Q:

You are given the polar curve r = cos(θ) + sin(θ)a) List all of the points (r,θ) where the tangent line is horizontal. In entering your answer, list the points starting with the smallest value of and limit yourself to (r > or equal to 0) and θ: [0, 2pi] If two or more points share the same value of r, list those starting with the smallest value of θ. Point 1 (r,θ): ?Point 2 (r,θ): ?Point 3 (r,θ): ?b) repeat this with a tangent vertical line

Accepted Solution

A:
Answer:   a) 1.30656∠(3π/8), 0.541196∠(15π/8)   b) 1.30656∠(π/8), 0.541196∠(5π/8)Step-by-step explanation:The critical points can be found in polar coordinates by considering ...   [tex]\displaystyle{dy\over dx}={dy/d\theta\over dx/d\theta}={r\cos\theta + r'\sin\theta\over -r\sin\theta + r'\cos\theta} \quad\text{where $r'=dr/d\theta$}[/tex]We can simplify the effort a little bit by rewriting r as …   [tex]r=\sqrt{2}\sin{(\theta+\pi /4)}[/tex]Then, filling in function and derivative values, we have …   [tex]\dfrac{dy}{dx}=\dfrac{\sqrt{2}(\sin{(\theta+\pi /4)}\cos{(\theta)}+\cos{(\theta+\pi /4)}\sin{(\theta)})}{\sqrt{2}(-\sin{(\theta+\pi /4)}\sin{(\theta)}+\cos{(\theta+\pi /4)}\cos{(\theta)})}\\\\=\dfrac{\sin{(2\theta+\pi /4)}}{\cos{(2\theta+\pi /4)}}\\\\\dfrac{dy}{dx}=\tan{(2\theta +\pi /4)}[/tex]__(a) For horizontal tangents, dy/dx = 0, so we have …  [tex]\tan{(2\theta+\pi /4)}=0\\\\2\theta+\dfrac{\pi}{4}=k\pi \quad\text{for some integer k}\\\\\theta=k\dfrac{\pi}{2}-\dfrac{\pi}{8}[/tex]We can use reference angles for the “r” expressions and write the two horizontal tangent point (r, θ) values of interest as …   [tex](\sqrt{2}\sin{\dfrac{3\pi}{8}},\dfrac{3\pi}{8})\ \text{and}\ (\sqrt{2}\sin{\dfrac{\pi}{8}},\dfrac{15\pi}{8})[/tex]__(b) For vertical tangents, dy/dx = undefined, so we have …  [tex]2\theta+\dfrac{\pi}{4}=k\pi +\dfrac{\pi}{2} \quad\text{for some integer k}\\\\\theta=k\dfrac{\pi}{2}+\dfrac{\pi}{8}[/tex]Again using reference angles for “r”, the two vertical tangent point values of interest are …   [tex](\sqrt{2}\sin{\dfrac{3\pi}{8}},\dfrac{\pi}{8})\ \text{and}\ (\sqrt{2}\sin{\dfrac{\pi}{8}},\dfrac{5\pi}{8})[/tex]__The attached graph shows the angle values in degrees and the radius values as numbers. The points of tangency are mirror images of each other across the line y=x.