Question

Determine the \(t\) intervals on which the curve is concave downward or concave upward. $$ x=t^{2}, \quad y=t^{3}-t $$

Short answer

The curve is concave upward on \( (-\infty, 0) \cup (0,\infty) \). There is no region where the curve is concave downward.

Step-by-step solution

Find \(y'\)

We know that \(x = t^2\) and \(y = t^3 - t\). So, we can find \(y'\) with respect to \(t\): \(dy/dt = 3t^2 -1\), and \(dx/dt = 2t\). Therefore, \(y' = dy/dt/dx/dt = (3t^2 -1)/(2t)\).

Find \(y''\)

To find \(y''\), we need to take the derivative of \(y'\) using quotient rule: \((y'') = (d/dt(y')/(dx/dt) = ((6t^2)(2t)-(2(3t^2-1)))/(2t)^2 = 2t^2/(2t)^2\).

Determine intervals of concavity

Concavity is determined by the sign of the second derivative. If \(y'' > 0\), the curve is concave upwards. If \(y'' < 0\), the curve is concave downwards. Looking at \(y'' = 2t^2/(2t)^2\), we see that \(y'' > 0\) when \( t != 0\), and \(y''\) is undefined when \( t = 0\). Therefore, the curve is concave upward on \( (-\infty, 0) \cup (0,\infty) \) and there's no region where the curve is concave downward.