

Answer 1

Is theta in radians or degrees? I will assume radians.

a)0 to (2*pi)/3
b)4. pi/3
c)pi/6, pi/2
d)its a circle...any type of symmetry

Answer 2

The equation r = -4cos3θ is a three petal flower that centers on the pole.

a) Find the range for theta so that the graph is traced exactly one time.

0 ≤ θ < π

b) Find the maximum value of r and where it occurs in the interval you wrote in part a.

The maximum value of r occurs when cos3θ = ±1. I am assuming you mean absolute value of r. That occurs when:

θ = 0, π/3, 2π/3

c) Where are the zeros of the graph?

The zeros occur when cos3θ = 0. That occurs when:

θ = π/6, π/2, 5π/6

d) What type of symmetry does this graph have?

The graph is a three petal flower centered on the pole. It has 120° symmetry about the pole.

Answer 3

nicely you recognize x=r cos? , y = r sin?. we favor to get rid of both r and ?. So use the substitutions: r = sqrt(x^2+y^2) cos? = x/r = x/sqrt(x^2+y^2) sec? = r/x = sqrt(x^2+y^2)/x So: r(4sec? + a million) = (8sec?) r(4 + cos?) = 8 sqrt(x^2+y^2) (4 + x/sqrt(x^2+y^2)) = 8 sq. both section: (x^2+y^2) (16 + 8x/sqrt(x^2+y^2) + x^2/(x^2+y^2)) = sixty 4 Multiply out: 16 (x^2+y^2) + 8x.sqrt(x^2+y^2) + x^2 = sixty 4 17x^2 + 16y^2 + 8x.sqrt(x^2+y^2) = sixty 4 no longer extremely: 15x^2 + 16y^2 + 16x - sixty 4 = 0 attempt 2: word also: r^2 = r^2 (cos^2 ? sin^2 ?) = x^2 + y^2 r = (8sec?) / (4sec? + a million) = 8/ (4 + cos?) r(4 + cos?) = 8 r^2(16 + 8cos? +cos^2 ?) = sixty 4 16r^2 + 8r^2 cos? + r^2 cos^2 ? = sixty 4 16(x^2 + y^2) + 8r (r cos?) + (r cos?)^2 = sixty 4 16(x^2 + y^2) + 8r (x) + x^2 = sixty 4 it truly is not any longer efficient, nonetheless have a element 8rx (= 8x.sqrt(x^2+y^2)). also the added x^2 time period is +a million no longer -a million? Dupinder has it, the trick replaced into to bypass the r cos? time period to the RHS to ward off getting the awkward 8x.sqrt(x^2+y^2) go-time period once you sq..