Question

If x = f(t) and y = g(t) are differentiable functions of t so that y is a differentiable function of x and `(dx)/(dt)` ≠ 0 then `(dy)/(dx) = ((dy)/(dt))/((dx)/(d"))`.
Hence find `(dy)/(dx)` if x = sin t and y = cost

Answer 1

x and y are differentiable functions of t.

Let there be a small increment δt in the value of t.

Correspondingly, there should be small increments δx, δy in the values of x and y, respectively.

As δt → 0, δx → 0, δy → 0

Consider, `(deltay)/(deltax) = ((deltay)/(deltat))/((deltax)/(deltat)), (deltax)/(deltat)` ≠ 0
Taking `lim_(deltat -> 0)` on both sides, we get

`lim_(deltat -> 0) (deltay)/(deltax) = (lim_(deltat -> 0)(deltay)/(deltat))/(lim_(deltat -> 0) (deltax)/(deltat))`

Since x and y are differentiable functions of t, `lim_(deltat -> 0) (deltay)/(deltat) = (dy)/(dt)` exists and is finite.

Also, `lim_(deltat -> 0) (deltax)/(deltat) = (dx)/(dt)` exists and is finite.

∴ `lim_(deltat -> 0) (deltay)/(deltax) = (((dy)/(dt))/((dx)/(dt)))`

As δt → 0, δx → 0

∴ `lim_(deltat -> 0) (deltay)/(deltax) = (((dy)/(dt))/((dx)/(dt)))`   .......(i)

Here, R.H.S. of (i) exist and are finite.

Hence, limits on L.H.S. of (i) also should exist and be finite.

∴ `lim_(deltat -> 0) (deltay)/(deltax) = (dy)/(dx)` exists and is finite.

∴ `(dy)/(dx) = (((dy)/(dt))/((dx)/(dt))), (dx)/(dt)` ≠ 0

Now, x = sin t and y = cos t

∴ `(dx)/(dt)` = cos t and `(dy)/(dt)` = –sin t

∴ `(dy)/(dx) = ((dy)/(dt))/((dx)/(dt)) = (-sin t)/cos t` = – tan t