Question

Parametric equations for a curve are given. Find \(\frac{d^{2} y}{d x^{2}},\) then determine the intervals on which the graph of the curve is concave up/down. \(x=\cos t, \quad y=\sin (2 t)\) on \([0,2 \pi]\)

Short answer

The curve has specific intervals of concavity on \([0, 2\pi]\) based on the sign of \(\frac{d^2y}{dx^2}\).

Step-by-step solution

Find the First Derivatives

First, find the derivatives of both parametric equations with respect to the parameter \(t\). For \(x = \cos t\), the derivative is \(\frac{dx}{dt} = -\sin t\). For \(y = \sin(2t)\), the derivative is \(\frac{dy}{dt} = 2\cos(2t)\).

Calculate \(\frac{dy}{dx}\)

Use the formula \(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\) to find \(\frac{dy}{dx}\). Substituting the derivatives from Step 1 gives \(\frac{dy}{dx} = \frac{2\cos(2t)}{-\sin t} = -\frac{2\cos(2t)}{\sin t}\).

Find the Second Derivative \(\frac{d^2y}{dx^2}\)

To find \(\frac{d^2y}{dx^2}\), we first need to differentiate \(\frac{dy}{dx}\) with respect to \(t\) and then divide by \(\frac{dx}{dt}\). The derivative of \(\frac{dy}{dx} = -\frac{2\cos(2t)}{\sin t}\) with respect to \(t\) is calculated using the quotient rule. This results in \(\frac{d}{dt}\left(-\frac{2\cos(2t)}{\sin t}\right) = \frac{2(2\sin(2t)\sin t + \cos(2t)\cos t)}{\sin^2 t}\). Simplify to get \(\frac{d^2y}{dx^2} = \frac{4\sin(t)}{\sin^3(t)} + \frac{2\cos(2t)\cos(t)}{\sin^3(t)}\). Finally, divide by \(\frac{dx}{dt} = -\sin t\) to obtain: \(\frac{d^2y}{dx^2} = \frac{4\sin(t) + 2\cos(2t)\cos(t)}{-\sin^4(t)}\).

Determine Concavity Intervals

The graph of the function is concave up where \(\frac{d^2y}{dx^2} > 0\) and concave down where \(\frac{d^2y}{dx^2} < 0\). Set \(\frac{d^2y}{dx^2} = 0\) and solve for \(t\) to find critical points. Analyzing the sign of \(\frac{d^2y}{dx^2}\) between these points on the interval \([0, 2\pi]\) will determine where the curve is concave up or down. Generally break the intervals at \(t = 0, \pi, \) and \(2\pi\) due to periodicity and symmetry.

Key concepts

Calculus

In parametric equations, calculus allows us to explore the behavior of curves beyond the standard Cartesian coordinates. By considering a curve defined by parametric equations, such as

• \(x = \cos t\)
• \(y = \sin (2t)\)

we can investigate its properties over an interval like \[0, 2\pi\]. Calculus provides the tools to find derivatives that describe how one variable changes concerning another, crucial for understanding the curve's behavior.

The process begins with finding the first derivatives of these parameterized expressions with respect to the parameter \(t\). This step is vital because it lays the groundwork for subsequent calculations, such as finding rates of change (slopes) and concavity (curvature) of the curve, helping us understand the shape and nature of the path traced by the curve.

Concavity

Concavity describes how the curve bends or curves outward and is linked with the second derivative of the function.

For parametric equations like

• \(x = \cos t\)
• \(y = \sin (2t)\)

we determine concavity by assessing the second derivative, \(\frac{d^2y}{dx^2}\).

A curve is said to be concave up where the second derivative is positive, indicating the curve is bending upwards like a cup. Conversely, if the second derivative is negative, the curve is bending downwards, described as concave down.

To find these intervals:

• Calculate the second derivative from the parametric functions.
• Set \(\frac{d^2y}{dx^2} = 0\) to identify potential points where the concavity changes, also known as inflection points.
• Assess the sign of \(\frac{d^2y}{dx^2}\) between these points over the interval \[0, 2\pi\].

Understanding where the curve is concave up or concave down helps visualize the graph's shape and provides insights into the function's behavior.

Derivative

Derivatives, particularly in parametric equations, capture the rate of change concerning a parameter rather than a specific axis.

The first derivative, \(\frac{dy}{dx}\), reveals the slope of the tangent to the parametric curve. For the given parametric equations:

• \(x = \cos t\)
• \(y = \sin (2t)\)

The first derivative is calculated by dividing the derivative of \(y\) with respect to \(t\) by the derivative of \(x\) with respect to \(t\), leading to \(\frac{dy}{dx} = -\frac{2\cos(2t)}{\sin t}\). This formula provides the slope of the curve at any point \(t\).

The second derivative, \(\frac{d^2y}{dx^2}\), is derived from the first and provides valuable information on the curve's concavity. Essentially, it tells us how the slope itself changes, offering insights into the curve's curvature.

In summary, derivatives are powerful in analyzing the directions and angles of a curve, understanding its inclination, and predicting its behavior by showing instantaneous changes at each point along the curve.