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)`.

Answer 1

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`.