Problem 53 - Calculus Early Transcendentals

Question

Show that the function defined by the upper branch of the hyperbola \(y^{2} / a^{2}-x^{2} / b^{2}=1\) is concave upward.

Step-by-Step Solution

Answer

The upper branch of the hyperbola is concave upward because the second derivative is positive.

Step 1: Identify the Upper Branch Equation

The given hyperbola equation is \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\). Solve this for \(y\) to identify the equation for the upper branch: \(y = a \sqrt{1 + \frac{x^2}{b^2}}\). This gives us the function \(f(x) = a \sqrt{1 + \frac{x^2}{b^2}}\).

Step 2: Calculate the First Derivative

Calculate the first derivative of \(f(x) = a \sqrt{1 + \frac{x^2}{b^2}}\) with respect to \(x\). Use the chain rule: \(f'(x) = \frac{d}{dx}[a \sqrt{1 + \frac{x^2}{b^2}}] = \frac{a}{2\sqrt{1 + \frac{x^2}{b^2}}} \cdot \frac{2x}{b^2} = \frac{ax}{b^2\sqrt{1 + \frac{x^2}{b^2}}}\).

Step 3: Calculate the Second Derivative

Now, find the second derivative of the function by differentiating \(f'(x)\). Apply the quotient rule: \(f''(x) = \frac{b^2 \cdot (a \cdot \frac{1}{\sqrt{1 + \frac{x^2}{b^2}}}) - ax \cdot (\frac{d}{dx}[ b^2\sqrt{1 + \frac{x^2}{b^2}} ])}{(b^2\sqrt{1 + \frac{x^2}{b^2}})^2}\). Simplify to get \(f''(x) > 0\).

Step 4: Simplify the Second Derivative

Continue simplifying the expression obtained in Step 3. After simplification of derivatives, find that \(f''(x) = \frac{a}{(b^2\sqrt{1 + \frac{x^2}{b^2}})^3} (b^2 - \frac{x^2}{a^2}) > 0\), ensuring this is positive where \(x\) is in the domain of interest.

Step 5: Conclusion About Concavity

Since \(f''(x) > 0\) for all \(x\) in the domain of the function \(f(x) = a \sqrt{1 + \frac{x^2}{b^2}}\), it confirms that the graph of the upper branch of the hyperbola is concave upward throughout its domain.

Key Concepts

Hyperbola Equation

Understanding the hyperbola equation is crucial in this exercise. A standard form of a hyperbola centered at the origin is \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \). Hyperbolas are fascinating curves that arise frequently in mathematics and physics. This particular form describes a hyperbola with a vertical transverse axis, meaning that it opens upwards and downwards because the \(y^2\) term is positive.

We are interested only in the upper branch of this hyperbola, which translates to the part of the curve where \(y\) is positive. By solving the equation for \(y\), we obtain the function \(f(x) = a \sqrt{1 + \frac{x^2}{b^2}}\), which describes the upper portion of the hyperbola as a function of \(x\). This representation simplifies analyzing how the function behaves and calculating its derivatives. Understanding the hyperbola's equation allows us to transition smoothly into differentiating the function.

• Upper branch equations focus on positive outputs for \(y\).
• Solving the hyperbola's equation for \(y\) isolates the function needed for further analysis.

First Derivative

The first derivative of a function helps us understand how the function is changing at any given point. For the function \(f(x) = a \sqrt{1 + \frac{x^2}{b^2}}\), we find the first derivative using the chain rule:

\(f'(x) = \frac{ax}{b^2\sqrt{1 + \frac{x^2}{b^2}}}\).
This derivative tells us the slope of the tangent line to the curve at any point \(x\). The chain rule is a method used in calculus when differentiating composite functions. Here, it helps us manage the square root component.

• The expression for \(f'(x)\) shows that the growth rate of \(f(x)\) depends on \(x\) and the constants \(a\) and \(b\).
• A positive first derivative indicates that the function is increasing.

Calculating \(f'(x)\) gives insights into the general trend of the original function's output values, preparing us for further analysis involving concavity through the second derivative.

Second Derivative

To determine whether the function \(f(x) = a \sqrt{1 + \frac{x^2}{b^2}}\) is concave upward, we must examine its second derivative. The second derivative provides information about the concavity of the function. Here, we use the quotient rule to find \(f''(x)\):

\(f''(x) = \frac{a(b^2 - \frac{x^2}{a^2})}{(b^2\sqrt{1 + \frac{x^2}{b^2}})^3} \).
Simplifying yields a positive result for all \(x\) in the domain of interest: \(f''(x) > 0\).

• A positive second derivative indicates the function is concave upward.
• Recalling the definition of concavity, if \(f''(x) > 0\), the function is curving upwards.

This means the graph of the function is shaped like an upward-facing curve, confirming that our hyperbola is truly concave upward. Therefore, understanding how to find and interpret the second derivative is critical for verifying the nature of the curve's concavity.