Question

Determine the \(t\) intervals on which the curve is concave downward or concave upward. $$ x=2 \cos t, \quad y=\sin t, \quad 0

Short answer

Detailed computations of the second derivative and determining the sign are required to find the intervals of concavity, which can be done following the steps outlined.

Step-by-step solution

Find the Derivatives

First, we need to find \(\frac{dy}{dt}\) and \(\frac{dx}{dt}\). Given:\(x=2 \cos t\) \(y=\sin t\)Differentiating both sides with respect to \(t\), we get:\(\frac{dx}{dt} =-2 \sin t\)\(\frac{dy}{dt} = \cos t\)

Find First Derivative of y with Respect to x

Next, find the derivative of \(y\) with respect to \(x\) (\(\frac{dy}{dx}\)) by dividing \(\frac{dy}{dt}\) by \(\frac{dx}{dt}\):\(\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = - \frac{\cos t}{2 \sin t}\)

Find Second Derivative (\(y\) with Respect to \(x\))

The second derivative (\(\frac{d^2y}{dx^2}\)) is given by \(\frac{d}{dt} (\frac{dy}{dx}) / \frac{dx}{dt}\) First, find \(\frac{d}{dt}(\frac{dy}{dx}): -\frac{d}{dt}(\frac{\cos t}{2 \sin t})\)Then, divide that by \( \frac{dx}{dt}\):\(\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}(\frac{\cos t}{2 \sin t})}{ \frac{dx}{dt} }\)

Determine Intervals of Concavity

Finally, determine the values of \(t\) where \(\frac{d^2y}{dx^2}\) is positive or negative. Recall that the curve is concave upward at values of \(t\) where \(\frac{d^2y}{dx^2}\) is positive, and concave downward where it is negative.