Question

Find `(d^2y)/(dx^2)` of the following : x = sinθ, y = sin³θ at θ = `pi/(2)`

Answer 1

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.