Q.If $x = a \sec\theta$, $y = b \tan\theta$ then $\dfrac{dy}{dx} =$

For a parametric curve, $\dfrac{dy}{dx} = \dfrac{dy/d\theta}{dx/d\theta}$.

$x = a\cos\theta \Rightarrow \dfrac{dx}{d\theta} = -a\sin\theta$

$y = a\sin\theta \Rightarrow \dfrac{dy}{d\theta} = a\cos\theta$

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$\dfrac{dy}{dx}=\dfrac{b}{a}\csc\theta$.

Concept. For a curve given parametrically as $x=f(\theta),\ y=g(\theta)$, $\dfrac{dy}{dx}=\dfrac{dy/d\theta}{dx/d\theta}$.

Steps.

• $x=a\sec\theta\Rightarrow \dfrac{dx}{d\theta}=a\sec\theta\tan\theta$.
• $y=b\tan\theta\Rightarrow \dfrac{dy}{d\theta}=b\sec^2\theta$.
• $\dfrac{dy}{dx}=\dfrac{b\sec^2\theta}{a\sec\theta\tan\theta}=\dfrac{b\sec\theta}{a\tan\theta}$. …

Differentiate x and y separately with respect to the parameter θ, then divide.

$x = a\sec\theta \Rightarrow \dfrac{dx}{d\theta} = a\sec\theta\tan\theta$

$y = b\tan\theta \Rightarrow \dfrac{dy}{d\theta} = b\sec^2\theta$

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Use parametric differentiation: $\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}$.

Given $x = 2at^2$, $y = at^4$.

$\frac{dx}{dt} = 4at, \qquad \frac{dy}{dt} = 4at^3$

So: …

Parametric differentiation gives $\frac{a\cos\theta}{-a\sin\theta}=-\cot\theta$.

$\frac{dx}{d\theta}=-a\sin\theta$, $\frac{dy}{d\theta}=a\cos\theta$.

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Parametric: $\frac{dy/dt}{dx/dt}=\frac{2a}{2at}=\frac1t$.

$\frac{dx}{dt}=2at$, $\frac{dy}{dt}=2a$.

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For parametric curves, first find dy/dx as a function of t, then differentiate that again with respect to t and divide by dx/dt once more.

$x=t^2 \Rightarrow \dfrac{dx}{dt}=2t$; $y=t^3 \Rightarrow \dfrac{dy}{dt}=3t^2$

$\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt} = \dfrac{3t^2}{2t} = \dfrac{3t}{2}$ …

second derivative of a parametric curve

$x=t^2\Rightarrow \dfrac{dx}{dt}=2t$; $\ y=t^3\Rightarrow \dfrac{dy}{dt}=3t^2$.

$\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}=\dfrac{3t^2}{2t}=\dfrac{3t}{2}$

Differentiate again with respect to $x$ (using $\dfrac{d}{dx}=\dfrac{1}{dx/dt}\dfrac{d}{dt}$):

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