Question

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

Short answer

The curve is concave upward for \(t > -1/3\) and concave downward for \(t < -1/3\)

Step-by-step solution

Find the first derivative

Differentiate the given function to find the first derivative. The derivative \(y'(t)\) is given by \(2t + 3t^{2}\)

Find the second derivative

Differentiate \(y'(t)\) to find the second derivative, \(y''(t)\). On differentiating, we find \(y''(t)\) to be \(2 + 6t\)

Determine intervals where second derivative is positive and negative

By setting \(y''(t)=0\), we can find the critical points. Solving \(2 + 6t = 0\), we get \( t = -1/3\). \(y''(t)\) is positive for \(t > -1/3\) and negative for \(t < -1/3\)

Interpret the results

The curve is concave upward when the second derivative is positive, and concave downward when the second derivative is negative. Thus, the curve is concave upward when \(t > -1/3\) and concave downward when \(t < -1/3\)