Problem 14 - 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? $$x=t+\ln t, \quad y=t-\ln t$$

Step-by-Step Solution

Answer

The curve is concave upward for all \(t \neq -1\).

Step 1: Differentiate x and y with respect to t

Given that \(x = t + \ln t\) and \(y = t - \ln t\). First, we find \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\). \[\frac{dx}{dt} = 1 + \frac{1}{t}\]\[\frac{dy}{dt} = 1 - \frac{1}{t}\]

Step 2: Find dy/dx

To find \(\frac{dy}{dx}\), use the chain rule by dividing \(\frac{dy}{dt}\) by \(\frac{dx}{dt}\):\[\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{1 - \frac{1}{t}}{1 + \frac{1}{t}}\]

Step 3: Simplify dy/dx

Simplify the expression for \(\frac{dy}{dx}\):Multiply numerator and denominator by \(t\):\[\frac{dy}{dx} = \frac{t(1 - \frac{1}{t})}{t(1 + \frac{1}{t})} = \frac{t - 1}{t + 1}\]

Step 4: Differentiate dy/dx to find d²y/dx²

Now differentiate \(\frac{dy}{dx} = \frac{t-1}{t+1}\) with respect to \(t\) using the quotient rule, where \(u = t-1\) and \(v = t+1\):\[\frac{d^2y}{dx^2} = \frac{(v)(\frac{du}{dt}) - (u)(\frac{dv}{dt})}{v^2}\]\[\frac{du}{dt} = 1, \quad \frac{dv}{dt} = 1\]\[\frac{d^2y}{dx^2} = \frac{(t+1)(1) - (t-1)(1)}{(t+1)^2} = \frac{t + 1 - t + 1}{(t+1)^2} = \frac{2}{(t+1)^2}\]

Step 5: Determine concavity of the curve

The curve is concave upward when \(\frac{d^2y}{dx^2} > 0\). Since \(\frac{d^2y}{dx^2} = \frac{2}{(t+1)^2}\), it is always positive since \((t+1)^2 > 0\) for all \(t eq -1\). Thus, the curve is concave upward for all \(t eq -1\).

Key Concepts

Understanding Concavity

Concavity is a property of a curve that tells us about its shape or direction. When a curve is said to be concave upward, it means that the curve opens upwards, similar to a cup or a bowl. This occurs when the second derivative of the curve, denoted as \( \frac{d^2y}{dx^2} \), is positive. It indicates that the slope of the tangent lines is increasing.

To determine the concavity of a curve described by parametric equations, like the given \( x = t + \ln t \) and \( y = t - \ln t \), we first find the second derivative \( \frac{d^2y}{dx^2} \). This involves differentiating \( \frac{dy}{dx} \) using techniques like the quotient rule. If the resulting function is positive for most or all values of \( t \), the curve is concave upward. In this case, \( \frac{d^2y}{dx^2} = \frac{2}{(t+1)^2} \), which is always greater than zero for \( t eq -1 \), confirming upward concavity for most of the curve.

The Chain Rule in Differentiation

The chain rule is a fundamental technique in calculus used to differentiate composite functions. When dealing with parametric equations, like \( x = t + \ln t \) and \( y = t - \ln t \), we often apply the chain rule to connect derivatives with respect to one variable to another. This is achieved by determining \( \frac{dy}{dx} \) indirectly through \( t \).

To apply the chain rule for parametric differentiation, we find \( \frac{dy}{dt} \) and \( \frac{dx}{dt} \) first. Then, we use the relation: \[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \] This converts derivatives with respect to \( t \) back to conventional derivatives with respect to \( x \), as needed to analyze the curve's behavior, like its slope at different points.

Using the Quotient Rule

The quotient rule is a method in calculus used to find the derivative of a quotient of two functions. When we have \( \frac{dy}{dx} = \frac{t-1}{t+1} \), and need to compute its derivative to find the second derivative \( \frac{d^2y}{dx^2} \), the quotient rule becomes necessary. The formula for the quotient rule is:

• \( \frac{d}{dt}\left( \frac{u}{v} \right) = \frac{v \frac{du}{dt} - u \frac{dv}{dt}}{v^2} \)

Here, \( u = t - 1 \) and \( v = t + 1 \). By applying the quotient rule, we find:

• \( \frac{d^2y}{dx^2} = \frac{(t+1)(1) - (t-1)(1)}{(t+1)^2} = \frac{2}{(t+1)^2} \)

This result helps us describe the concavity of the curve by evaluating whether the second derivative is positive or negative across the domain of the curve.

Parametric Equations and their Derivatives

Parametric equations offer a way to describe curves by expressing the coordinates \( (x, y) \) as functions of an independent parameter \( t \). Instead of using an equation relating \( x \) and \( y \) directly, the curve is defined by equations like \( x = f(t) \) and \( y = g(t) \).

For these equations, we derive \( \frac{dy}{dx} \) - the rate of change of \( y \) with respect to \( x \) - by using the derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \). This approach is quite powerful, particularly in complex shapes or curves that aren't simple functions of \( x \). With the example \( x = t + \ln t \) and \( y = t - \ln t \), we find:

• \( \frac{dx}{dt} = 1 + \frac{1}{t} \)
• \( \frac{dy}{dt} = 1 - \frac{1}{t} \)

By dividing these, \( \frac{dy}{dx} = \frac{t-1}{t+1} \), simplifies our study of how \( y \) changes with \( x \), crucial for analyzing the curve's features like its slope and concavity.