Question

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

Answer 1

`f'(x)=3`

Explanation:

So we have `(t+1,t^3)`
This really means that:
`x=t+1` and `y=t^3`

We find `t` in terms of `x`.
`x=t+1`
`=>t=x-1`

We now plug this value in our function `y=t^3`.
`y=(x-1)^3` We simplify this.
`y=(x-1)^3=>y=x^3-3x^2+3x-1`
We now find the derivative of this using the power rule.
`f'(x)=3x^2-6x+3`

Now, when `t=1`, we see that:
`x=t+1`
=>`x=2`
We plug this into our derivative:
`f'(x)=3(2)^2-6(2)+3`
=>`f'(x)=12-12+3`
`f'(x)=12-12+3`
`f'(x)=3` That is our answer!

Answer 2

`"slope "=3`

Explanation:

`•color(white)(x)m_(color(red)"tangent")=dy/dx" at t = 1"`

`color(red)(bar(ul(|color(white)(2/2)color(black)(dy/dx=(dy)/(dt)xx(dt)/(dx))color(white)(2/2)|)))`

`x=t+1rArrdx/dt=1rArrdt/dx=1`

`y=t^3rArrdy/dt=3t^2`

`rArrdy/dx=3t^2`

`t=1tody/dx=3`