Parametric Differentiation
If \(x\) and \(y\) are given as functions of a parameter \(t\), then:
\[\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\]
Find the equation of the normal at the point \(p\) where \(\theta=\frac{\pi}{6}\) to the curve with parametric equations \(x=3\sin\theta\) and \(y=5\cos\theta\)
- When \(\theta=\frac{\pi}{6}\), \(x=3\sin\left( \frac{\pi}{6} \right)=\frac{3}{2}\) and \(y=5\cos\left( \frac{\pi}{6} \right)=\frac{5\sqrt{ 3 }}{2}\)
- \(\frac{dx}{d\theta}=3\cos\theta\)
- \(\frac{dy}{d\theta}=-5\sin\theta\)
- \(\frac{dy}{dx}=-\frac{5\sin\theta}{3\cos\theta}=-\frac{3}{5}\tan\theta\)
- When \(\theta=\frac{\pi}{6}\), \(\frac{dy}{dx}=-\frac{5}{3}\tan\left( \frac{\pi}{6} \right)=-\frac{5\sqrt{ 3 }}{9}\)
- \(\left( y-\frac{5\sqrt{ 3 }}{2} \right)=-\frac{9}{5\sqrt{ 3 }}\left( x-\frac{3}{2} \right)\)
- \(\left( y-\frac{5\sqrt{ 3 }}{2} \right)=-\frac{3\sqrt{ 3 }}{5}\left( x-\frac{3}{2} \right)\)
\(x=\sqrt{ 3 }\sin2t\), \(y=4\cos^2t\), \(0\leq t\leq \pi\)
Show that \(\frac{dy}{dx}=k\sqrt{ 3 }\tan2t\)
- \(\frac{dy}{dt}4(\cos t)^2=8\cos t\times-\sin t=-8\cos t\sin t\)
- \(\frac{dx}{dt}=2\sqrt{ 3 }\cos2t\)
- \(\frac{dy}{dx}=-\frac{8\cos t\sin t}{2\sqrt{ 3 }\cos2t}\)