Question

Find $\frac{dy}{dx}$, if x = 3 cos θ - 2 cos³ θ, y = 3 sin θ - 2 sin³ θ at θ = π/4

Answer 1

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Answer 2

Answer:

$\rm \text{At }\theta = \dfrac \pi 4,\\\dfrac{dy}{dx}=1.$

Step-by-step explanation:

Given:

• x = 3 cos θ - 2 cos³ θ.
• y = 3 sin θ - 2 sin³ θ.

We know,

$\rm \dfrac{dy}{dx}=\dfrac{\dfrac{dy}{d\theta}}{\dfrac{dx}{d\theta}}.$

Therefore,

$\rm \dfrac{dy}{d\theta}=\dfrac{d}{d\theta}\left (3\sin \theta - 2\sin^3\theta \right )\\=3\dfrac{d(\sin\theta)}{d\theta}-2\dfrac{d(\sin^3\theta)}{d\theta}\\=3\cos\theta-2\times 3\sin^2\theta\dfrac{d(\sin\theta)}{d\theta}\\=3\cos\theta -6\sin^2\theta \cos\theta.$

$\rm \dfrac{dx}{d\theta}=\dfrac{d}{d\theta}\left (3\cos \theta-2\cos^3\theta \right )\\=3\dfrac{d(\cos\theta)}{d\theta}-2\dfrac{d(\cos^3\theta)}{d\theta}\\=-3\sin\theta-2\times 3\cos^2\theta\dfrac{d(\cos\theta)}{d\theta}\\=-3\sin\theta +6\cos^2\theta \sin\theta.$

Using these two values,

$\rm \dfrac{dy}{dx}=\dfrac{\dfrac{dy}{d\theta}}{\dfrac{dx}{d\theta}}=\dfrac{3\cos\theta -6\sin^2\theta \cos\theta}{-3\sin\theta +6\cos^2\theta \sin\theta}=\dfrac{3\cos\theta(1-2\sin^2\theta)}{3\sin\theta(-1+2\cos^2\theta) }.\\=\dfrac{3\cos\theta(\cos(2\theta))}{3\sin\theta(\cos(2\theta))}\ \ \ \ \ \ \ \ \ \ \ (\ \because\  \cos(2\theta)=1-2\sin^2\theta =2\cos^2\theta-1.)\\\Rightarrow \dfrac{dy}{dx}=\cot\theta.\\\\\text{At }\theta = \dfrac \pi 4,\\\dfrac{dy}{dx}=\cot \left (\dfrac \pi 4 \right )=1.$