Question-177456

Question Number 177456 by cortano1 last updated on 05/Oct/22

Answered by a.lgnaoui last updated on 05/Oct/22

p o s o n s x = \sin t a l o r s y = \cos t

\frac{1}{\sin t} + \frac{4}{\cos t} = \frac{\cos t + 4 \sin t}{\sin t \cos t}

l e r a p o o r t e s t m i n i m a l e s i scos t + 4 \sin t m n i m a l e)

{\begin{cases} \cos t + 4 \sin t = 0 \\ \sin t \cos t \neq 0 \end{cases}

4 \sin t = - \cos t \tan t = - \frac{1}{4} = \frac{\sin t}{\cos t}

\sin^{2} t + 16 \sin^{2} t = 1 \sin t = \frac{\sqrt{17}}{17} t = \sin^{- 1} (\frac{\sqrt{17}}{17}) \Rightarrow x = \frac{\sqrt{17}}{17}

y = \sqrt{1 - \frac{1}{17}} = \frac{4 \sqrt{17}}{17}

\frac{1}{x} + \frac{4}{y} = \frac{\sqrt{17}}{17} + (4 \times \frac{17}{4 \sqrt{17}}) = \frac{18 \sqrt{17}}{17}

\frac{1}{x} + \frac{4}{y} = 4, 365

Commented by mr W last updated on 05/Oct/22

{i t}^{'} s w r o n g s i r!

x = \cos t > 0

y = \sin t > 0

\cos t + 4 \sin t > 0!

y o u {c a n}^{'} t g e t \cos t + 4 \sin t = 0!

Answered by mr W last updated on 05/Oct/22

x^{2} + y^{2} = 1 a n d x, y > 0

\Rightarrow x = \cos θ, y = \sin θ w i t h 0 < θ < \frac{π}{2}

k = \frac{1}{x} + \frac{4}{y} = \frac{1}{\cos θ} + \frac{4}{\sin θ}

\frac{d k}{d θ} = \frac{\sin θ}{\cos^{2} θ} - \frac{4 \cos θ}{\sin^{2} θ} = 0

\Rightarrow \tan^{3} θ = 4

\Rightarrow \tan θ = \sqrt[3]{4}

\sin θ = \frac{\sqrt[3]{4}}{\sqrt{1 + 2 \sqrt[3]{2}}}

\cos θ = \frac{1}{\sqrt{1 + 2 \sqrt[3]{2}}}

k_{m i n} = \sqrt{1 + 2 \sqrt[3]{2}} + \frac{4 \sqrt{1 + 2 \sqrt[3]{2}}}{\sqrt[3]{4}}

= (1 + \frac{4}{\sqrt[3]{4}}) \sqrt{1 + 2 \sqrt[3]{2}}

= {(1 + 2 \sqrt[3]{2})}^{\frac{3}{2}} \approx 6.604

Commented by Strengthenchen last updated on 05/Oct/22

h o w c a n g e t d i f f e r e n t i a l i s 0 ?

b e c a u s e i t i s a t r i g o n o m e t r i c v a l u e ?

g e t d (c) / d x = 0 ?

Commented by Tawa11 last updated on 05/Oct/22

Great sirs

Commented by Strengthenchen last updated on 06/Oct/22

t h a n k a l o t, {i t}^{'} s p e r f e c t t o u s e c h a r a c t e r o f f u n c t i o n