Question

What is the slope of `f(t) = (t^2-t,t+3)` at `t =1`?

Answer 1

The slope of `f(t)` is `1` when `t=1`

Explanation:

Let `f(t) = {(x(t)=t^2-t,=> dx/dt=2t-1),(y(t)=t+3,=>dy/dt=1) :}`

Then By the chain rule; `dy/dx=dy/dt.dt/dx`

Alternatively, `dy/dx=(dy/dt)/(dx/dt)`
`:. dy/dx = (1)/(2t-1)`

When `t=1 => dy/dx = (1)/(2-1) = 1`

Hence, the slope of `f(t)` is `1` when `t=1`