Question

If x = 3 sin t − sin 3t, y = 3 cos t − cos 3t find `(dy)/(dx)` and prove that `(d^2y)/(dx^2) = (-cosec^3 2t * cosec t)/3`.

Answer 1

Given:

x = 3 sin t − sin 3t, y = 3 cos t − cos 3t

Differentiate with respect to t:

`(dx)/(dt) = 3 cos t - 3 cos 3t`

`(dy)(dt) = -3 sin t + 3 sin 3t`

So,

`(dy)/(dx) = ((dy)/(dt))/((dx)/(dt))`

= `(-3 sin t + 3 sin 3t)/(3 cos t - 3 cos 3t)`

= `(sin 3t - sint)/(cos t - cos 3t)`

Using identities:

sin 3t − sin t = 2 cos 2t sin t

cos t − cos 3t = 2 sin 2t sin t

Thus,

`(dy)/(dx) = (2 cos 2t sin t)/(2 sin 2t sin t)`

`(dy)/(dx) = cot 2t`

Now,

`(d^2y)/(dx^2) = d/(dx) ((dy)/(dx))`

= `(d/(dt) (cot 2t))/((dx)/(dt))`

Here,

`d/(dt) (cot 2t) = - 2 csc^2 2t`

`(dx)/(dt) = 3(cos t - cos 3t)`

= 3(2 sin 2t sin t)

= 6 sin ⁡2t sin ⁡t

Therefore,

`(d^2y)/(dx^2) = (- 2 csc^2 2t)/(6 sin ⁡2t sin ⁡t)`

= `- 1/3 * 1/(sin^3 2t) * 1/(sin t)`

`(d^2y)/(dx^2) = - (csc^2 2t * csc t)/3`

Hence proved.