Question

$ifx=\sqrt{a^{si{n}^{-1}t}},y=\sqrt{a^{co{s}^{-1}t}},thenshowthat\frac{dy}{dx}=-\frac{y}{x}.$

Answer 1

Given, $x=\sqrt{a^{si{n}^{-1}t}},y=\sqrt{a^{co{s}^{-1}t}}$

$ie..,x=a^{\frac{1}{2}si{n}^{-1}t}andy=a^{\frac{1}{2}co{s}^{-1}t}[\because (a^b{)}^c=a^{bc}]Differentiatingw.r.t.t,weget\frac{dx}{dt}=a^{\frac{1}{2}si{n}^{-1}t}loga\frac{d}{dt}\left(\frac{1}{2}si{n}^{-1}t\right)[\because \frac{d}{dx}{a}^x=a^{x}loga]=a^{\frac{1}{2}si{n}^{-1}t}loga\left(\frac{1}{2\sqrt{1-t^2}}\right)=\frac{a^{\frac{1}{2}si{n}^{-1}t}loga}{2\sqrt{1-t^2}}and\frac{dy}{dt}=a^{\frac{1}{2}co{s}^{-1}t}loga\frac{d}{dt}\left(\frac{1}{2}co{s}^{-1}t\right)(Usingchainrule)=a^{\frac{1}{2}co{s}^{-1}t}loga\left(\frac{-1}{2\sqrt{1-t^2}}\right)=\frac{-a^{\frac{1}{2}co{s}^{-1}t}loga}{2\sqrt{1-t^2}}\Rightarrow \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{-a^{\frac{1}{2}co{s}^{-1}t}}{a^{\frac{1}{2}si{n}^{-1}t}}=-\frac{\sqrt{a^{co{s}^{-1}}t}}{\sqrt{a^{si{n}^{-1}}t}}=-\frac{y}{x}$

Alternate~method

$x=\sqrt{a^{si{n}^{-1}}t}Takinglogonbothsideslogx=log(a^{si{n}^{-1}t}{)}^{1/2}\Rightarrow logx=\frac{1}{2}logasi{n}^{-1}\Rightarrow logx=\frac{1}{2}si{n}^{-1}tloga\Rightarrow logx=\frac{1}{2}logasi{n}^{-1}tDifferentiatingw.r.t.t,weget\frac{1}{x}.\frac{dx}{dt}=\frac{1}{2}loga\frac{d}{dt}(si{n}^{-1t})=\frac{1}{2}loga\frac{1}{\sqrt{1-t^2}}\Rightarrow \frac{dx}{dt}=\frac{xloga}{2}\times \frac{1}{\sqrt{1-t^2}}Again,y=\sqrt{a^{co{s}^{-1}t}}Takinglogonbothsides,logy=log(a^{co{s}^{-1}t}{)}^{1/2}\Rightarrow logy=\frac{1}{2}log{a}^{co{s}^{-1}t}\Rightarrow logy=\frac{1}{2}co{s}^{-1}tlogaDifferentiatingw.r.t.t,weget\frac{1}{y}.\frac{dy}{dt}=\frac{1}{2}loga\times \left[\frac{-1}{\sqrt{1-t^2}}\right]\Rightarrow \frac{dy}{dt}=-\frac{yloga}{2}\times \frac{1}{\sqrt{1-t^2}}Weknowthat\frac{dy}{dx}=\frac{dy}{dt}\times \frac{dt}{dx}\Rightarrow \frac{dy}{dx}=\frac{-yloga}{2}\times \frac{1}{\sqrt{1-t^2}}\times \frac{2\times \sqrt{1-t^2}}{xloga}\Rightarrow \frac{dy}{dx}=\frac{-y}{x}$

Another method

$x=\sqrt{a^{si{n}^{-1}t}}........\left(i\right)andy=\sqrt{a^{co{s}^{-1}t}}.......\left(ii\right)MultiplyingEqs.\left(i\right)and\left(ii\right),weget\Rightarrow xy=\sqrt{a^{si{n}^{-1}t}}\times \sqrt{a^{co{s}^{-1}t}}\Rightarrow xy=\sqrt{a^{si{n}^{-1}t}.a^{co{s}^{-1}t}}\Rightarrow xy=\sqrt{a^{si{n}^{-1}t+co{s}^{-1}t}}[\because si{n}^{-1}x+co{s}^{-1}x=\frac{\pi }{2}]\Rightarrow xy=\sqrt{a^{\frac{\pi }{2}}}Differentiatingw.r.t.x,wegetx\frac{dy}{dx}+y=0\Rightarrow x\frac{dy}{dx}=-y\Rightarrow \frac{dy}{dx}=\frac{-y}{x}\left(\frac{d}{dx}\right(constant)=0)$