The length of the normal to the curve x=a(the | vedclass

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

Question

(B) 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

$a \sqrt{2}$

Vedclass · Answer

(B) 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)$$\frac{dy}{d	heta} = a(\sin 	heta)$Now,the slope of the tangent $m = \frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta} = \frac{a \sin 	heta}{a(1 + \cos 	heta)} = \frac{\sin 	heta}{1 + \cos 	heta} = 	an(\frac{	heta}{2})$.At $	heta = \frac{\pi}{2}$,the slope of the tangent $m = 	an(\frac{\pi}{4}) = 1$.The slope of the normal is $m_n = -\frac{1}{m} = -1$.The coordinates of the point at $	heta = \frac{\pi}{2}$ are $x = a(\frac{\pi}{2} + 1)$ and $y = a(1 - 0) = a$.The length of the normal is given by the formula $|y \sqrt{1 + m^2}| = |a \sqrt{1 + (1)^2}| = |a \sqrt{2}| = a \sqrt{2}$.

The length of the normal to the curve $x=a(\theta+\sin \theta), y=a(1-\cos \theta)$ at $\theta=\frac{\pi}{2}$ is

Similar Questions

Find the point on the curve $y = \sqrt{4x - 3} - 1$ where the tangent has a slope of $\frac{2}{3}$.

The equation of the tangent to the curve $y=1-e^{\frac{x}{3}}$ at the point of intersection with the $Y$-axis is

Let $n \in (0, \infty)$. If all the curves $y = x^n \log x$ for distinct values of $n$ always have $y = x - 1$ as the tangent drawn at a fixed point $(\alpha, \beta)$,then $\alpha + \beta =$

What is the equation of the tangent to the curve $y = 2 \cos x$ at $x = \pi / 4$?

The slope of the tangent to the curve $y = \int_0^x \frac{1}{1+t^3} dt$ at the point where $x = 1$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module