Need Help?

  • Notes
  • Comments & Questions

Find the slope of the tangent line of the curve \(r=1+\sin \theta\) when \(\theta=\frac{\pi}{3}\)

slope \(=\frac{d y}{d x}=\frac{\frac{d r}{d \theta} \sin \theta+r \cos \theta}{\frac{d r}{d \theta} \cos \theta-r \sin \theta}\)

but \(\frac{d r}{d \theta}=\cos \theta\)

slope \(=\frac{\cos \theta \sin \theta+(1+\sin \theta) \cos \theta}{\cos \theta \cdot \cos \theta-(1+\sin \theta) \sin \theta}=\frac{\cos \theta(\sin \theta+(1+\sin \theta))}{\cos ^{2} \theta-sin \theta-\sin ^{2} \theta}\)

but \(\cos ^{2} \theta=1-\sin ^{2} \theta\)

slope \(=\frac{\cos \theta(\sin \theta+(1+\sin \theta))}{\left(1-\sin ^{2} \theta\right)-\sin \theta-\sin ^{2} \theta}=\frac{\cos \theta(2 \sin \theta+1)}{1-2 \sin ^{2} \theta-\sin \theta}\)

\(\left.\frac{d y}{d x}\right|_{\theta=\frac{\pi}{3}}=\frac{\cos \left(\frac{\pi}{3}\right)\left(2 \sin \frac{\pi}{3}+1\right)}{1-2 \sin ^{2}\left(\frac{\pi}{3}\right)-\sin \left(\frac{\pi}{3}\right)}\)

\(=\frac{\frac{1}{2}\left(2 \cdot \frac{\sqrt{3}}{2}+1\right)}{1-2\left(\frac{\sqrt{3}}{2}\right)^{2}-\frac{\sqrt{3}}{2}}=\frac{\sqrt{3}+\frac{1}{2}}{1-2\left(\frac{3}{4}\right)-\frac{\sqrt{3}}{2}}\)

\(\left.\frac{d y}{d x}\right|_{\theta=\frac{\pi}{3}}=\frac{\frac{1}{2}(\sqrt{3}+1)}{-\frac{1}{2}-\frac{\sqrt{3}}{2}}=\frac{\frac{1}{2}(\sqrt{3}+1)}{\frac{1}{2}(-1-\sqrt{3})}=\frac{\sqrt{3}+1}{-(\sqrt{3+1})}=-1\)

Let \(r=1+\sin \theta \: , \: 0 \leq \theta \leq \frac{\pi}{2}\) Find the points at which the tangent is horizontal or vertical

tangent is horizontal \(\rightarrow \frac{d y}{d x}=0\)

\(x=r \cos \theta=(1+\sin \theta) \cos \theta \rightarrow \)

\(\frac{d x}{d \theta}=(\cos \theta)(\cos \theta)+-\sin \theta(1+\sin \theta)=\cos ^{2} \theta-\sin \theta(1+\sin \theta)\)

\(y=r \sin \theta=(1+\sin \theta) \sin \theta \rightarrow \frac{d y}{d \theta }=(\cos \theta)(\sin \theta)+\cos \theta(1+\sin \theta)\)

\(\frac{d y}{d x}=\frac{d y / d \theta}{d x / d \theta}=\frac{\cos \theta \sin \theta+\cos \theta+\cos \theta \sin \theta}{\cos ^{2} \theta-\sin \theta(1+\sin \theta)}\)

\(=\frac{\cos \theta(\sin \theta+1+\sin \theta)}{\cos ^{2} \theta-\sin \theta-\sin ^{2} \theta}\)

\(\cos \theta(1+2 \sin \theta)=0 \longrightarrow \cos \theta=0 \Rightarrow \theta=\frac{\pi}{2} \in\left[0, \frac{\pi}{2}\right]\)

\(\theta=3 \pi / 2 \notin [ o, \frac{\pi}{2} ]\)

\(\rightarrow 1+2 \sin \theta=0 \rightarrow 2 \sin \theta=-1\)

\(\rightarrow \frac{2 \sin \theta}{2}=\frac{-1}{2} \rightarrow \sin \theta=\frac{-1}{2} \rightarrow \theta=\sin ^{-1}\left(\frac{1}{2}\right)=-\pi \notin [0,\frac {\pi}{2}]\)

tangent is horizontal when \(\theta =\frac{\pi}{2}\)

\(\theta=\frac{\pi}{2} \Rightarrow x=\left(1+\sin \left(\frac{\pi}{2}\right)\right) \cos \frac{\pi}{2}=0\)

\(\Rightarrow y=\left(1+\sin \left(\frac{\pi}{2}\right) \sin \frac{\pi}{2}=(1+1)(1)=2(1)=2\right.\)

point \((x, y)=(0,2)\)

Tangent is vertical when \(\rightarrow \frac{d y}{d x} \rightarrow \infty\)

\(\frac{d y}{d x}=\frac{d y / d \theta}{d x / d \theta}\)

\(\frac{d x}{d \theta}=0 \Rightarrow \cos ^{2} \theta-\sin \theta-\sin ^{2} \theta=\)0

\(\cos 2 \theta=\cos ^{2} \theta-\sin ^{2} \theta\)

\(\cos ^{2} \theta-\sin ^{2} \theta-\sin \theta=0\)

\(\cos 2 \theta-\sin \theta=0\)

\(\cos 2 \theta=\sin \theta\)

\(\sin \frac{\pi}{6}=\frac{1}{2}\)

\(\cos \frac{\pi}{3}=\frac{1}{2}\)

\(\cos (2 \cdot {\frac {\pi}{6}})=\sin (\frac{\pi}{6}) \Rightarrow \cos (\frac{\pi}{3})=\sin (\frac{\pi}{6})\)

\(\theta=\frac{\pi}{6}\)

\(x=(1+\sin \theta) \cos \theta=\left(1+\sin \frac{\pi}{6}\right) \cos \frac{\pi}{6} =\left(1+\frac{1}{2}\right) \frac{\sqrt{3}}{2}=\frac{3 \sqrt{3}}{4}\)

\(y=(1+\sin \theta) \sin \theta=\left(1+\sin \frac{\pi}{6}\right) \sin \frac{\pi}{6}=\left(1+\frac{1}{2}\right) \frac{1}{2}=\frac{3}{4}\)

Tangent is vertical when \(\theta=\frac{\pi}{6}\) at point \(\left(\frac{3 \sqrt{3}}{4}, \frac{3}{4}\right)\)

No comments yet

Join the conversation

Join Notatee Today!