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Question Number 120463 by bramlexs22 last updated on 31/Oct/20

F i n d t h e s l o p e o f t h e l i n e t a n g e n t

t o t h e g r a p h r = 3 \cos^{2} (2 θ) a t

θ = \frac{π}{6} .

Answered by mr W last updated on 31/Oct/20

\frac{d r}{d θ} = - 12 \cos (2 θ) \sin (2 θ) = - 6 \sin (4 θ)

x = r \cos θ

\frac{d x}{d θ} = - r \sin θ + \frac{d r}{d θ} \cos θ

= - 3 \cos^{2} (2 θ) \sin θ - 6 \sin (4 θ) \cos θ

y = r \sin θ

\frac{d y}{d θ} = r \cos θ + \frac{d r}{d θ} \sin θ

= 3 \cos^{2} (2 θ) \cos θ - 6 \sin (4 θ) \sin θ

\frac{d y}{d x} = \frac{\frac{d y}{d θ}}{\frac{d x}{d θ}} = \frac{3 \cos^{2} (2 θ) \cos θ - 6 \sin (4 θ) \sin θ}{- 3 \cos^{2} (2 θ) \sin θ - 6 \sin (4 θ) \cos θ}

= \frac{\cos^{2} (2 θ) \cos θ - 2 \sin (4 θ) \sin θ}{- \cos^{2} (2 θ) \sin θ - 2 \sin (4 θ) \cos θ}

= - \frac{\cos^{2} (2 θ) - 2 \sin (4 θ) \tan θ}{\cos^{2} (2 θ) \tan θ + 2 \sin (4 θ)}

= - \frac{\cos^{2} (\frac{π}{3}) - 2 \sin (\frac{2 π}{3}) \tan \frac{π}{6}}{\cos^{2} (\frac{π}{3}) \tan \frac{π}{6} + 2 \sin (\frac{2 π}{3})}

= - \frac{\frac{1}{4} - 2 \times \frac{\sqrt{3}}{2} \times \frac{1}{\sqrt{3}}}{\frac{1}{4 \sqrt{3}} + 2 \times \frac{\sqrt{3}}{2}}

= \frac{3 \sqrt{3}}{13}

Commented by mr W last updated on 31/Oct/20

Answered by john santu last updated on 01/Nov/20

T h e s l o p e o f t h e t a n g e n t t o t h e g r a p h

o f r = f (θ) a t (r, θ) g i v e n b y

m = \frac{d y}{d x} = \frac{\tan θ . \frac{d r}{d θ} + r}{\frac{d r}{d θ} - r \tan θ} .

H e r e f (θ) = 3 \cos^{2} θ \to f^{'} (θ) = - 6 \sin 4 θ .

n o w f (π / 6) = 3 \cos^{2} (π / 6) = \frac{3}{4} a n d

f^{'} (π / 6) = - 3 \sqrt{3} . T h u s t h e f o r m u l a

g i v e s m = \frac{d y}{d x} = \frac{- 3 \sqrt{3} \tan (π / 6) + \frac{3}{4}}{- 3 \sqrt{3} - (\frac{3}{4}) \tan (π / 6)}

= \frac{3 \sqrt{3}}{13}