मराठी

Let x(t) = 22costsin2t and y(t) = π22sintsin2t,t∈(0,π2).Then 1+(dydx)2d2ydx2 at t = ππ4 is equal to ______. -

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प्रश्न

Let x(t) = `2sqrt(2) cost sqrt(sin2t)` and y(t) = `2sqrt(2) sint sqrt(sin2t), t ∈ (0, π/2)`. Then `(1 + (dy/dx)^2)/((d^2y)/(dx^2)` at t = `π/4` is equal to ______.

पर्याय

  • `(-2sqrt(2))/3`

  • `2/3`

  • `1/3`

  • `(-2)/3`

MCQ
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उत्तर

Let x(t) = `2sqrt(2) cost sqrt(sin2t)` and y(t) = `2sqrt(2) sint sqrt(sin2t), t ∈ (0, π/2)`. Then `(1 + (dy/dx)^2)/((d^2y)/(dx^2)` at t = `π/4` is equal to `underlinebb((-2)/3)`.

Explanation:

x = `2sqrt(2) cost sqrt(sin2t)` 

Differentiate with respect to t

`(dx)/(dt) = (2sqrt(2)cos3t)/sqrt(sin2t)`

y = `2sqrt(2) sint sqrt(sin2t)`

Differentiate with respect to t

`(dy)/(dt) = (2sqrt(2)sin3t)/sqrt(sin2t)`

`\implies (dy)/(dx) = (dy//dt)/(dx//dt)` = tan 3t

`(dy)/(dx)` = –1 at t = `π/4`

`\implies (d^2y)/(dx^2) = 3/(2sqrt(2)) sec^3 3t.sqrt(sin 2t)` = –3 at t = `π/4`

∴ `(1 + (dy/dx)^2)/((d^2y)/(dx^2)) = (1 + 1)/(-3) = - 2/3`

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