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

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

Question

(C) The length of the normal is given by the formula: $L = |y| \sqrt{1 + (dy/dx)^2}$.First,we find the derivatives with respect to $	heta$:$dx/d	het

Vedclass · Accepted Answer

$\sqrt{2}a$

Vedclass · Answer

(C) The length of the normal is given by the formula: $L = |y| \sqrt{1 + (dy/dx)^2}$.First,we find the derivatives with respect to $	heta$:$dx/d	heta = a(1 + \cos 	heta)$$dy/d	heta = a \sin 	heta$Thus,$dy/dx = (dy/d	heta) / (dx/d	heta) = (a \sin 	heta) / (a(1 + \cos 	heta)) = \sin 	heta / (1 + \cos 	heta)$.Using trigonometric identities,$\sin 	heta = 2 \sin(	heta/2) \cos(	heta/2)$ and $1 + \cos 	heta = 2 \cos^2(	heta/2)$.So,$dy/dx = 	an(	heta/2)$.At $	heta = \pi/2$,$dy/dx = 	an(\pi/4) = 1$.Also,at $	heta = \pi/2$,$y = a(1 - \cos(\pi/2)) = a(1 - 0) = a$.Substituting these values into the formula:$L = |a| \sqrt{1 + (1)^2} = a \sqrt{2}$.

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

Similar Questions

The angle of intersection of the curves $r = \sin \theta + \cos \theta$ and $r = 2 \sin \theta$ is equal to:

If the curves $x=y^{4}$ and $xy=k$ cut at right angles,then $(4k)^{6}$ is equal to ..... .

The length of the subtangent to the curve $x^2 y^2 = a^4$ at the point $(-a, a)$ is

The perpendicular distance from the origin to the normal drawn at any point on the curve $x=a(\cos \theta+\theta \sin \theta), y=a(\sin \theta-\theta \cos \theta)$ is

At what points of the curve $y = \frac{2}{3}x^3 + \frac{1}{2}x^2$ does the tangent make an equal angle with the axes?

Vedclass Test Series

Exam Paper Generator

Online Exam Module