If $x=2\sin \theta-\sin 2\theta$ and $y=2\cos \theta-\cos 2\theta$ , $\theta \in[0,2\pi]$ , then $(d^{2}y)/(dx^{2})$ at $\theta=\pi$ is :

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Q. If

x=2\sin \theta-\sin 2\theta

and

y=2\cos \theta-\cos 2\theta

\theta \in[0,2\pi]

, then

(d^{2}y)/(dx^{2})

\theta=\pi

is :

Find Expressions for $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$ : First, we need to find the expressions for $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$ . To do this, we will differentiate $x$ and $y$ with respect to $\theta$ and then use the chain rule to find $\frac{dy}{dx}$ as $\frac{dy}{d\theta}$ divided by $\frac{dx}{d\theta}$ .
Differentiate $x$ and $y$ with respect to $\theta$ : Differentiate $x$ with respect to $\theta$ :
$x = 2\sin(\theta) - \sin(2\theta)$
$\frac{dx}{d\theta} = \frac{d}{d\theta} [2\sin(\theta) - \sin(2\theta)]$
Using the chain rule and trigonometric identities, we get:
$\frac{dx}{d\theta} = 2\cos(\theta) - 2\cos(2\theta)$
Find $\frac{dy}{dx}$ by dividing $\frac{dy}{d\theta}$ by $\frac{dx}{d\theta}$ : Differentiate $y$ with respect to $\theta$ :
$y = 2\cos(\theta) - \cos(2\theta)$
$\frac{dy}{d\theta} = \frac{d}{d\theta} [2\cos(\theta) - \cos(2\theta)]$
Using the chain rule and trigonometric identities, we get:
$\frac{dy}{d\theta} = -2\sin(\theta) + 2\sin(2\theta)$
Differentiate $\frac{dy}{dx}$ with respect to $\theta$ : Now we find $\frac{dy}{dx}$ by dividing $\frac{dy}{d\theta}$ by $\frac{dx}{d\theta}$ :
$\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}$
$\frac{dy}{dx} = \frac{-2\sin(\theta) + 2\sin(2\theta)}{2\cos(\theta) - 2\cos(2\theta)}$
Apply Quotient Rule to find $d^2y/dx^2$ : Next, we differentiate $dy/dx$ with respect to $\theta$ to find $d^2y/dx^2$ . This requires using the quotient rule: $\newline$ $d^2y/dx^2 = \frac{d/d\theta [dy/dx]}{(dx/d\theta)}$
Differentiate $\frac{dy}{d\theta}$ and $\frac{dx}{d\theta}$ with respect to $\theta$ : Apply the quotient rule: $\newline$ $\frac{d^2y}{dx^2} = \frac{\left(\frac{dx}{d\theta}\right)\left(\frac{d}{d\theta}\right)\left(\frac{dy}{d\theta}\right) - \left(\frac{dy}{d\theta}\right)\left(\frac{d}{d\theta}\right)\left(\frac{dx}{d\theta}\right)}{\left(\frac{dx}{d\theta}\right)^2}$
Evaluate expression at $\theta = \pi$ : We need to differentiate $\frac{dy}{d\theta}$ and $\frac{dx}{d\theta}$ with respect to $\theta$ :
$\frac{d}{d\theta} \left(\frac{dy}{d\theta}\right) = \frac{d}{d\theta} \left(-2\sin(\theta) + 2\sin(2\theta)\right)$
$\frac{d}{d\theta} \left(\frac{dx}{d\theta}\right) = \frac{d}{d\theta} \left(2\cos(\theta) - 2\cos(2\theta)\right)$
Substitute values into expression: Differentiate both expressions: $\newline$ $\frac{d}{d\theta} \left(\frac{dy}{d\theta}\right) = -2\cos(\theta) + 4\cos(2\theta)$ $\newline$ $\frac{d}{d\theta} \left(\frac{dx}{d\theta}\right) = -2\sin(\theta) + 4\sin(2\theta)$
Simplify the expression: Substitute these derivatives back into the expression for $\frac{d^2y}{dx^2}$ : $\frac{d^2y}{dx^2} = \frac{[(2\cos(\theta) - 2\cos(2\theta))(-2\cos(\theta) + 4\cos(2\theta)) - (-2\sin(\theta) + 2\sin(2\theta))(-2\sin(\theta) + 4\sin(2\theta))]}{(2\cos(\theta) - 2\cos(2\theta))^2}$
Simplify the expression: Substitute these derivatives back into the expression for $\frac{d^2y}{dx^2}$ : $\frac{d^2y}{dx^2} = \frac{[(2\cos(\theta) - 2\cos(2\theta))(-2\cos(\theta) + 4\cos(2\theta)) - (-2\sin(\theta) + 2\sin(2\theta))(-2\sin(\theta) + 4\sin(2\theta))]}{(2\cos(\theta) - 2\cos(2\theta))^2}$ Now we need to evaluate this expression at $\theta = \pi$ : $\text{At } \theta = \pi, \text{ we have:}$ $\sin(\pi) = 0, \sin(2\pi) = 0$ $\cos(\pi) = -1, \cos(2\pi) = 1$
Simplify the expression: Substitute these derivatives back into the expression for $d^2y/dx^2$ : $d^2y/dx^2 = \frac{[(2\cos(\theta) - 2\cos(2\theta))(-2\cos(\theta) + 4\cos(2\theta)) - (-2\sin(\theta) + 2\sin(2\theta))(-2\sin(\theta) + 4\sin(2\theta))]}{(2\cos(\theta) - 2\cos(2\theta))^2}$ Now we need to evaluate this expression at $\theta = \pi$ : $\text{At } \theta = \pi, \text{ we have:}$ $\sin(\pi) = 0, \sin(2\pi) = 0$ $\cos(\pi) = -1, \cos(2\pi) = 1$ Substitute these values into the expression for $d^2y/dx^2$ : $d^2y/dx^2 = \frac{[(2(-1) - 2(1))(-2(-1) + 4(1)) - (-2(0) + 2(0))(-2(0) + 4(0))]}{(2(-1) - 2(1))^2}$
Simplify the expression: Substitute these derivatives back into the expression for $d^2y/dx^2$ : $d^2y/dx^2 = \frac{[(2\cos(\theta) - 2\cos(2\theta))(-2\cos(\theta) + 4\cos(2\theta)) - (-2\sin(\theta) + 2\sin(2\theta))(-2\sin(\theta) + 4\sin(2\theta))]}{(2\cos(\theta) - 2\cos(2\theta))^2}$ Now we need to evaluate this expression at $\theta = \pi$ : At $\theta = \pi$ , we have: $\sin(\pi) = 0, \sin(2\pi) = 0$ $\cos(\pi) = -1, \cos(2\pi) = 1$ Substitute these values into the expression for $d^2y/dx^2$ : $d^2y/dx^2 = \frac{[(2(-1) - 2(1))(-2(-1) + 4(1)) - (-2(0) + 2(0))(-2(0) + 4(0))]}{(2(-1) - 2(1))^2}$ Simplify the expression: $d^2y/dx^2 = \frac{((-2 - 2)(2 + 4))}{((-2 - 2)^2)}$ $d^2y/dx^2 = \frac{(-4)(6)}{(-4)^2}$ $d^2y/dx^2 = \frac{-24}{16}$ $d^2y/dx^2 = -\frac{3}{2}$