Problem 99 - Calculus Volume 3

Question

Determine the concavity of the curve \(x=2 t+\ln t, y=2 t-\ln t\).

Step-by-Step Solution

Answer

The curve is concave upwards for all \( t > 0 \).

Step 1: Find the first derivatives

The parametric equations are given as \( x(t) = 2t + \ln t \) and \( y(t) = 2t - \ln t \). Find the derivatives \( x'(t) \) and \( y'(t) \) with respect to \( t \).. * For \( x(t) \): \( x'(t) = \frac{d}{dt}[2t + \ln t] = 2 + \frac{1}{t} \).* For \( y(t) \): \( y'(t) = \frac{d}{dt}[2t - \ln t] = 2 - \frac{1}{t} \).

Step 2: Calculate the second derivatives

Find the second derivatives, \( x''(t) \) and \( y''(t) \).* Since \( x'(t) = 2 + \frac{1}{t} \), \( x''(t) = \frac{d}{dt}[2 + \frac{1}{t}] = 0 - \frac{1}{t^2} = -\frac{1}{t^2} \).* Similarly, since \( y'(t) = 2 - \frac{1}{t} \), \( y''(t) = \frac{d}{dt}[2 - \frac{1}{t}] = 0 + \frac{1}{t^2} = \frac{1}{t^2} \).

Step 3: Use the formula for the curvature of parametric curves

For parametric curves, the concavity is often analyzed using the second derivative of \( y \) with respect to \( x \), which relies on the formula:\[\frac{d^2 y}{dx^2} = \frac{y''(t)x'(t) - y'(t)x''(t)}{(x'(t))^3}\]Substitute \( x'(t) = 2 + \frac{1}{t} \), \( y'(t) = 2 - \frac{1}{t} \), \( x''(t) = -\frac{1}{t^2} \), \( y''(t) = \frac{1}{t^2} \) into the equation.

Step 4: Compute \( \frac{d^2 y}{dx^2} \)

Substitute the derivatives into our formula:\[\frac{d^2 y}{dx^2} = \frac{\left(\frac{1}{t^2}\right)\left(2 + \frac{1}{t}\right) - \left(2 - \frac{1}{t}\right)\left(-\frac{1}{t^2}\right)}{\left(2 + \frac{1}{t}\right)^3}\]Simplifying the numerator:\[= \frac{2}{t^2} + \frac{1}{t^3} + \frac{2}{t^2} - \frac{1}{t^3} = \frac{4}{t^2}\]Therefore, \[\frac{d^2 y}{dx^2} = \frac{\frac{4}{t^2}}{\left(2 + \frac{1}{t}\right)^3}\]

Step 5: Determine the sign of \( \frac{d^2 y}{dx^2} \)

Since \( t > 0 \) and the denominator \( \left(2 + \frac{1}{t}\right)^3 > 0 \), the sign of \(\frac{d^2 y}{dx^2}\) is determined by the numerator \( \frac{4}{t^2} \), which is always positive for \( t > 0 \).

Step 6: Conclusion: Concavity

Because \( \frac{d^2 y}{dx^2} > 0 \) for all \( t > 0 \), the curve is concave upwards for the given parameter range.

Key Concepts

Parametric Equations

Parametric equations are a set of equations that express the coordinates of the points that make up a curve as functions of a variable, known as a parameter. Instead of describing a curve explicitly as a relationship between x and y, parametric equations allow for a more flexible description.
For our exercise, we have two parametric equations:

• \(x(t) = 2t + \ln t\)
• \(y(t) = 2t - \ln t\)

Here, both x and y are expressed in terms of the parameter \(t\). This provides a way to describe the curve as \(t\) varies. By manipulating \(t\), we can analyze the movement and shape of the curve in the plane. Parametric equations are particularly useful for describing curves that are challenging to illustrate with a single function \(y = f(x)\).
To better understand a curve, it's often essential to explore its derivatives, which we will discuss more in the next sections.

Second Derivative

The second derivative in the context of parametric curves helps us understand the behavior of the curve beyond just its direction. The first derivatives \(x'(t)\) and \(y'(t)\) tell us about the slope or direction at any point.
When analyzing concavity, we rely on the second derivatives \(x''(t)\) and \(y''(t)\), which indicate how the slopes of the tangents to the curve change. For the given functions:

• \(x'(t) = 2 + \frac{1}{t}\)
• \(y'(t) = 2 - \frac{1}{t}\)
• \(x''(t) = -\frac{1}{t^2}\)
• \(y''(t) = \frac{1}{t^2}\)

These derivatives tell us how fast and in what manner the change is occurring. Second derivatives are critical in determining curvature, as they reflect whether a curve is bending upwards or downwards.

Concave Upwards

A curve is said to be concave upwards if it bends upwards like a cup. This characteristic is determined by the sign of the second derivative. For parametric curves, the analysis involves calculating \(\frac{d^2 y}{dx^2}\).
In the exercise, we computed \(\frac{d^2 y}{dx^2}\) as:\[\frac{d^2 y}{dx^2} = \frac{\frac{4}{t^2}}{(2 + \frac{1}{t})^3}\]Since \(\frac{d^2 y}{dx^2} > 0\) for \(t > 0\), it confirms that the curve is concave upwards for the domain of interest.
Concave upwards curves will appear as though they can hold water like a basin. This property is significant in predicting the behavior and stability of trajectories, optimizing paths, and more.

Curvature Analysis

Curvature analysis helps describe how a curve bends in space. The curvature is closely connected to the concept of concavity and is expressed through the second derivative in parametric equations.
In our example, the formula used for evaluating curvature is:\[\frac{d^2 y}{dx^2} = \frac{y''(t)x'(t) - y'(t)x''(t)}{(x'(t))^3}\]This formula combines first and second derivatives to measure how sharply a curve is turning at any given point.
An important aspect of this analysis is understanding the sign of the second derivative. Positive values indicate concave upward behavior, while negative values suggest concave downward behavior.
Curvature analysis is essential for applications in physics, engineering, and computer graphics, where understanding the shape and changes of curves is crucial.