At any two points of the curve represented pa | vedclass

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

Question

(A) Given the parametric equations: $x = a(2 \cos t - \cos 2t)$ and $y = a(2 \sin t - \sin 2t)$.To find the slope of the tangent,we calculate $\frac{d

Vedclass · Accepted Answer

$2\pi / 3$

Vedclass · Answer

(A) Given the parametric equations: $x = a(2 \cos t - \cos 2t)$ and $y = a(2 \sin t - \sin 2t)$.To find the slope of the tangent,we calculate $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$.$\frac{dy}{dt} = a(2 \cos t - 2 \cos 2t) = 2a(\cos t - \cos 2t)$.$\frac{dx}{dt} = a(-2 \sin t + 2 \sin 2t) = 2a(\sin 2t - \sin t)$.Thus,$\frac{dy}{dx} = \frac{2a(\cos t - \cos 2t)}{2a(\sin 2t - \sin t)} = \frac{\cos t - \cos 2t}{\sin 2t - \sin t}$.For the tangent to be parallel to the $x$-axis,$\frac{dy}{dx} = 0$,which implies $\cos t = \cos 2t$.Using the identity $\cos A = \cos B \implies A = 2n\pi \pm B$,we have $2t = 2n\pi \pm t$.Case $1$: $2t = 2n\pi + t \implies t = 2n\pi$.Case $2$: $2t = 2n\pi - t \implies 3t = 2n\pi \implies t = \frac{2n\pi}{3}$.For $n=1$,$t = 2\pi/3$. For $n=2$,$t = 4\pi/3$.The difference between these values is $\frac{4\pi}{3} - \frac{2\pi}{3} = \frac{2\pi}{3}$.

At any two points of the curve represented parametrically by $x = a(2 \cos t - \cos 2t)$ and $y = a(2 \sin t - \sin 2t)$,the tangents are parallel to the $x$-axis. The values of the parameter $t$ corresponding to these points differ from each other by:

Similar Questions

If $x = a(t - \frac{1}{t})$ and $y = a(t + \frac{1}{t})$,then $\frac{dy}{dx} = $

If $x = \sqrt{a^{\sin^{-1} t}}$ and $y = \sqrt{a^{\cos^{-1} t}},$ then $\frac{dy}{dx} = . . . . . .$

The slope of the tangent to the curve $x=t^{2}+3t-8$,$y=2t^{2}-2t-5$ at the point $(2,-1)$ is

If $x = a (t - 1/t)$ and $y = a (t + 1/t)$,where $t$ is the parameter,then $dy/dx = ?$

The slope of the tangent to the curve $x = t^2 + 3t - 8$,$y = 2t^2 - 2t - 5$ at the point $(2, -1)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module