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 (cost + t sin t) y = a (sin t – tcost)

Now, 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

and y = a (sin t - t cos t)

On differentiating with respect to t,

`dy/dx = 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`