Question

Find $\frac{dy}{dx}$, if x and y are connected parametrically by the equations given in questions without eliminating the parameter.

$x=\frac{si{n}^{3}t}{\sqrt{cost2t}},y=\frac{co{s}^{3}t}{\sqrt{cos2t}}$

Answer 1

Given, $x=\frac{si{n}^{3}t}{\sqrt{cost2t}},y=\frac{co{s}^{3}t}{\sqrt{cos2t}}$

Differentiating w.r.t. t, we get

$\therefore \frac{dx}{dt}=\frac{d}{dt}\left(\frac{si{n}^{3}t}{\sqrt{cos2t}}\right)=\frac{\sqrt{cos2t}\left(3si{n}^{2}tcost\right)-si{n}^{3}t\left(\frac{-2sin2t}{x\sqrt{cos2t}}\right)}{(\sqrt{cos2t}{)}^2}(Usingquotientrule,\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v\frac{du}{dx}-u\frac{du}{dx}}{v^2})=\frac{3\left(cos2t\right)si{n}^{2}tcost+sin2tsi{n}^{3}t}{cos2t\sqrt{cos2t}}=\frac{3(1-2si{n}^{2}t)si{n}^{2}tcost+\left(2sintcost\right)si{n}^{3}t}{cos2t\sqrt{cos2t}}(\because cos2t=1-2si{n}^{2}tandsin2t=2sintcost)=\frac{3si{n}^{2}tcost-4si{n}^{4}tcost}{cos2t\sqrt{cos2t}}and\frac{dy}{dt}=\frac{d}{dt}\left(\frac{co{s}^{3}t}{\sqrt{cos2t}}\right)=\frac{\sqrt{cos2t}(-3co{s}^{2}tsint)-co{s}^{3}t\left(\frac{-2sin2t}{2\sqrt{cos2t}}\right)}{(\sqrt{cos2t}{)}^2}\left(Usingquotientrule\right)=\frac{-3\left(cos2t\right)co{s}^{2}tsint+sin2tco{s}^{3}t}{cos2t\sqrt{cos2t}}=\frac{-3(2co{s}^{2}t-1)co{s}^{2}tsint+co{s}^{3}t\left(2sintcost\right)}{cos2t\sqrt{cos2t}}\left[\begin{array}{cc}\because cos2t=2co{s}^{2}t-1& \\ sin2t=2sintcost& \end{array}\right]=\frac{3co{s}^{2}tsint-4co{s}^{4}tsint}{cos2t\sqrt{cos2t}}\Rightarrow \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{dy}{dt}\times \frac{dt}{dx}=\frac{3co{s}^{2}tsint-4co{s}^{4}tsint}{3si{n}^{2}tcost-4si{n}^{4}tcost}=\frac{co{s}^{2}tsint(3-4co{s}^{2}t)}{si{n}^{2}tcost(3-4si{n}^{2}t)}=\frac{cost(3-4co{s}^{2}t)}{sint(3-4si{n}^{2}t)}=\frac{3cost-4co{s}^{3}t}{3sint-4si{n}^{3}t}=\frac{-3cos3t}{sin3t}=-cot3t\left[\begin{array}{cc}\because & cos3t=4co{s}^{3}t-3cost\\ & sin3t=3sint-4si{n}^{3}t\end{array}\right]$