Question

If x = a sin t - b cos t, y = a cos t+ b sin t, then `"y"^3 ("d"^2"y")/"dx"^2 + x^2 + y^2` = ?

पर्याय

• 2

• 1

• 0

• - 1

Answer 1

0

Explanation:

We have,

x = a sin t - b cos t    ...(i)

y = a cos t + b sin t   ....(ii)

Squaring and adding Eqs. (i) and (ii), we get

x² + y² = a² + b²

On differentiating w.r.t. 'x', we get

2x + 2y = `"dy"/"dx"` = 0

`"dy"/"dx" = (- x)/"y"`

Again differentiating w.r.t. 'x', we get

`("d"^2y)/"dx"^2 = - (("y" - x "dy"//"dx")/("y"^2))   ...[because "dy"/"dx" = - x/y]`

`=> "y"^2 ("d"^2"y")/"dx"^2 = - ("y" + x/y)`

`=> "y"^3 ("d"^2"y")/"dx"^2 + x^2 + "y"^2` = 0