Question

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

Short answer

The curve cannot be classified as concave upward or concave downward for any interval of \(t\), because the concavity function \(K(t)\) equals \(0\) for all \(t\).

Step-by-step solution

Compute the first derivatives

Differentiation of the given functions for \(x\) and \(y\) with respect to \(t\) will give us first-order derivatives: \(x'(t) = 2 + \frac{1}{t}\), \(y'(t) = 2 - \frac{1}{t}\)

Compute the second derivatives

Differentiation of the first-order derivatives will provide the second-order derivatives: \(x''(t) = -\frac{1}{t^2}\), \(y''(t) = \frac{1}{t^2}\)

The Concavity function

The concavity \(K(t)\) can be computed based on \(K(t) = \frac{x'(t)y''(t) - y'(t)x''(t)}{(x'(t)^2 + y'(t)^2)^\frac{3}{2}}\). After substituting the values, we will find that \(K(t)\) equals \(0\) for all \(t\). So, it can't be classified as concave upward or concave downward.