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

Find all points on the curve $y = 4x^3 - 2x^5$ at which the tangent passes through the origin.

If all the normals drawn to the curve $y=\frac{1+3x^2}{3+x^2}$ at the points of intersection of $y=\frac{1+3x^2}{3+x^2}$ and $y=1$ pass through the point $(\alpha, \beta)$,then $3\alpha+2\beta=$

The abscissae of the points,where the tangent to the curve $y=x^{3}-3x^{2}-9x+5$ is parallel to the $x$-axis,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 length of the subtangent to the curve $x^2 y^2 = a^4$ at the point $(-a, a)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module