Advertisements
Advertisements
प्रश्न
Find `(d^2y)/(dx^2)` of the following : x = a(θ – sin θ), y = a(1 – cos θ)
बेरीज
Advertisements
उत्तर
x = a(θ – sin θ), y = a(1 – cos θ)
Differentiating x and y w.r.t. θ, we get
`"dx"/"dθ" = a"d"/"dθ"(θ - sin θ)`
= a(1 – cos θ) ...(1)
and
`"dy"/"dθ" = a"d"/"dθ"(1 - cos θ)`
= a[0 – (– sin θ)]
= a sin θ
∴ `"dy"/"dx" = (("dy"/"dθ"))/(("dx"/"dθ")`
= `"a sin θ"/"a(1 - cos θ)"`
= `(2sin(θ/2).cos(θ/2))/(2sin^2(θ/2)) = cot(θ/2)`
and
`(d^2y)/(dx^2) = "d"/"dx"[cot(θ/2)]`
= `"d"/"dx"[cot(θ/2)].("d"θ/2)/"dx"]`
= `-"cosec"^2(θ/2)."d"/"dθ"(θ/2) xx (1)/(("dx"/"dθ")`
= `-"cosec"^2(θ/2) xx (1)/(2) xx (1)/(a(1 - cosθ)` ...[by (1)]
= `-(1)/(2a)"cosec"^2(θ/2) xx (1)/(2sin^2(θ/2)`
=`-(1)/(4a)."cosec"^4(θ/2)`.
shaalaa.com
या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
