If x and y are connected parametrically by th | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given equations are $x=a\left(\cos t+\log 	an \frac{t}{2}ight)$ and $y=a \sin t$.First,differentiate $x$ with respect to $t$:$\frac{dx}{dt} = a

Vedclass · Accepted Answer

$	an t$

Vedclass · Answer

(A) Given equations are $x=a\left(\cos t+\log 	an \frac{t}{2}ight)$ and $y=a \sin t$.First,differentiate $x$ with respect to $t$:$\frac{dx}{dt} = a \left[ \frac{d}{dt}(\cos t) + \frac{d}{dt}(\log 	an \frac{t}{2}) ight]$$= a \left[ -\sin t + \frac{1}{	an \frac{t}{2}} \cdot \sec^2 \frac{t}{2} \cdot \frac{1}{2} ight]$$= a \left[ -\sin t + \frac{\cos \frac{t}{2}}{\sin \frac{t}{2}} \cdot \frac{1}{\cos^2 \frac{t}{2}} \cdot \frac{1}{2} ight]$$= a \left[ -\sin t + \frac{1}{2 \sin \frac{t}{2} \cos \frac{t}{2}} ight]$$= a \left[ -\sin t + \frac{1}{\sin t} ight] = a \left( \frac{1 - \sin^2 t}{\sin t} ight) = a \frac{\cos^2 t}{\sin t}$.Next,differentiate $y$ with respect to $t$:$\frac{dy}{dt} = a \frac{d}{dt}(\sin t) = a \cos t$.Now,find $\frac{dy}{dx}$ using the chain rule:$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{a \cos t}{a \frac{\cos^2 t}{\sin t}} = \frac{\cos t \cdot \sin t}{\cos^2 t} = \frac{\sin t}{\cos t} = 	an t$.

If $x$ and $y$ are connected parametrically by the equations,without eliminating the parameter,find $\frac{dy}{dx}$ for $x=a\left(\cos t+\log \tan \frac{t}{2}\right), y=a \sin t$.

Similar Questions

If $x$ and $y$ are connected parametrically by the equations,without eliminating the parameter,find $\frac{dy}{dx}$ for $x=2at^{2}$ and $y=at^{4}$.

If $x=\sec \theta-\cos \theta$ and $y=\sec ^n \theta-\cos ^n \theta$,then $\frac{d y}{d x}=$

If $x = \frac{1 + t}{t^3}$ and $y = \frac{3}{2t^2} + \frac{2}{t}$,then $x \left( \frac{dy}{dx} \right)^3 - \frac{dy}{dx}$ is equal to (where $t$ is a real parameter).

If $\frac{dy}{dx} = 4$ and $\frac{d^2y}{dx^2} = -3$ at a point $P$ on the curve $y = f(x)$,then $\left(\frac{d^2x}{dy^2}\right)_P = $

If $x=e^\theta(\sin \theta-\cos \theta)$ and $y=e^\theta(\sin \theta+\cos \theta)$,then $\frac{dy}{dx}$ at $\theta=\frac{\pi}{4}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module