Problem 12 - Calculus Early Transcendentals

Question

Find \(d y / d x\) and \(d^{2} y / d x^{2} .\) For which values of \(t\) is the curve concave upward? $$x=t^{3}-12 t, \quad y=t^{2}-1$$

Step-by-Step Solution

Answer

The curve is concave upward for \(-2 < t < 2\).

Step 1: Find dx/dt

To find \( \frac{dy}{dx} \), we first need \( \frac{dx}{dt} \). Given \( x = t^3 - 12t \), we take the derivative of \( x \) with respect to \( t \):\[\frac{dx}{dt} = 3t^2 - 12\]

Step 2: Find dy/dt

Next, we find \( \frac{dy}{dt} \). Given \( y = t^2 - 1 \), we take the derivative of \( y \) with respect to \( t \):\[\frac{dy}{dt} = 2t\]

Step 3: Find dy/dx

We use the chain rule: \[\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt}\]Substitute in the derivatives:\[\frac{dy}{dx} = \frac{2t}{3t^2 - 12}\]

Step 4: Find d^2y/dx^2

To find \( \frac{d^2y}{dx^2} \), we differentiate \( \frac{dy}{dx} \) with respect to \( t \) and then divide by \( \frac{dx}{dt} \):First, find the derivative of \( \frac{dy}{dx} \) with respect to \( t \): \[\text{Let } u = 2t \,\,\text{and } v = 3t^2 - 12.\]Using the quotient rule:\[\frac{d}{dt}\left(\frac{u}{v}\right) = \frac{v\frac{du}{dt} - u\frac{dv}{dt}}{v^2}\]Where \( \frac{du}{dt} = 2 \) and \( \frac{dv}{dt} = 6t. \)Calculate:\[\frac{d}{dt}\left(\frac{2t}{3t^2 - 12}\right) = \frac{(3t^2 - 12)\cdot 2 - 2t\cdot 6t}{(3t^2 - 12)^2}\]Simplify:\[= \frac{6t^2 - 24 - 12t^2}{(3t^2 - 12)^2} = \frac{-6t^2 - 24}{(3t^2 - 12)^2}\]Then divide by \( \frac{dx}{dt} \):\[\frac{d^2y}{dx^2} = \frac{-6t^2 - 24}{(3t^2 - 12)^3}\]

Step 5: Determine when the curve is concave upward

The curve is concave upward when \( \frac{d^2y}{dx^2} > 0. \) From our expression:\[\frac{d^2y}{dx^2} = \frac{-6t^2 - 24}{(3t^2 - 12)^3} > 0\]Solving the inequality:1. The numerator \(-6t^2 - 24\) simplifies to \(-6(t^2 + 4)\), which is always negative because \(t^2 + 4\) is always positive.2. For the fraction to be positive, the denominator \((3t^2 - 12)^3\) must also be negative.3. Notice that \(3t^2 - 12\) must be negative for \((3t^2 - 12)^3\) to be negative: \(3t^2 < 12 \implies t^2 < 4 \implies -2 < t < 2\).Thus, the curve is concave upward for \(-2 < t < 2\).

Key Concepts

Derivatives of Parametric Equations

Parametric equations are a powerful tool in calculus that allow us to describe curves in terms of a third variable, usually denoted as \( t \). When dealing with such equations, understanding their derivatives is crucial. To find the first derivative \( \frac{dy}{dx} \) of parametric equations, we need the derivatives of both \( x \) and \( y \) with respect to \( t \). We use the chain rule for derivatives:

• Calculate \( \frac{dx}{dt} \), which is the rate of change of \( x \) with respect to \( t \).
• Calculate \( \frac{dy}{dt} \), which is the rate of change of \( y \) with respect to \( t \).
• The first derivative \( \frac{dy}{dx} \) is then obtained using: \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \).

This provides the slope of the tangent to the curve at any point \( t \). This process is important for analyzing the behavior of the curve, such as finding where it is increasing or decreasing.

Concave Upward

When a curve is concave upward, it resembles a cup that's open upwards. In calculus terms, a curve is concave upward at a region where its second derivative \( \frac{d^2y}{dx^2} \) is positive. This indicates that the slope of the tangent line is increasing.For parametric equations, we first find \( \frac{d^2y}{dx^2} \) using the chain rule and quotient rule as necessary. Once the second derivative is calculated, we determine intervals where this derivative is greater than zero. This provides us insight into where the curve bends upwards, which is crucial for understanding the overall shape of the graph.

Second Derivative Test

The second derivative test in a parametric context helps to determine the concavity of a curve. After finding \( \frac{d^2y}{dx^2} \), we need to examine its sign to conclude about the curve's concavity.

• If \( \frac{d^2y}{dx^2} > 0 \), the curve is concave upward at that interval.
• If \( \frac{d^2y}{dx^2} < 0 \), the curve is concave downward there.
• A helpful method is to identify the intervals of \( t \) where these inequalities hold.

In our exercise, to find when the curve is concave upward, we solved the inequality \( \frac{d^2y}{dx^2} > 0 \). By analyzing the sign of both the numerator and denominator of the second derivative, we identified the range of \( t \) values where the curve captured this property, revealing the symmetries and movements of the trajectory.