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प्रश्न
Find `(d^2y)/(dx^2)` of the following : x = sinθ, y = sin3θ at θ = `pi/(2)`
उत्तर
x = sinθ, y = sin3θ
Differentiating x and y w.r.t. θ, we get
`"dx"/"dθ" = "d"/"dθ"(sinθ)` = cosθ ...(1)
and
`"dy"/"dθ" = "d"/"dθ"(sinθ)^3`
= `3(sinθ)^2."d"/"dθ"(sinθ)`
= 3sin2θ.cosθ
∴ `"dy"/"dx" = (("dy"/"dθ"))/(("dx"/"dθ"))`
= `(3sin^2θcosθ)/"cosθ"`
= 3sin2θ
and
`(d^2y)/(dx^2) = 3"d"/"dx"(sinθ)^2`
= `3"d"/"dθ"(sinθ)^2 xx "dθ"/"dx"`
= `3 xx 2sinθ"d"/"dθ"(sinθ) xx (1)/(("dx"/"dθ")`
= `6sinθ.cosθ xx (1)/"cosθ"` ...[By (1)]
= 6sinθ
∴ `((d^2y)/(dx^2))_("at" θ = pi/(2)`
= `6sin pi/(2)`
= 6 x 1
= 6.
Alternative Method :
x = sinθ, y = sin3θ
∴ y = x3
∴ `"dy"/"dx" = "d"/"dx"(x^3)` = 3x2
∴ `(d^2y)/(dx^2) = 3"d"/"dx"(x^2)`
= 3 x 2x
= 6x
If θ = `pi/(2), "then" x = sin pi/(2)` = 1
∴ `((d^2y)/(dx^2))_("at" θ = pi/(2)`
= `((d^2y)/(dx^2))_("at" x = 1)`
= 6(1)
= 6.
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