Question

How do you find the slope of the tangent line to the graph of the function `f(t)=3t-t^2` at (0,0)?

Answer 1

`y=3x`

Explanation:

To find tangent at a point on curve `y=f(x)`, where `x=x_0`, we have the point on the curve, which is `(x_0,f(x_0))` and the slope of the tangent at that point, which is given by value of `(df)/(dx)` at `x=x_0`.

The difference here is thatv we have `f(t)=3t-t^2`, i.e. `f(t)` a function of `t` in place of `x`. Note that `f(0)=0` and hence `(0,0)` is on te curve.

The slope of line `f(t)=3t-t^2` is given by differential of `f(t)` at that point. As `(df)/(dt)=3-2t` and at `t=0`, `(dF)/(dt)=3`

Hence, the equation of line is `(y-0)=3(t-0)` or `y=3t`.

graph{(3x-x^2-y)(y-3x)=0 [-5, 5, -2.32, 2.68]}