Question

find d²y/dx² if x=sin theta and y=sin³theta at theta=π/2​

Answer 1

EXPLANATION.

⇒ x = sinθ. - - - - - (1).

⇒ y = sin³θ. - - - - - (2).

at x = π/2.

As we know that,

From equation (1), we get.

⇒ dx/dθ = d(sinθ)/dθ.

⇒ dx/dθ = cosθ. - - - - - (a).

From equation (2), we get.

⇒ y = sin³θ.

⇒ dy/dθ = 3sin²θ.(cosθ). - - - - - (b).

⇒ dy/dx = dy/dθ/dx/dθ.

⇒ dy/dx = 3sin²θ.(cosθ)/cosθ.

⇒ dy/dx = 3sin²θ.

Again differentiate the function, we get.

⇒ d²y/dx² = 3sin²θ.

⇒ d²y/dx² = 6sinθ.cosθ/cosθ.

⇒ d²y/dx² = 6sinθ.

⇒ (d²y/dx²) at θ = π/2.

Put the value of θ = π/2 in the equation, we get.

⇒ d²y/dx² = 6sin(π/2).

⇒ d²y/dx² = 6.

                                                                                                                         

MORE INFORMATION.

(1) = d(sin x)/dx = cos x.

(2) = d(cos x)/dx = - sin x.

(3) = d(tan x)/dx = sec²x.

(4) = d(cot x)/dx = - cosec²x.

(5) = d(sec x)/dx = sec x tan x.

(6) = d(cosec x)/dx = - cosec x cot x.

Answer 2

Step-by-step explanation:

$\longrightarrow \: \tt \: x = sin \: θ, y = { sin }^{3} θ at \: θ  = \frac{\pi}{2} \\ \longrightarrow \tt \: \therefore \: y = {x}^{3} \\ \longrightarrow \therefore \: \tt \frac{dy}{dx} = 3 \frac{d}{dx} ( {x}^{3} ) \\ \longrightarrow = \tt 3 \times 2x \\ \longrightarrow = \tt \: 6x \\ \\\longrightarrow \tt If \: θ = \frac{\pi}{2} ,then \: x = sin \frac{\pi}{2} = 1 \\ \longrightarrow \therefore \: \tt \: ( \frac{ {d}^{2}y }{d {x}^{2} } ) \: at \: θ = \frac{\pi}{2} \\ \longrightarrow \tt \: ( \frac{ {d}^{2}y }{d {x}^{2} } ) \: at \: x = 1 \\ \longrightarrow \: \tt 6(1) \\ \longrightarrow \: \tt \: \blue{6}$