(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Find the points at which the following polar curves have a horizontal and vertical tangent line.

(a) r = 3 + 6 cos(theta)

2. Relevant equations

3. The attempt at a solution

x = r cos(theta) = (3 + 6 cos(theta)))cos(theta) = 3cos(theta) + 6 cos(theta)^2

y = rsin(theta) = (3 + 6cos(theta) )sin(theta) = 3 sin(theta) + 6 cos(theta)sin(theta)

dy/dtheta = 3 cos(theta) + 6[cos(theta)^2 - sin(theta)^2]

dx/dtheta = -3 sin(theta) - 12 cos(theta)sin(theta)

dy/dx = - [cos(theta) + 2[cos(theta)^2 - sin(theta)^2]]/[sin(theta) - 4cos(theta)sin(theta)]

for vertical tangent

sin(theta) = 4 cos(theta)sin(theta)

1 = 4 cos(theta)

theta = arccos(1/4)

for horizontal tangent

cos(theta) + 2cos(theta)^2 = 2sin(theta)^2 = 2(1-cos(theta)^2) = 2 - 2cos(theta)^2

cos(theta) + 2cos(theta)^2 + 2cos(theta)^2 - 2 = 0

4cos(theta)^2 + cos(theta) - 2 = 0

cos(theta) = (-1 +/- sqrt(1 - 4(-2)(4) ))/(2(4)) = (-1 +/- sqrt(33) )/8

theta = arccos( (-1 +/- sqrt(33) )/8 )

I don't see what I'm doing wrong, the answer key says that the vertical tangent occurs at theta = arcsin(1/4) and the horizontal tangents occur at arcsin( (-1 +/- sqrt(33) )/8 )

Thanks for any help!

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# Homework Help: Calculus II - Calculus in Polar Coordinates

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