Question

If x = a (cos t + t sin t) and y = a (sin t – t cos t), find `(d^2y)/dx^2`.

Answer 1

Here, x = a (cos t + t sin t)

On differentiating with respect to t,

`dx/dt` = a (−sin t + t · cos t + sin t)

= at cos t

y = a (sin t − t cos t)

On differentiating with respect to t,

`dy/dt` = a [cos t − {t (−sin t) + cos t}]

= a {cos t + t sin t − cos t}

= at sin t

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

= `(at sin t)/(at cos t)`

= tan t

Differentiating again with respect to x,

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

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

= `d/dt (tan t) xx dt/dx`

= `sec^2 t xx 1/(at cos t)   ...[because dx/dt = at cos t]`

= `1/at sec^3 t`

∴ `(d^2y)/dx^2 = (sec^3 t)/(at), 0 <t <pi/2`