If x = sin theta, y = sin^3 theta then frac{d | vedclass

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

Question

(B) Given $x = \sin 	heta$ and $y = \sin^3 	heta$.First,find the derivatives with respect to $	heta$:$\frac{dx}{d	heta} = \cos 	heta$$\frac{dy}{d

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(B) Given $x = \sin 	heta$ and $y = \sin^3 	heta$.First,find the derivatives with respect to $	heta$:$\frac{dx}{d	heta} = \cos 	heta$$\frac{dy}{d	heta} = 3 \sin^2 	heta \cos 	heta$Now,find $\frac{dy}{dx}$ using the chain rule:$\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta} = \frac{3 \sin^2 	heta \cos 	heta}{\cos 	heta} = 3 \sin^2 	heta$Next,differentiate $\frac{dy}{dx}$ with respect to $x$:$\frac{d^2 y}{dx^2} = \frac{d}{dx}(3 \sin^2 	heta) = \frac{d}{d	heta}(3 \sin^2 	heta) \cdot \frac{d	heta}{dx}$$\frac{d^2 y}{dx^2} = (6 \sin 	heta \cos 	heta) \cdot \frac{1}{\cos 	heta} = 6 \sin 	heta$Finally,evaluate at $	heta = \frac{\pi}{2}$:$\frac{d^2 y}{dx^2} = 6 \sin(\frac{\pi}{2}) = 6(1) = 6$.

If $x = \sin \theta, y = \sin^3 \theta$ then $\frac{d^2 y}{d x^2}$ at $\theta = \frac{\pi}{2}$ is . . . . . .

Similar Questions

If $x=f(t)$ and $y=g(t)$ are differentiable functions of $t$,then $\frac{d^2 y}{d x^2}$ is

If $x$ and $y$ are connected parametrically by the equations,without eliminating the parameter,find $\frac{dy}{dx}$ for $x = a \sec \theta$ and $y = b \tan \theta$.

If $x = 2 \sin \theta - \sin 2 \theta$ and $y = 2 \cos \theta - \cos 2 \theta$ where $\theta \in [0, 2 \pi]$,then find the value of $\frac{d^{2} y}{dx^{2}}$ at $\theta = \pi$.

If $x = a(\cos t + \log \tan \frac{t}{2})$ and $y = a \sin t$,then $\frac{dy}{dx} = $

If $x=5(1-\sin t)$ and $y=5(t+\cos t)$, then $\frac{dx}{dy}=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module