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If x=f(t), y=g(t) are differentiable fun...

If `x=f(t), y=g(t)` are differentiable functions of parameter 't' then prove that y is a differentiable function of 'x' and
`(dy)/(dx)=((dy)/(dt))/((dx)/(dt)),(dx)/(dt) ne 0`
Hence find `(dy)/(dx)` if `x=a cos t, y= a sin t.`

Text Solution

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Let `delta t ` be the increment in t.
` therefore ` The corresponding increments in x and y are are ` delta x `
and ` deltay` respectively.
Consider , `(deltay)/(deltax) `
`therefore (deltay)/(deltax) = ((deltay)/(deltat))/((deltax)/(deltat)) , delta t ne 0 `
Taking limit as ` deltat to 0 ` , on both sides ,
`underset(delta t to0) (lim)((deltay)/(delx)) = underset(delta t to 0)lim ((deltay//deltat)/(deltax//deltat)) `
` = (underset(deltat to 0 ) lim (deltay)/(deltat))/(underset(deltat to 0 ) lim (deltax)/(deltat))` ...(i)
Since y and x are differentiable functions of t, ` underset(delta t to 0) lim(deltay)/(deltat)`
and ` underset(delta t to 0) lim (deltax)/(deltat)` exist, and `underset(delta t to 0) lim(deltay)/(deltat)= dy/dt and underset(delta t to 0) lim(deltax)/(deltat)=dx/dt`.
We further assume that ` dx/dt ne 0 `.
` therefore ` Equation (i) becomes , `underset(deltat to0)lim (deltay)/(deltax)= (deltay//deltat)/(dx//dt)`
` therefore ` R.H.S. of equation (i) exists .
As ` delta t to 0 , deltax to 0 ` .
` therefore ` L.H.S of equation (i) also exists.
` therefore y ` is differentiable function of x .
`underset(deltat to0)lim (deltay)/(deltax)= dy/dx= (((dy)/(dt)))/(((dx)/(dt))) , dx/dt ne 0 `
Now , x ` = a cos^(2) t and y = a sin^(2) t `
Differentiating x and y w.r.t.t,
` therefore dx/dt = 2a cos t. (-sin t)`
` = - 2a sin t . cos t `
and ` (dy)/(dt) = 2a sin t . cos t` .
`therefore dy/dt = (dy/dt)/(dx/dt) `
` = (2a sin t, cos t)/(- 2a sin t . cos t ) = - 1 `
` therefore (dy)/(dx) = - 1` .
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