Respuesta :
Answer:
Tangent line is horizontal at points : [tex](2,\frac{3\pi}{2}),(\frac{1}{2},\frac{\pi}{6}),(\frac{1}{2},\frac{5\pi}{6})[/tex] and vertical at points : [tex](0,\frac{\pi}{2}),(\frac{3}{2},\frac{7\pi}{6}),(\frac{3}{2},\frac{11\pi}{6})[/tex] .
Step-by-step explanation:
Given,
[tex]r=1-\sin\theta[/tex] where [tex]0\leq \theta \leq \frac{\pi}{2}[/tex]
Let, [tex]x=r\cos\theta[/tex] and [tex]y=r\sin\theta[/tex]. Then,
[tex]x=(1-\sin\theta)\cos\theta=\cos\theta-\frac{1}{2}\sin 2\theta[/tex]
[tex]y=(1-\sin\theta)\sin\theta=\sin\theta-\sin^{2}\theta[/tex]
Now, to find tangent [tex]\frac{dy}{dx}[/tex], we have to find,
[tex]\frac{dx}{d\theta}=-\sin\theta-\cos 2\theta[/tex]
[tex]\frac{dy}{d\theta}=\cos\theta-2\sin\theta\cos\theta=\cos\theta-\sin 2\theta[/tex]
Therefore,
[tex]\frac{dy}{dx}=\frac{dy}{d\theta}\times\frac{d\theta}{dx}=\frac{\frac{dy}{d\theta}}{\frac{d\theta}{dx}}=\frac{\sin 2\theta-\cos\theta}{\cos 2\theta+\sin\theta}[/tex]
The numerator of [tex]\frac{dy}{dx}[/tex] is 0 implies the tangent line is horizontal. That is,
[tex] \sin 2 \theta-\cos\theta=0[/tex]
[tex]\implies 2\sin\theta\cos\theta-\cos\theta=0[/tex]
[tex]\implies \cos(2\sin\theta-1)=0[/tex]
[tex]\cos\theta=0[/tex] or, [tex]2\sin\theta-1=0[/tex]
[tex]\theta=\frac{\pi}{2},\frac{3\pi}{2}[/tex] or, [tex] \theta=\frac{\pi}{6},\frac{5\pi}{6}[/tex]
Now, [tex]\lim_limits_{\theta\to \frac{\pi}{2}}\frac{\sin 2\theta-\cos\theta}{\cos 2\theta+\sin\theta}\to \infty[/tex] implies that there exist a vertical tangent at [tex]\frac{\pi}{2}[/tex].
And the denominator of [tex]\frac{dy}{dx}[/tex] is 0 implies the tangent line is vertical. That is,
[tex]\sin\theta+\cos\theta=0[/tex]
[tex]\implies (\sin\theta-1)(2\sin\theta+1)=0[/tex]
[tex]\sin\theta=1\implies \theta=\frac{\pi}{2}[/tex]
[tex]\sin\theta=-\frac{1}{2}\implies\theta=\frac{7\pi}{6},\frac{11\pi}{6}[/tex]
When,
[tex]\theta=\frac{\pi}{2},r=1-\sin\frac{\pi}{2}=0[/tex]
[tex]\theta=\frac{3\pi}{2},r=1-\sin\frac{3\pi}{2}=1-(-1)=2[/tex]
[tex]\theta=\frac{\pi}{6},r=1-\sin\frac{\pi}{6}=1-\frac{1}{2}-\frac{1}{2}[/tex]
[tex]\theta=\frac{5\pi}{6},r=1-\sin\frac{5\pi}{6}=1-\frac{1}{2}=\frac{1}{2}[/tex]
[tex]\theta=\frac{7\pi}{6},r=1-\sin\frac{7\pi}{6}=\frac{3}{2}[/tex]
[tex]\theta=\frac{11\pi}{6},r=1-\sin\frac{11\pi}{6}=\frac{3}{2}[/tex]
Hence, tangent line is,
horizontal at points : [tex](2,\frac{3\pi}{2}),(\frac{1}{2},\frac{\pi}{6}),(\frac{1}{2},\frac{5\pi}{6})[/tex]
vertical at points : [tex](0,\frac{\pi}{2}),(\frac{3}{2},\frac{7\pi}{6}),(\frac{3}{2},\frac{11\pi}{6})[/tex]
