Question

The curve given by the parametric equations `x=16 - t^2`, `y= t^3 - 1 t` is symmetric about the x-axis. At which x value is the tangent to this curve horizontal?

Answer 1

`x=15 2/3`

Explanation:

Slope of the tangent at a point on the cuurve, is the value of first derivative at that point.

Hence the tangent will be hrizontal when first derivative `(dy)/(dx)` is zero.

`(dy)/(dx)` of a parametric equation is given by `((dy)/(dt))/((dx)/(dt))`

as `(dy)/(dt)=3t^2-1` and `(dx)/(dt)=-2t`

`(dy)/(dx)=-(3t^2-1)/(2t)` and this is zero when `3t^2=1` ad `t!=0`

i.e. `t=+-1/sqrt3`

i.e. `x=16-1/3=47/3=15 2/3`

Note that for same value of `x`, there are two values of `y`, where tangent will be hrizontal, depending on sign of `t`.
graph{(16-x)(15-x)^2-y^2=0 [1.34, 21.34, -5.12, 4.88]}