If x and y are connected parametrically by th | vedclass

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Question

(A) Given equations are $x = a \sec 	heta$ and $y = b 	an 	heta$. Differentiating $x$ with respect to $	heta$: $\frac{dx}{d	heta} = a \sec 	heta

Vedclass · Accepted Answer

$\frac{b}{a} \csc 	heta$

Vedclass · Answer

(A) Given equations are $x = a \sec 	heta$ and $y = b 	an 	heta$. Differentiating $x$ with respect to $	heta$: $\frac{dx}{d	heta} = a \sec 	heta 	an 	heta$. Differentiating $y$ with respect to $	heta$: $\frac{dy}{d	heta} = b \sec^2 	heta$. Using the chain rule for parametric differentiation: $\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta} = \frac{b \sec^2 	heta}{a \sec 	heta 	an 	heta}$. Simplifying the expression: $\frac{dy}{dx} = \frac{b}{a} \cdot \frac{\sec 	heta}{	an 	heta} = \frac{b}{a} \cdot \frac{1/\cos 	heta}{\sin 	heta / \cos 	heta} = \frac{b}{a} \cdot \frac{1}{\sin 	heta} = \frac{b}{a} \csc 	heta$.

If $x$ and $y$ are connected parametrically by the equations,without eliminating the parameter,find $\frac{dy}{dx}$ for $x = a \sec \theta$ and $y = b \tan \theta$.

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