Question and Answers Forum

All Questions      Topic List

Differentiation Questions

Previous in All Question      Next in All Question      

Previous in Differentiation      Next in Differentiation      

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

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

$$posonsx=\sin talorsy=\cos t$$$$\frac{1}{\sin t}+\frac{4}{\cos t}=\frac{\cos t+4\sin t}{\sin t\cos t}$$$$lerapoortestminimalesi\mathrm{scos}t+4\sin tmnimale)$$$$\{\begin{array}{l}\cos t+4\sin t=0\\ \sin t\cos t\ne 0\end{array}$$$$4\sin t=-\cos t\tan t=-\frac{1}{4}=\frac{\sin t}{\cos t}$$$$\sin {}^{2}t+16\sin {}^{2}t=1\sin t=\frac{\sqrt{17}}{17}t={\sin }^{-1}\left(\frac{\sqrt{17}}{17}\right)\Rightarrow x=\frac{\sqrt{17}}{17}$$$$y=\sqrt{1-\frac{1}{17}}=\frac{4\sqrt{17}}{17}$$$$\frac{1}{x}+\frac{4}{y}=\frac{\sqrt{17}}{17}+\left(4\times \frac{17}{4\sqrt{17}}\right)=\frac{18\sqrt{17}}{17}$$$$\frac{1}{x}+\frac{4}{y}=4,365$$$$$$

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

$${it}^{\prime }swrongsir!$$$$x=\cos t>0$$$$y=\sin t>0$$$$\cos t+4\sin t>0!$$$$you{can}^{\prime }tget\cos t+4\sin t=0!$$

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

$$x^2+y^2=1andx,y>0$$$$\Rightarrow x=\cos \theta ,y=\sin \theta with0<\theta <\frac{\pi }{2}$$$$k=\frac{1}{x}+\frac{4}{y}=\frac{1}{\cos \theta }+\frac{4}{\sin \theta }$$$$\frac{dk}{d\theta }=\frac{\sin \theta }{{\cos }^2\theta }-\frac{4\cos \theta }{{\sin }^2\theta }=0$$$$\Rightarrow {\tan }^3\theta =4$$$$\Rightarrow \tan \theta =\sqrt[3]{4}$$$$\sin \theta =\frac{\sqrt[3]{4}}{\sqrt{1+2\sqrt[3]{2}}}$$$$\cos \theta =\frac{1}{\sqrt{1+2\sqrt[3]{2}}}$$$$k_{min}=\sqrt{1+2\sqrt[3]{2}}+\frac{4\sqrt{1+2\sqrt[3]{2}}}{\sqrt[3]{4}}$$$$=(1+\frac{4}{\sqrt[3]{4}})\sqrt{1+2\sqrt[3]{2}}$$$$={(1+2\sqrt[3]{2})}^{\frac{3}{2}}\approx 6.604$$

Commented by Strengthenchen last updated on 05/Oct/22

$$howcangetdifferentialis0?$$$$becauseitisatrigonometricvalue?$$$$getd\left(c\right)/dx=0?$$

Commented by Tawa11 last updated on 05/Oct/22

$$\mathrm{Great}\mathrm{sirs}$$

Commented by Strengthenchen last updated on 06/Oct/22

$$thankalot,{it}^{\prime }sperfecttousecharacteroffunction$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com