Question

If x = f(t) and y = g(t) are differentiable functions of t, then prove that:

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

Hence, find `dy/dx` if x = a cot θ, y = b cosec θ.

Answer 1

Given: x = f(t) and y = g(t)

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

Since x and y are differentiable functions of t,

`dx/dt = lim_(δt→0) (δx)/(δt)` and `dy/dt = lim_(δt -> 0) (δy)/(δt)`  ...(1)

Also, as δt `→` 0, δx `→` 0  ...(2)

Now, `(δy)/(δx) = ((δy//δt))/((δx//δt))`    ...[δ ≠ 0]

Taking limits as δt `→` 0, we get

`lim_(δt → 0) (δy)/(δx) = lim_(δt → 0) ((δy//δt))/((δx//δt))`

`lim_(δx → 0) (δy)/(δx) = (lim_(δt → 0)  (δy//δt))/(lim_(δt -> 0)  (δx//δt)) = ((dy//dt))/((dy//dt))`      ...[By (1) and (2)]

∵ The limits in RHS exist

∴ `lim_(δx -> 0) (δy)/(δx)` exists and is equal to `dy/dx`

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

To find `bb(dy/dx)` if x = a cot θ, y = b cosec θ:

x = a cot θ, y = b cosec θ

Differentiating x and y w.r.t.θ, we get

`dx/(dθ) = a d/(dθ) (cot θ)` = a(– cosec²θ) = – a cosec²θ

and `dy/(dθ) = b d/(dθ)("cosec"  θ)` = b(– cosec θ cot θ)

= – b cosec θ cot θ

∴ `dy/dx = ((dy//dθ))/((dx//dθ))`

= `(- b  "cosec"  θ cot θ)/(- a  "cosec"^2θ)`

= `b/a.(cot θ)/("cosec"  θ)`

= `b/a xx cosθ/sinθ xx sinθ`

= `(b/a) cos θ`.