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Question
If x= a cos θ, y = b sin θ, show that `a^2[y(d^2y)/(dx^2) + (dy/dx)^2] + b^2` = 0.
Solution
x= a cos θ, y = b sin θ
Differentiating x and y w.r.t. θ, we get
`"dx"/"dθ" = a"d"/"dθ"(cosθ)` = a(– sinθ) = – sinθ ...(1)
and
`"dy"/"dθ" = (("dy"/"dθ"))/(("dx"/"dθ")`
= `(bcosθ)/(-asinθ)`
= `(-b/a)cotθ`
∴ `(d^2y)/(dx^2) = "d"/"dx"[(-b/a)cotθ]`
= `(-b/a)."d"/"dθ"(cotθ)."dθ"/"dx"`
= `(-b/a).(-"cosec"^2θ) xx (1)/(("dx"/"dθ")`
= `(-b/a)"cosec"^2θ xx (1)/(-asinθ)` ...[By (1)]
= `(-b/a^2)"cosec"^3θ`
∴ `a^2[y(d^2y)/(dx^2) + (dy/dx)^2] + b^2`
= `a^2[bsinθ.(-b/a^2)"cosec"^3θ + {(-b/a)cotθ}^2] + b^2`
= `a^2[-b^2/a^2"cosec"^2θ + b^2/a^2cot^2θ] + b^2`
= `a^2(-b^2/a^2)("cosec"^2θ - cot^2θ) + b^2`
= – b2 + b2 ...[∵ cosec2θ – cot2θ = 1]
∴ `a^2[y(d^2y)/(dx^2) + (dy/dx)^2] + b^2` = 0.
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