r(theta)=3+8sin thetar(θ)=3+8sinθ
by converting into parametric equations,
Rightarrow{(x(theta)=r(theta)cos theta=(3+8sin theta)cos theta),(y(theta)=r(theta)sin theta=(3+8sin theta)sin theta):}
By differentiating with respect to theta,
x'(theta)=8cos theta cdotcos theta+(3+8sin theta)cdot(-sin theta)
=8(cos^2theta-sin^2theta)-3sin theta
=8cos2theta-3sin theta
by evaluating at theta=pi/6,
Rightarrow x'(pi/6)=8cos(pi/3)-3sin(pi/6)=4-3/2=5/2
By differentiating with respect to theta,
y'(theta)=8cos theta cdot sin theta+(3+8sin theta)cdotcos theta
=16sin theta cos theta+3cos theta
=8sin2theta+3cos theta
by evaluating at theta=pi/6,
Rightarrow y'(pi/6)=8sin(pi/3)+3cos(pi/6)=4sqrt{3}+{3sqrt{3}}/2={11sqrt{3}}/2
So, the slope m can be found by
m={dy}/{dx}|_{theta=pi/6}={{dy}/{d theta}|_{theta = pi /6}}/{{dx]/{d theta}|_{theta = pi /6}}={y'(pi/6)}/{x'(pi/6)}={{11sqrt{3}}/{2}}/{{5}/{2}}={11sqrt{3}}/5
The graph along with its tangent line at theta=pi/6 looks like:
![enter image source here]()
I hope that this was helpful.
