Question

If x = f(t) and y = g(t) are differentiable functions of t, so that y is function of x and `(dx)/dt ≠ 0` then prove that `dy/(dx) = (dy/dt)/((dx)/dt)`. Hence find `dy/(dx)`, if x = at², y = 2at.

Answer 1

Given x = f(t) and y = g(t) are differentiable functions of t.

Let δx and δy be the small increments in x and y respectively corresponding to the increment δt in t.

Consider the incrementary ratio `(delta y)/(delta x)` and note that δx → 0 ⇒ δt → 0.

Consider `(delta y)/(delta x) = ((delta y)/(delta t))/((delta x)/(delta t))`, since `(delta x)/(delta t) ne 0`

Taking the limit as δt → 0 on both sides, we get,

`lim _(delta t -> 0) ((delta y)/(delta x)) = lim_(delta t -> 0) (((delta y)/(delta t))/((delta x)/(delta t)))`

As δt → 0, δx → 0

∴ `lim_(delta x -> 0) ((delta y)/(delta x)) = (lim _ (delta t -> 0)(((delta y)/(delta t))/((delta x)/(delta t))))/(lim _ (delta t -> 0)(((delta x)/(delta t))/((delta x)/(delta t))))`    ...(i)

Since x and y are differentiable functions of t, we have,

`lim_(delta t -> 0) (delta x)/(delta t) = dx/dt` and

`lim_(delta t -> 0) (delta y)/(delta t) = dy/dt`    ...(ii)

exist and are finite

From (i) and (ii),

`lim_(delta x -> 0) (delta y)/(delta x) = (dy/dt)/(dx/dt)`    ...(iii)

The R.H.S. of (iii) exists and finite implies L.H.S. of (iii) also exists and finite.

∴ `lim_(delta x -> 0) ((delta y)/(delta x)) = dy/dx`

Thus the equation (iii) becomes,

`dy/dx = (dy/dt)/(dx/dt), dx/dt ne 0`

Now, x = at² and y = 2at

⇒ `dx/dt = 2at` and

`dy/dt = 2a`

∴ `dy/dx = (dy/dt)/(dx/dt) = (2 a)/(2 at) = 1/t`