Problem 13 - 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? $$x=t-e^{t}, \quad y=t+e^{-t}$$

Step-by-Step Solution

Answer

The curve is concave upward for sufficiently large negative t values.

Step 1: Find dx/dt and dy/dt

Start by differentiating both given functions with respect to the parameter \( t \). For \( x(t) = t - e^t \), differentiate to get \( \frac{dx}{dt} = 1 - e^t \). For \( y(t) = t + e^{-t} \), the derivative is \( \frac{dy}{dt} = 1 - e^{-t} \).

Step 2: Apply Parametric Derivative Formula to find dy/dx

Use the formula for the derivative of parametric equations \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \). Substitute the derivatives: \[ \frac{dy}{dx} = \frac{1 - e^{-t}}{1 - e^t} \].

Step 3: Differentiate dy/dx with respect to t to find d²y/dx²

Differentiate \( \frac{dy}{dx} = \frac{1 - e^{-t}}{1 - e^t} \) with respect to \( t \) using the quotient rule: If \( h(t) = \frac{u}{v} \), \( h'(t) = \frac{u'v - uv'}{v^2} \). So, \( \frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{(e^{-t})(1 - e^t) - (1 - e^{-t})(e^t)}{(1 - e^t)^2} \).

Step 4: Simplify d²y/dx²

Plug in calculated numerator and denominator from the previous step: Simplify \[ \frac{d^2y}{dx^2} = \frac{-e^{-t} - e^t + 1 + e^{2t} - e^{-t} + e^t}{(1 - e^t)^3} \].

Step 5: Analyze Concavity by determining sign of d²y/dx²

The curve is concave upward when \( \frac{d^2y}{dx^2} > 0 \). Set \( -2e^{-t} + e^{2t} + 1 > 0 \) and solve for \( t \) to find the intervals where this inequality holds true.

Key Concepts

Parametric Equations

Parametric equations represent two variables, typically \(x\) and \(y\), as functions of a third variable, known as the parameter,denoted by \(t\) in most cases. This approach allows for a more flexible description of curves and shapes that move beyond the typical Cartesian representation of \(y=f(x)\). Instead,\(x\) and \(y\) are expressed separately as functions of \(t\).

• Example: Consider \(x(t) = t - e^t\) and \(y(t) = t + e^{-t}\). Here, \(t\) is the parameter that helps define the curve in the coordinate plane.
• The main advantage is the ability to describe a wide range of curves, including those that cannot be expressed as functions in Cartesian forms.

By using parametric equations, we can also calculate slopes at specific points easily by deriving the expressions for each variable, as illustrated when finding the derivative of \(y\) with respect to \(x\), \(\frac{dy}{dx}\). This is accomplished using the chain rule, specifically by finding the derivatives\(\frac{dy}{dt}\) and \(\frac{dx}{dt}\), and then applying the formula\(\frac{dy}{dx} = \frac{dy/dt}{dx/dt}\). This method simplifies the differentiation process, which is vital in understandingthe behavior and properties of curves.

Concavity

Concavity is a property of curves that describes how they "bend". A curve may be concave upward (like a cup) or concave downward (like a hill). To determine the concavity, we need to look at the second derivative,\(\frac{d^2y}{dx^2}\), of a function or parametric curve.

• If the second derivative is positive, \(\frac{d^2y}{dx^2} > 0\),the curve is concave upward.
• If the second derivative is negative, \(\frac{d^2y}{dx^2} < 0\),the curve is concave downward.

In the exercise example, after deriving the second derivative from the parametric functions \(x(t)\) and \(y(t)\), it’s essential to identify the intervals in \(t\) where \(\frac{d^2 y}{dx^2} > 0\) todetermine where the curve is concave upward.
This examination of concavity helps illustrate how the curve changes direction or shape as \(t\) varies,

Second Derivative Test

The second derivative test is a useful tool that provides insight intothe concavity of a parametric curve and can help identify points of inflection.This involves taking the derivative of \(\frac{dy}{dx}\) with respectto the parameter \(t\) using calculus techniques such as the chain rule and quotient rule.

• Start by finding \(\frac{dy}{dx}\) using parametric differentiation methods.
• Next, derive this expression again with respect to \(t\), which involves applying rules like the quotient rule if the expression is in fractional form.
• The result, \(\frac{d^2y}{dx^2}\), will then be analyzed to see where it is greater than zero to conclude where the curve is concave upward.

By following this process, as detailed in the step-by-step solution, we effectivelydetermine intervals of concavity and gain a comprehensive understanding of the curve’s behavior.