The length of the subtangent at t on the curv | vedclass

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

Question

(A) Given,$x=a(t+\sin t)$ and $y=a(1-\cos t)$.First,we find the derivatives with respect to $t$:$\frac{dx}{dt} = a(1+\cos t) = 2a \cos^2 \frac{t}{2}$$

Vedclass · Accepted Answer

$a \sin t$

Vedclass · Answer

(A) Given,$x=a(t+\sin t)$ and $y=a(1-\cos t)$.First,we find the derivatives with respect to $t$:$\frac{dx}{dt} = a(1+\cos t) = 2a \cos^2 \frac{t}{2}$$\frac{dy}{dt} = a \sin t = 2a \sin \frac{t}{2} \cos \frac{t}{2}$Now,the slope of the tangent is given by $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2a \sin \frac{t}{2} \cos \frac{t}{2}}{2a \cos^2 \frac{t}{2}} = 	an \frac{t}{2}$.The length of the subtangent is defined as $\left| \frac{y}{dy/dx} ight|$.Substituting the values,we get:Length of subtangent $= \frac{a(1-\cos t)}{	an \frac{t}{2}} = \frac{2a \sin^2 \frac{t}{2}}{	an \frac{t}{2}} = \frac{2a \sin^2 \frac{t}{2}}{\sin \frac{t}{2} / \cos \frac{t}{2}} = 2a \sin \frac{t}{2} \cos \frac{t}{2} = a \sin t$.

The length of the subtangent at $t$ on the curve $x=a(t+\sin t), y=a(1-\cos t)$ is

Similar Questions

The angle between the curves $y=\sin 2x$ and $y=\cos 2x$ is

The line $x + y = 0$ touches the curve $y^2 = ax^3 + b$ at the point $(1, -1)$. Find the values of $a$ and $b$.

The equation of the tangent to the curve $y=x^3-2x+7$ at the point $(1,6)$ is

Find the length of the tangent and the sub-tangent for the curve $y^2 = 4ax$ at the point $(at^2, 2at)$.

For the curve $xy = c^2$,the subnormal at any point varies as

Vedclass Test Series

Exam Paper Generator

Online Exam Module