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प्रश्न
If x = at2 and y = 2at, then show that `xy(d^2y)/(dx^2) + a` = 0.
उत्तर
x = at2 and y = 2at ...(1)
Differentiating x and y w.r.t. t, we get
`(dx)/(dt) = (d)/(dt)(at^2)`
= `a d/(dt)(t^2)`
= a × 2t
= 2at ...(2)
and
`(dy)/(dt) = d/(dt)(2at)`
= `2a d/(dt)(t)`
= 2a × 1
= 2a
∴ `(dy)/(dx) = ((dy/dt))/((dx/dt)`
= `(2a)/(2at) = (1)/t`
and
`(d^2y)/(dx^2) = d/(dx)(1/t)`
= `d/(dt)(t^-1).(dt)/(dx)`
= `(-1)t^-2.(1)/((dx/dt)`
= `(-1)/t^2 xx (1)/(2at)` ...[By (2)]
= `(-1)/(2at^3)`
∴ `(d^2y)/(dx^2) = (-1)/(2at^3)`
= `(-1)/(yt^2) = (-1)/(y xx x/a)` ...[Using (1)]
∴ `xy(d^2y)/(dx^2)` = – a
∴ `xy(d^2y)/(dx^2) + a` = 0.
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