Find the point on the curve x = a(theta + sin | vedclass

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

Question

(C) Given the parametric equations of the curve are $x = a(	heta + \sin 	heta)$ and $y = a(1 - \cos 	heta)$.First,we find the derivatives with resp

Vedclass · Accepted Answer

$\left( a\left( \frac{\pi }{2} + 1 \right), a \right)$

Vedclass · Answer

(C) Given the parametric equations of the curve are $x = a(	heta + \sin 	heta)$ and $y = a(1 - \cos 	heta)$.First,we find the derivatives with respect to $	heta$:$\frac{dx}{d	heta} = a(1 + \cos 	heta) = a(2 \cos^2 \frac{	heta}{2})$$\frac{dy}{d	heta} = a \sin 	heta = 2a \sin \frac{	heta}{2} \cos \frac{	heta}{2}$The slope of the tangent is given by $\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta} = \frac{2a \sin \frac{	heta}{2} \cos \frac{	heta}{2}}{2a \cos^2 \frac{	heta}{2}} = 	an \frac{	heta}{2}$.Since the tangent makes an angle of $\pi / 4$ with the $x$-axis,the slope is $	an(\pi / 4) = 1$.Therefore,$	an \frac{	heta}{2} = 1$,which implies $\frac{	heta}{2} = \frac{\pi}{4}$,so $	heta = \frac{\pi}{2}$.Now,substitute $	heta = \frac{\pi}{2}$ into the equations for $x$ and $y$:$x = a(\frac{\pi}{2} + \sin \frac{\pi}{2}) = a(\frac{\pi}{2} + 1)$$y = a(1 - \cos \frac{\pi}{2}) = a(1 - 0) = a$Thus,the required point is $\left( a\left( \frac{\pi}{2} + 1 ight), a ight)$.

Find the point on the curve $x = a(\theta + \sin \theta)$,$y = a(1 - \cos \theta)$ where the tangent makes an angle of $\pi / 4$ with the $x$-axis.

Similar Questions

If $y=4x-5$ is tangent to the curve $y^2=px^3+q$ at $(2,3)$,then

The lengths of tangent, subtangent, normal and subnormal for the curve $y=x^2+x-1$ at $(1,1)$ are $A, B, C$ and $D$ respectively, then their increasing order is

The points on the curve $y=\frac{2}{3} x^3+\frac{1}{2} x^2$,where the tangents make equal angles with coordinate axes are

Find the equation of the tangent line to the curve $y=x^{2}-2x+7$ which is parallel to the line $2x-y+9=0$.

The equation of the tangent to the curve $y=\pi e^{\frac{-x}{\pi}}$ at the point where it crosses the $y$-axis is

Vedclass Test Series

Exam Paper Generator

Online Exam Module