Question

For the parametric equations x = t^2 1 y = t^2 t for which values of t is the curve concave upward?

Answer 1

I assume you have

`\begin{cases} x(t) = t^2 + 1 \\ y(t) = t^2 + t \end{cases}`

Compute the derivatives
`(dy)/(dx)` and
`(d^2y)/(dx^2)` using the chain rule.

`(dy)/(dx) = (dy)/(dt) \cdot (dt)/(dx) = ((dy)/(dt))/((dx)/(dt))`

`(dx)/(dt) = 2t`

`(dy)/(dt) = 2t + 1`

`\implies (dy)/(dx) = (2t + 1)/(2t) = 1 + \frac1{2t}`

For the second derivative, we have by the chain rule

`(d^2y)/(dx^2) = (d)/(dx)\left[(dy)/(dx)\right] = (d)/(dt)\left[(dy)/(dx)\right] \cdot (dt)/(dx) = \frac{\frac d{dt}\left[(dy)/(dx)\right]}{(dx)/(dt)}`

`(d)/(dt)\left[(dy)/(dx)\right] = -\frac1{2t^2}`

`\implies (d^2y)/(dx^2) = \frac{-\frac1{2t^2}}{2t} = -\frac1{4t^3}`

The curve is concave upward when the second derivative has positive sign. This happens for

`-\frac1{4t^3} > 0 \implies \frac1{4t^3} < 0 \implies \boxed{t<0}`