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Finding a Second Derivative Calculate the second derivative \(d^{2} y / d x^{2}\) for the plane curve defined by the parametric equations \(x(t)=t^{2}-3, y(t)=2 t-1,-3 \leq t \leq 4\)

Short Answer

Expert verified
The second derivative is \(\frac{d^2y}{dx^2} = -\frac{1}{2t^3}\).

Step by step solution

01

Find the First Derivatives

First, we find the derivatives of the parametric equations with respect to the parameter \(t\). For \(x(t) = t^2 - 3\), the derivative with respect to \(t\) is \(\frac{dx}{dt} = 2t\). For \(y(t) = 2t - 1\), the derivative is \(\frac{dy}{dt} = 2\).
02

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

To find \(\frac{dy}{dx}\), use the chain rule for parametric equations: \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt}\). Substitute the derivatives we found: \(\frac{dy}{dx} = \frac{2}{2t} = \frac{1}{t}\) for \(t eq 0\).
03

Find the Derivative of \(\frac{dy}{dx}\) with Respect to \(t\)

Differentiate \(\frac{dy}{dx} = \frac{1}{t}\) with respect to \(t\). This gives \(\frac{d}{dt}\left(\frac{1}{t}\right) = -\frac{1}{t^2}\).
04

Use the Chain Rule to Find \(\frac{d^2y}{dx^2}\)

The second derivative \(\frac{d^2y}{dx^2}\) is found using the formula \(\frac{d}{dt}\left( \frac{dy}{dx} \right) \div \frac{dx}{dt}\). Substitute the results from the previous steps: \(\frac{d^2y}{dx^2} = \frac{-1/t^2}{2t} = -\frac{1}{2t^3}\).
05

Simplify if Necessary

The expression for the second derivative is \(-\frac{1}{2t^3}\). This is the simplest form of the second derivative for any valid \(t\) within the given range \(-3 \leq t \leq 4\), except when \(t = 0\) where the derivative is undefined.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Parametric Equations
Parametric equations allow us to define a curve using a third variable, typically denoted as the parameter, often labeled as \( t \). Instead of describing \( x \) and \( y \) directly in terms of each other, parametric equations describe both \( x \) and \( y \) in terms of \( t \). This can be especially useful for curves that are difficult to express as a single equation in \( x \) and \( y \).

In the exercise, the parametric equations are \( x(t) = t^2 - 3 \) and \( y(t) = 2t - 1 \). Here, \( t \) is the parameter that traces the curve in the plane as it varies from \(-3\) to \(4\). By using parametric equations, we gain more flexibility:
  • They can describe more complex curves, like loops or openings.
  • They provide a way to calculate properties of curves, such as finding derivatives.
  • They can be applied in a variety of fields, from physics to computer graphics.
As \( t \) changes, both \( x(t) \) and \( y(t) \) change, outlining the shape of the curve. Analyzing such equations is the foundation to understanding complex curves.
Chain Rule
The chain rule is a fundamental concept used in calculus to differentiate composite functions. It's especially useful when dealing with parametric equations, where we express derivatives with respect to a parameter. The basic premise is:

When you have a function composed of other functions, to find the derivative, you multiply the derivatives of the inner functions.

For parametric equations, the chain rule helps us transition from a parameter-based view to a more familiar context. When calculating \( \frac{dy}{dx} \) of parametric lines, the chain rule yields:
  • \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \)
This means that you take the derivative of \( y \) with respect to \( t \) and divide it by the derivative of \( x \) with respect to \( t \).

Through simple division, this operation transforms the ratio of derivatives into the slope of the curve, much like traditional derivatives for Cartesian equations. Understanding how the chain rule connects parametric derivatives back to \( x \) and \( y \) domains is vital for further calculus applications, such as finding the second derivative.
Plane Curve
A plane curve is simply a curve that lies on a two-dimensional plane, described using parametric or explicit equations. These are significant in calculus as they offer insights into the behavior of functions and their graphical representation.

The discussed problem focuses on a plane curve defined by parametric equations, refining our ability to analyze motion and change across a plane.
  • Plane curves can exhibit simple lines or complex loops and turns, depending on their equation.
  • They are often analyzed through derivatives to understand their tangents and curvature.
  • The second derivative offers insight into the curve's concavity and the rate at which the slope changes.
Being part of a foundational concept in geometry, plane curves enhance depth in understanding 2D shapes and optimizing paths, making them integral in robotics or trajectories, for example. The exercise exploits this idea by first calculating the first derivative for quick slopes and transitions to the second derivative for deeper insight into how the curve evolves.

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