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A curve is defined by the parametric equations: $x=\cos2t, y=\sin2t, 0<t<π.$

a) Use parametric equations to find $\frac{dy}{dx}$. Hence find the equation of the tangent when $t=\frac{π}{8}$.

b) Obtain an expression for $\frac{d^2y}{dx^2}$ and hence show that: $\sin2t\left(\frac{d^2y}{dx^2}\right)+\left(\frac{dy}{dx}\right)^2=k$, where $k$ is an integer. State the value of $k$.

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  • $\begingroup$ For part a) $\frac{dy}{dx}=\frac{-1}{tan2t}$. And the value of $\frac{dy}{dx}=-1$. The equation of the tangent is: $y-\sqrt{2}/{2}=-1(x-\sqrt{2}/{2})$ $\endgroup$ – James786 Sep 22 '15 at 19:52
  • $\begingroup$ For part a) $\frac{dy}{dx}=\frac{-1}{tan2t}$. And the value of $\frac{dy}{dx}=-1$ when $t=\frac{π}{8}$. The equation of the tangent is: $y-\sqrt{2}/{2}=-1(x-\sqrt{2}/{2})$. $\endgroup$ – James786 Sep 22 '15 at 19:58
  • $\begingroup$ What is required for part b). $\endgroup$ – James786 Sep 22 '15 at 19:59
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Using $$\displaystyle \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$

So here $x=\cos 2t\;,$ Then $$\displaystyle \frac{dx}{dt} = -2\sin 2t$$ and $y=\sin 2t\;,$ Then $$\displaystyle \frac{dy}{dt} = 2\cos 2t$$

So $$\displaystyle \frac{dy}{dx} = -\frac{2\cos 2t}{2\sin 2t} = -\cot 2t\;,$$ Now $$\displaystyle \left(\frac{dy}{dx}\right)_{t=\frac{\pi}{8}} = -\left[\cot 2t \right]_{t=\frac{\pi}{8}} = -1$$

Now $$\displaystyle \frac{d}{dx}\left(\frac{dy}{dx}\right) = -\frac{d}{dx}(\cot 2t) =-\frac{d}{dt}(\cot 2t)\cdot \frac{dt}{dx} = 2\csc^2 2t\cdot \frac{1}{-2\sin 2t} $$

So $$\displaystyle \frac{d^2y}{dx^2} = -\frac{1}{\sin^3 2t}$$

Now $$\displaystyle \sin 2t\cdot \frac{d^2y}{dx^2}+\left(\frac{dy}{dx}\right)^2=-\frac{1}{\sin^2 2t}+\frac{\cos^2 2t}{\sin^2 2t} = -\frac{1}{\sin^2 2t}(1-\cos^2 2t)$$

So we get $$\displaystyle \sin 2t\cdot \frac{d^2y}{dx^2}+\left(\frac{dy}{dx}\right)^2 = -\frac{\sin^2 2t}{\sin^2 2t} = -1=k$$(Given)

So we get $$k=-1$$

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  • $\begingroup$ (+1) However, consider voting up the question if you think it's worth a long-form answer... $\endgroup$ – Zach466920 Sep 22 '15 at 20:25
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Notice, we have $$x=\cos 2t\implies \frac{dx}{dt}=-2\sin 2t$$ $$y=\sin 2t\implies \frac{dy}{dt}=2\cos 2t$$

a) $$\frac{dy}{dx}=\frac{\left(\frac{dy}{dt}\right)}{\left(\frac{dx}{dt}\right)}=\frac{2\cos 2t}{-2\sin 2t}=\color{}{-\cot 2t}$$ $$\left(\frac{dy}{dx}\right)_{t=\pi/8}=-\cot\frac{\pi}{4}=\color{red}{-1}$$

b) From the above result, we have $$\frac{dy}{dx}=-\frac{\cos 2t}{\sin 2t}=-\frac{x}{y}$$

$$\frac{d}{dx}\left(\frac{dy}{dx}\right)=-\frac{d}{dx}\left(\frac{x}{y}\right)$$ $$\frac{d^2y}{dx^2}=-\frac{y-x\frac{dy}{dx}}{y^2}$$ $$y\frac{d^2y}{dx^2}=\frac{x}{y}\frac{dy}{dx}-1$$ $$\sin 2t\frac{d^2y}{dx^2}=-\frac{dy}{dx}\frac{dy}{dx}-1$$ $$\sin 2t\frac{d^2y}{dx^2}=-\left(\frac{dy}{dx}\right)^2-1$$ $$\sin 2t\frac{d^2y}{dx^2}+\left(\frac{dy}{dx}\right)^2=-1$$ Comparing with $\sin 2t\frac{d^2y}{dx^2}+\left(\frac{dy}{dx}\right)^2=k$ we get $$\color{red}{k=-1}$$

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