Question

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

Short answer

The curve is concave downward for all \(t > 0\).

Step-by-step solution

Find first derivative

Start with the parametric definitions \(x = t^2\) and \(y = \ln t\). To get dy/dx, we need to find dx/dt and dy/dt first. The derivative of \(x\) with respect to \(t\) is \(dx/dt = 2t\) while the derivative of \(y\) with respect to \(t\) is \(dy/dt = 1/t\). Divide \(dy/dt\) by \(dx/dt\) to get: \[dy/dx = \frac{1/t}{2t} = \frac{1}{2t^2}\].

Find second derivative

The second derivative \(d^2y/dx^2\) is given by differentiating dy/dx with respect to 't' and then dividing by \(dx/dt = 2t\). The derivative of \(\frac{1}{2t^2}\) with respect to 't' is \(-\frac{1}{t^3}\). Therefore, \[d^2y/dx^2 = \frac{-1/t^3}{2t} = -\frac{1}{2t^4}\].

Determine the intervals of concavity

The curve is concave upward wherever the second derivative \(d^2y/dx^2 = -1/2t^4 > 0\), and concave downward wherever the second derivative is less than zero. But here, the second derivative is always negative for all real positive t (excluding 0), thus the curve is concave downward for all \(t > 0\).