Given that [tex]r=\theta[/tex], then [tex]r'=1[/tex]
The slope of a tangent line in the polar coordinate is given by:
[tex]m= \frac{r'\sin\theta+r\cos\theta}{r'\cos\theta-r\sin\theta} [/tex]
Thus, we have:
[tex]m= \frac{\sin\theta+\theta\cos\theta}{\cos\theta-\theta\sin\theta} [/tex]
Part A:
For horizontal tangent lines, m = 0.
Thus, we have:
[tex]\sin\theta+\theta\cos\theta=0 \\ \\ \theta\cos\theta=-\sin\theta \\ \\ \theta=- \frac{\sin\theta}{\cos\theta} =-\tan\theta[/tex]
Therefore, the values of θ on the polar curve r = θ, with 0 ≤ θ ≤ 2π, such that the tangent lines are horizontal are:
θ = 0
θ = 2.02875783811043
θ = 4.91318043943488
Part B:
For vertical tangent lines, [tex] \frac{1}{m} =0[/tex]
Thus, we have:
[tex]\cos\theta-\theta\sin\theta=0 \\ \\ \Rightarrow\theta\sin\theta=\cos\theta \\ \\ \Rightarrow\theta= \frac{\cos\theta}{\sin\theta} =\sec\theta[/tex]
Therefore, the values of θ on the polar curve r = θ, with 0 ≤ θ ≤ 2π, such that the tangent lines are vertical are:
θ = 4.91718592528713
