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 $k$.

  • $\begingroup$ 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

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 $$k=-1$$

  • $\begingroup$ (+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

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 $$\color{red}{k=-1}$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.