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If x=2sin⁡θ−sin⁡2θx=2\sin \theta-\sin 2\theta and y=2cos⁡θ−cos⁡2θy=2\cos \theta-\cos 2\theta, θ∈[0,2π]\theta \in[0,2\pi], then (d2y)/(dx2)(d^{2}y)/(dx^{2}) at θ=π\theta=\pi is :

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Q. If x=2sin⁡θ−sin⁡2θx=2\sin \theta-\sin 2\theta and y=2cos⁡θ−cos⁡2θy=2\cos \theta-\cos 2\theta, θ∈[0,2π]\theta \in[0,2\pi], then (d2y)/(dx2)(d^{2}y)/(dx^{2}) at θ=π\theta=\pi is :
  1. Find Expressions for dydx\frac{dy}{dx} and d2ydx2\frac{d^2y}{dx^2}: First, we need to find the expressions for dydx\frac{dy}{dx} and d2ydx2\frac{d^2y}{dx^2}. To do this, we will differentiate xx and yy with respect to θ\theta and then use the chain rule to find dydx\frac{dy}{dx} as dydθ\frac{dy}{d\theta} divided by dxdθ\frac{dx}{d\theta}.
  2. Differentiate xx and yy with respect to θ\theta: Differentiate xx with respect to θ\theta:
    x=2sin⁡(θ)−sin⁡(2θ)x = 2\sin(\theta) - \sin(2\theta)
    dxdθ=ddθ[2sin⁡(θ)−sin⁡(2θ)]\frac{dx}{d\theta} = \frac{d}{d\theta} [2\sin(\theta) - \sin(2\theta)]
    Using the chain rule and trigonometric identities, we get:
    dxdθ=2cos⁡(θ)−2cos⁡(2θ)\frac{dx}{d\theta} = 2\cos(\theta) - 2\cos(2\theta)
  3. Find dydx\frac{dy}{dx} by dividing dydθ\frac{dy}{d\theta} by dxdθ\frac{dx}{d\theta}: Differentiate yy with respect to θ\theta:
    y=2cos⁡(θ)−cos⁡(2θ)y = 2\cos(\theta) - \cos(2\theta)
    dydθ=ddθ[2cos⁡(θ)−cos⁡(2θ)]\frac{dy}{d\theta} = \frac{d}{d\theta} [2\cos(\theta) - \cos(2\theta)]
    Using the chain rule and trigonometric identities, we get:
    dydθ=−2sin⁡(θ)+2sin⁡(2θ)\frac{dy}{d\theta} = -2\sin(\theta) + 2\sin(2\theta)
  4. Differentiate dydx\frac{dy}{dx} with respect to θ\theta: Now we find dydx\frac{dy}{dx} by dividing dydθ\frac{dy}{d\theta} by dxdθ\frac{dx}{d\theta}:
    dydx=dydθdxdθ\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}
    dydx=−2sin⁡(θ)+2sin⁡(2θ)2cos⁡(θ)−2cos⁡(2θ)\frac{dy}{dx} = \frac{-2\sin(\theta) + 2\sin(2\theta)}{2\cos(\theta) - 2\cos(2\theta)}
  5. Apply Quotient Rule to find d2y/dx2d^2y/dx^2: Next, we differentiate dy/dxdy/dx with respect to θ\theta to find d2y/dx2d^2y/dx^2. This requires using the quotient rule:\newlined2y/dx2=d/dθ[dy/dx](dx/dθ)d^2y/dx^2 = \frac{d/d\theta [dy/dx]}{(dx/d\theta)}
  6. Differentiate dydθ\frac{dy}{d\theta} and dxdθ\frac{dx}{d\theta} with respect to θ\theta: Apply the quotient rule:\newlined2ydx2=(dxdθ)(ddθ)(dydθ)−(dydθ)(ddθ)(dxdθ)(dxdθ)2\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}
  7. Evaluate expression at θ=π\theta = \pi: We need to differentiate dydθ\frac{dy}{d\theta} and dxdθ\frac{dx}{d\theta} with respect to θ\theta:
    ddθ(dydθ)=ddθ(−2sin⁡(θ)+2sin⁡(2θ))\frac{d}{d\theta} \left(\frac{dy}{d\theta}\right) = \frac{d}{d\theta} \left(-2\sin(\theta) + 2\sin(2\theta)\right)
    ddθ(dxdθ)=ddθ(2cos⁡(θ)−2cos⁡(2θ))\frac{d}{d\theta} \left(\frac{dx}{d\theta}\right) = \frac{d}{d\theta} \left(2\cos(\theta) - 2\cos(2\theta)\right)
  8. Substitute values into expression: Differentiate both expressions:\newlineddθ(dydθ)=−2cos⁡(θ)+4cos⁡(2θ)\frac{d}{d\theta} \left(\frac{dy}{d\theta}\right) = -2\cos(\theta) + 4\cos(2\theta)\newlineddθ(dxdθ)=−2sin⁡(θ)+4sin⁡(2θ)\frac{d}{d\theta} \left(\frac{dx}{d\theta}\right) = -2\sin(\theta) + 4\sin(2\theta)
  9. Simplify the expression: Substitute these derivatives back into the expression for d2ydx2\frac{d^2y}{dx^2}:d2ydx2=[(2cos⁡(θ)−2cos⁡(2θ))(−2cos⁡(θ)+4cos⁡(2θ))−(−2sin⁡(θ)+2sin⁡(2θ))(−2sin⁡(θ)+4sin⁡(2θ))](2cos⁡(θ)−2cos⁡(2θ))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}
  10. Simplify the expression: Substitute these derivatives back into the expression for d2ydx2\frac{d^2y}{dx^2}:d2ydx2=[(2cos⁡(θ)−2cos⁡(2θ))(−2cos⁡(θ)+4cos⁡(2θ))−(−2sin⁡(θ)+2sin⁡(2θ))(−2sin⁡(θ)+4sin⁡(2θ))](2cos⁡(θ)−2cos⁡(2θ))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 θ=π\theta = \pi:At θ=π, we have:\text{At } \theta = \pi, \text{ we have:}sin⁡(π)=0,sin⁡(2π)=0\sin(\pi) = 0, \sin(2\pi) = 0cos⁡(π)=−1,cos⁡(2π)=1\cos(\pi) = -1, \cos(2\pi) = 1
  11. Simplify the expression: Substitute these derivatives back into the expression for d2y/dx2d^2y/dx^2:d2y/dx2=[(2cos⁡(θ)−2cos⁡(2θ))(−2cos⁡(θ)+4cos⁡(2θ))−(−2sin⁡(θ)+2sin⁡(2θ))(−2sin⁡(θ)+4sin⁡(2θ))](2cos⁡(θ)−2cos⁡(2θ))2d^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 θ=π\theta = \pi:At θ=π, we have:\text{At } \theta = \pi, \text{ we have:}sin⁡(π)=0,sin⁡(2π)=0\sin(\pi) = 0, \sin(2\pi) = 0cos⁡(π)=−1,cos⁡(2π)=1\cos(\pi) = -1, \cos(2\pi) = 1Substitute these values into the expression for d2y/dx2d^2y/dx^2:d2y/dx2=[(2(−1)−2(1))(−2(−1)+4(1))−(−2(0)+2(0))(−2(0)+4(0))](2(−1)−2(1))2d^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}
  12. Simplify the expression: Substitute these derivatives back into the expression for d2y/dx2d^2y/dx^2:d2y/dx2=[(2cos⁡(θ)−2cos⁡(2θ))(−2cos⁡(θ)+4cos⁡(2θ))−(−2sin⁡(θ)+2sin⁡(2θ))(−2sin⁡(θ)+4sin⁡(2θ))](2cos⁡(θ)−2cos⁡(2θ))2d^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 θ=π\theta = \pi: At θ=π\theta = \pi, we have:sin⁡(π)=0,sin⁡(2π)=0\sin(\pi) = 0, \sin(2\pi) = 0cos⁡(π)=−1,cos⁡(2π)=1\cos(\pi) = -1, \cos(2\pi) = 1Substitute these values into the expression for d2y/dx2d^2y/dx^2:d2y/dx2=[(2(−1)−2(1))(−2(−1)+4(1))−(−2(0)+2(0))(−2(0)+4(0))](2(−1)−2(1))2d^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:d2y/dx2=((−2−2)(2+4))((−2−2)2)d^2y/dx^2 = \frac{((-2 - 2)(2 + 4))}{((-2 - 2)^2)}d2y/dx2=(−4)(6)(−4)2d^2y/dx^2 = \frac{(-4)(6)}{(-4)^2}d2y/dx2=−2416d^2y/dx^2 = \frac{-24}{16}d2y/dx2=−32d^2y/dx^2 = -\frac{3}{2}

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