Question

What is the instantaneous velocity of an object moving in accordance to `f(t)= (t^3/(t-2),1/t)` at `t=1`?

Answer 1

The instantaneous velocity of the object is 1/4 (distance unit)/(time unit)

Explanation:

Define `x(t), y(t)` as follows

`{ (x(t)=t^3/(t-2)), (y(t)=1/t) :} => f(t) = (x(t), y(t))`

Differentiating `x(t)` wrt `t` (using quotient rule) we get:

`dot x(t) = dx/dt = {(t-2)(d/dtt^3) - (t^3)(d/dt(t-2))} / (t-2)^2`
`:. dx/dt = {(t-2)(3t^2) - (t^3)(1)} / (t-2)^2`
`:. dx/dt = {3t^3-6t^2 - t^3} / (t-2)^2`
`:. dx/dt = (2t^3-6t^2) / (t-2)^2`

Differentiating `y(t)` wrt `t` we get:

`dot y(t) = dy/dt=-1/t^2`

Then the velocity of `f(t)` is given by applying the chain rule;
`f'(t) = dy/dx = dy/dt*dt/dx = (dy/dt)/(dx/dt)`
`:. f'(t) = ( -1/t^2 ) / ( (2t^3-6t^2) / (t-2)^2 )`
`:. f'(t) = -1/t^2 * (t-2)^2/ (2t^3-6t^2)`

So when `t=1` we get:
`:. f'(1) = -1/1 * (-1)^2/ (2-6) = 1/4`

Hence, The instantaneous velocity of the object is 1/4 (distance unit)/(time unit)