If x=a(t+sin t) and y=a(1-cos t),then find fr | vedclass

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

Question

(B) Given $x=a(t+\sin t)$ and $y=a(1-\cos t)$.First,find the derivatives with respect to $t$:$\frac{d x}{d t}=a(1+\cos t)$ and $\frac{d y}{d t}=a \sin

Vedclass · Accepted Answer

$\frac{1}{4 a \cos ^4(\frac{t}{2})}$

Vedclass · Answer

(B) Given $x=a(t+\sin t)$ and $y=a(1-\cos t)$.First,find the derivatives with respect to $t$:$\frac{d x}{d t}=a(1+\cos t)$ and $\frac{d y}{d t}=a \sin t$.Now,find $\frac{d y}{d x}$:$\frac{d y}{d x}=\frac{d y / d t}{d x / d t}=\frac{a \sin t}{a(1+\cos t)}=\frac{2 \sin(t/2) \cos(t/2)}{2 \cos^2(t/2)}=	an(t/2)$.Next,differentiate $\frac{d y}{d x}$ with respect to $x$:$\frac{d^2 y}{d x^2}=\frac{d}{d x}(	an(t/2))=\sec^2(t/2) \cdot \frac{1}{2} \cdot \frac{d t}{d x}$.Since $\frac{d x}{d t}=a(1+\cos t)=2a \cos^2(t/2)$,then $\frac{d t}{d x}=\frac{1}{2a \cos^2(t/2)}$.Substituting this back:$\frac{d^2 y}{d x^2}=\sec^2(t/2) \cdot \frac{1}{2} \cdot \frac{1}{2a \cos^2(t/2)}=\frac{1}{4a \cos^4(t/2)}$.

If $x=a(t+\sin t)$ and $y=a(1-\cos t)$,then find $\frac{d^2 y}{d x^2}$.

Similar Questions

$A$ curve is given by the equations $x = a \cos \theta + \frac{1}{2}b \cos 2\theta$ and $y = a \sin \theta + \frac{1}{2}b \sin 2\theta$. The points for which $\frac{d^2y}{dx^2} = 0$ are given by:

If $y=a \sin ^3 t$ and $x=a \cos ^3 t$,then $\frac{d y}{d x}$ at $t=\frac{3 \pi}{4}$ is

If $x=\sec \theta-\cos \theta$ and $y=\sec ^n \theta-\cos ^n \theta$,then $\frac{d y}{d x}=$

If $x = a \sin t - b \cos t$ and $y = a \cos t + b \sin t$,then $y^{3} \frac{d^{2} y}{d x^{2}} + x^{2} + y^{2} = $

The derivative of $f(\tan x)$ with respect to $g(\sec x)$ at $x = \frac{\pi}{4}$,where $f^{\prime}(1) = 2$ and $g^{\prime}(\sqrt{2}) = 4$,is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module