Question

For the following exercises, find \(d^{2} y / d x^{2}\) at the given point without eliminating the parameter.\(x=\frac{1}{2} t^{2}, \quad y=\frac{1}{3} t^{3}, \quad t=2\)

Short answer

The second derivative \(\frac{d^2y}{dx^2}\) at \(t=2\) is \(\frac{1}{2}\).

Step-by-step solution

Differentiate x with respect to t

Given that \(x = \frac{1}{2}t^2\), differentiate \(x\) with respect to \(t\). We get \(\frac{dx}{dt} = t\).

Differentiate y with respect to t

Given \(y = \frac{1}{3}t^3\), differentiate \(y\) with respect to \(t\). This gives \(\frac{dy}{dt} = t^2\).

Find the first derivative dy/dx

The first derivative \(\frac{dy}{dx}\) is found by dividing \(\frac{dy}{dt}\) by \(\frac{dx}{dt}\). So, \(\frac{dy}{dx} = \frac{t^2}{t} = t\).

Differentiate dy/dx with respect to t

Differentiate the expression for \(\frac{dy}{dx} = t\) with respect to \(t\). We get \(\frac{d}{dt}\left(\frac{dy}{dx}\right) = 1\).

Find the second derivative d^2y/dx^2

The second derivative \(\frac{d^2y}{dx^2}\) can be found using the formula \(\frac{d}{dt}\left(\frac{dy}{dx}\right) \div \frac{dx}{dt}\). Thus, \(\frac{d^2y}{dx^2} = \frac{1}{t}\).

Evaluate the second derivative at t=2

Plug in \(t = 2\) into \(\frac{d^2y}{dx^2} = \frac{1}{t}\). This results in \(\frac{d^2y}{dx^2} = \frac{1}{2}\).

Key concepts

Second Derivative

The second derivative, often denoted as \(\frac{d^2y}{dx^2}\), provides information about the curvature of a graph. It tells us how the rate of change (slope) is changing, or in simple terms, it informs us about the graph's concavity. If the second derivative is positive, the graph is concave up, like a smile. If it is negative, the graph is concave down, similar to a frown. The second derivative can indicate points of inflection, where the graph may change from being concave up to concave down, or vice versa.
To find the second derivative in parametric equations, we follow a multi-step process:

• Find the first derivatives: Differentiate both \(x\) and \(y\) with respect to the parameter (often \(t\)).
• Express \(\frac{dy}{dx}\): Divide \(\frac{dy}{dt}\) by \(\frac{dx}{dt}\).
• Differentiation of \(\frac{dy}{dx}\): Differentiate \(\frac{dy}{dx}\) with respect to the parameter.
• Calculate \(\frac{d^2y}{dx^2}\): Use \(\frac{d}{dt}\left(\frac{dy}{dx}\right) \div \frac{dx}{dt}\) to find the second derivative.

By systematically following these steps, we can successfully analyze the curvature and behavior of parametric equations.

Parametric Equations

Parametric equations are a way of defining a mathematical relationship using parameters rather than traditional \(x\) and \(y\) coordinates alone. This method uses a separate variable, usually \(t\), to define both \(x\) and \(y\) equations.
For this problem, we had:

• \(x = \frac{1}{2} t^2\)
• \(y = \frac{1}{3} t^3\)

These equations express both \(x\) and \(y\) in terms of the parameter \(t\), allowing us to describe paths that may not be functions in the traditional \(y = f(x)\) sense. This is particularly useful when dealing with paths that loop back on themselves or represent more complex curves.
Parametric equations make it easier to analyze the motion and change in complex systems, offering a clearer picture of how \(x\) and \(y\) evolve as \(t\) changes.

Differentiation with Respect to a Parameter

When working with parametric equations, we frequently perform differentiation with respect to the parameter \(t\). This means finding how fast \(x\) and \(y\) change as \(t\) changes.
The steps for differentiation in a parametric context involve:

• Identifying derivatives \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) by differentiating \(x\) and \(y\) separately with respect to \(t\).
• Using these derivatives to find \(\frac{dy}{dx}\), which is \(\frac{dy}{dt} \div \frac{dx}{dt}\).

This process illustrates how \(y\) changes with \(x\), even though both are expressed in terms of \(t\).
The procedure highlights the sensitivity to change provided by differentiating with respect to a parameter, making it invaluable for understanding the detailed behavior of parametric curves. In our example, the straightforward calculations demonstrated how the second derivative, based on these initial steps, can be determined, further enhancing our analysis of the curve's geometry.