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If x = a secθ, y = b tanθ, then d2y/dx2 at θ = π/6 is :(a) \(\frac{-3\sqrt 3b}{a^2}\)(b) \(\frac{-2\sqrt 3b}{a}\)(c) \(\frac{-3\sqrt 3b}{a}\)(d) \(\frac{-b}{3\sqrt 3a^2}\) |
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Answer» Option : (a) x = a secθ ⇒ \(\frac{dx}{d\theta}\) = a tanθ secθ y = b tanθ ⇒ \(\frac{dy}{d\theta}\) = b sec2θ ∴ \(\frac{dy}{dx}\) = \(\frac{b}{a}\) cosecθ ⇒ \(\frac{d^2y}{dx^2}\) = \(\frac{-b}{a}\)cosecθ.cotθ.\(\frac{d\theta}{dx}\) = \(\frac{-b}{a^2}\) cot3θ ∴ \(\frac{d^2y}{dx^2}\)]θ = \(\frac{\pi}{6}\) = \(\frac{-3\sqrt 3b}{a^2}\) |
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