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Question
If x = a sin t – b cos t, y = a cos t + b sin t, show that `(d^2y)/(dx^2) = -(x^2 + y^2)/(y^3)`.
Solution
x = a sin t – b cos t, y = a cos t + b sin t
Differentiating x and y w.r.t. t, we get
`"dx"/"dt" = a"d"/"dx" (sint) - b"d"/"dt"(cost)`
= a cos t – b( – sin t)
= a cos t + b sin t
and
`"dy"/"dt" = a"d"/"dx"(cos t) - b"d"/"dt"(sint)`
= a(– sin t) + b cos t
= – a sin t + b cos t
∴ `"dy"/"dx" = (("dy"/"dt"))/(("dx"/"dt")`
= `(-asint + bcost)/(acost + bsint)`
= `-((asint - bcost)/(acost + bsint))`
∴ `"dy"/"dx" = -x/y` ...(1)
∴ `(d^2y)/(dx^2) = -"d"/"dx"(x/y)`
= `-[(y"d"/"dx"(x) - x"dy"/"dx")/y^2]`
= `-[(y xx 1 - x(-x/y))/y^2]` ...[By (1)]
= `-[(y^2 + x^2)/y^3]`
∴ `(d^2y)/(dx^2) = -(x^2 + y^2)/y^3`.
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