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 40277 by Raj Singh last updated on 18/Jul/18

Answered by MJS last updated on 18/Jul/18

(d/dθ)[sin^p θ cos^q θ]=sin^(p−1) θ cos^(q−1) θ (pcos^2 θ −qsin^2 θ) zeros at sin θ =0 ⇒ θ=2nπ cos θ =0 ⇒ θ=(2n+1)(π/2) pcos^2 θ −qsin^2 θ =0 pc^2 −qs^2 =0 p(1−s^2 )−qs^2 =0 p−ps^2 −qs^2 =0 sin^2 θ =(p/(p+q)) ⇒ cos^2 θ =1−(p/(p+q))=(q/(p+q)) ⇒ tan^2 θ =(p/q) ⇒ θ=arctan (√(p/q))

$$\frac{{d}}{{d}\theta}\left[\mathrm{sin}^{{p}} \:\theta\:\mathrm{cos}^{{q}} \:\theta\right]=\mathrm{sin}^{{p}−\mathrm{1}} \:\theta\:\mathrm{cos}^{{q}−\mathrm{1}} \:\theta\:\left({p}\mathrm{cos}^{\mathrm{2}} \:\theta\:−{q}\mathrm{sin}^{\mathrm{2}} \:\theta\right) \\ $$$$\mathrm{zeros}\:\mathrm{at} \\ $$$$\mathrm{sin}\:\theta\:=\mathrm{0}\:\Rightarrow\:\theta=\mathrm{2}{n}\pi \\ $$$$\mathrm{cos}\:\theta\:=\mathrm{0}\:\Rightarrow\:\theta=\left(\mathrm{2}{n}+\mathrm{1}\right)\frac{\pi}{\mathrm{2}} \\ $$$$ \\ $$$${p}\mathrm{cos}^{\mathrm{2}} \:\theta\:−{q}\mathrm{sin}^{\mathrm{2}} \:\theta\:=\mathrm{0} \\ $$$${pc}^{\mathrm{2}} −{qs}^{\mathrm{2}} =\mathrm{0} \\ $$$${p}\left(\mathrm{1}−{s}^{\mathrm{2}} \right)−{qs}^{\mathrm{2}} =\mathrm{0} \\ $$$${p}−{ps}^{\mathrm{2}} −{qs}^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{sin}^{\mathrm{2}} \:\theta\:=\frac{{p}}{{p}+{q}} \\ $$$$\Rightarrow\:\mathrm{cos}^{\mathrm{2}} \:\theta\:=\mathrm{1}−\frac{{p}}{{p}+{q}}=\frac{{q}}{{p}+{q}} \\ $$$$\Rightarrow\:\mathrm{tan}^{\mathrm{2}} \:\theta\:=\frac{{p}}{{q}} \\ $$$$\Rightarrow\:\theta=\mathrm{arctan}\:\sqrt{\frac{{p}}{{q}}} \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 18/Jul/18

y=sin^p θcos^q θ lny=plnsinθ+qlncosθ (1/y)(dy/dθ)=pcotθ−qtanθ for max/min (dy/dθ)=0 pcotθ−qtanθ=0 qtanθ=pcotθ tan^2 θ=(p/q) so tanθ=(√(p/q)) (dy/dθ)=sin^p θcos^q θ(pcotθ−qtanθ) (dy/dθ)=y(pcotθ−qtanθ) (d^2 y/dθ^2 )=(dy/dθ)(pcotθ−qtanθ)+y(−pcosec^2 θ−qsec^2 θ) (d^2 y/dθ^2 )=(dy/dθ)(p.(√(q/p)) −q.(√(p/q)) )−y{p(1+(q/p))+q(1+(p/q) (d^2 y/dθ^2 )=0−2y(p+q) =−2y(p+q) at tanθ=(√(p/q)) (d^2 y/dθ^2 ) is −ve so at tanθ=(√(p/q)) sin^p θcos^q θ is maximum

$${y}={sin}^{{p}} \theta{cos}^{{q}} \theta \\ $$$${lny}={plnsin}\theta+{qlncos}\theta \\ $$$$\frac{\mathrm{1}}{{y}}\frac{{dy}}{{d}\theta}={pcot}\theta−{qtan}\theta \\ $$$${for}\:{max}/{min}\:\frac{{dy}}{{d}\theta}=\mathrm{0} \\ $$$${pcot}\theta−{qtan}\theta=\mathrm{0} \\ $$$${qtan}\theta={pcot}\theta \\ $$$${tan}^{\mathrm{2}} \theta=\frac{{p}}{{q}}\:\:\:\:{so}\:{tan}\theta=\sqrt{\frac{{p}}{{q}}} \\ $$$$\frac{{dy}}{{d}\theta}={sin}^{{p}} \theta{cos}^{{q}} \theta\left({pcot}\theta−{qtan}\theta\right) \\ $$$$\frac{{dy}}{{d}\theta}={y}\left({pcot}\theta−{qtan}\theta\right) \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{d}\theta^{\mathrm{2}} }=\frac{{dy}}{{d}\theta}\left({pcot}\theta−{qtan}\theta\right)+{y}\left(−{pcosec}^{\mathrm{2}} \theta−{qsec}^{\mathrm{2}} \theta\right) \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{d}\theta^{\mathrm{2}} }=\frac{{dy}}{{d}\theta}\left({p}.\sqrt{\frac{{q}}{{p}}}\:−{q}.\sqrt{\frac{{p}}{{q}}}\:\right)−{y}\left\{{p}\left(\mathrm{1}+\frac{{q}}{{p}}\right)+{q}\left(\mathrm{1}+\frac{{p}}{{q}}\right.\right. \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{d}\theta^{\mathrm{2}} }=\mathrm{0}−\mathrm{2}{y}\left({p}+{q}\right)\:=−\mathrm{2}{y}\left({p}+{q}\right) \\ $$$${at}\:{tan}\theta=\sqrt{\frac{{p}}{{q}}}\:\:\:\:\frac{{d}^{\mathrm{2}} {y}}{{d}\theta^{\mathrm{2}} }\:{is}\:−{ve} \\ $$$${so}\:{at}\:{tan}\theta=\sqrt{\frac{{p}}{{q}}}\:\:{sin}^{{p}} \theta{cos}^{{q}} \theta\:\:{is}\:{maximum} \\ $$$$ \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com