If x - a 2 -sin 2 - and y - a 1-cos 2 - find d - d x when -3-

X=a(2 θ sin 2 θ) dx/d θ=a(2 cos 2 θ·2) = 2a(1 cos 2 θ) y=a(1 cos 2 θ) dy/d θ=a(sin 2 θ·2) = 2a·sin 2 θ Now, dy/dx=dy/d θ/dx/d θ=2a.sin 2 θ/2a(1 cos 2 θ)=s ...

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Standard XII

Mathematics

Sin(A+B)Sin(A-B)

If x = a 2 θ-...

Question

If $x = a (2 θ - sin 2 θ)$ and $y = a (1 - cos 2 θ)$ , find $\frac{d y}{d x}$ when $θ = \frac{π}{3}$ .

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Solution

$x = a (2 θ - sin 2 θ)$

$\frac{d x}{d θ} = a (2 - cos 2 θ \cdot 2) = 2 a (1 - cos 2 θ)$

$y = a (1 - cos 2 θ)$

$\frac{d y}{d θ} = a (sin 2 θ \cdot 2) = 2 a \cdot sin 2 θ$

Now, $\frac{d y}{d x} = \frac{\frac{d y}{d θ}}{\frac{d x}{d θ}} = \frac{2 a . sin 2 θ}{2 a (1 - cos 2 θ)} = \frac{sin 2 θ}{1 - cos 2 θ}$

$\frac{d y}{d x}]_{θ = \frac{π}{3}} = \frac{sin \frac{2 π}{3}}{1 - cos \frac{2 π}{3}} = \frac{\frac{\sqrt{3}}{2}}{1 + \frac{1}{2}} = \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}}$

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If x=a(2θ−sin 2θ) and y=a(1−cos 2θ), find dydx when θ=π3.

x=a(2θ−sin 2θ) dxdθ=a(2−cos 2θ⋅2)=2a(1−cos 2θ) y=a(1−cos 2θ) dydθ=a(sin 2θ⋅2)=2a⋅sin 2θ Now, dydx=dydθdxdθ=2a.sin 2θ2a(1−cos 2θ)=sin 2θ1−cos 2θ dydx]θ=π3=sin 2π31−cos2π3=√321+12=√33=1√3

If $x = a (2 θ - sin 2 θ)$ and $y = a (1 - cos 2 θ)$ , find $\frac{d y}{d x}$ when $θ = \frac{π}{3}$ .