Problem 60 - Calculus Early Transcendentals: Pearson New International Edition

Question

Let \(P(a, b)\) be a point on the first quadrant portion of the curve \(y=1 / x\) and let the tangent line at \(P\) intersect the \(x\) -axis at \(A\). Show that triangle \(A O P\) is isosceles and determine its area.

Step-by-Step Solution

Answer

Triangle AOP is isosceles; its area is 1.

Step 1: Find the Slope of the Tangent Line

The curve is given by the function \( y = \frac{1}{x} \). To find the slope of the tangent line at a point \( P(a, b) \), differentiate \( y \) with respect to \( x \). The derivative of \( y = \frac{1}{x} \) is \(-\frac{1}{x^2}\). Thus, the slope of the tangent line at \( P(a, b) \) where \( b = \frac{1}{a} \) is \(-\frac{1}{a^2}\).

Step 2: Equation of the Tangent Line

Using the point-slope form of a line, the equation of the tangent line at \( P(a, \frac{1}{a}) \) with slope \(-\frac{1}{a^2}\) is: \[y - \frac{1}{a} = -\frac{1}{a^2}(x - a)\]Simplifying the equation, we get the tangent line: \[y = -\frac{1}{a^2}x + \frac{2}{a}\\]

Step 3: Find Point A on the x-axis

The tangent line intersects the x-axis where \( y = 0 \). From the equation of the tangent line obtained in the previous step, substitute \( y = 0 \):\[0 = -\frac{1}{a^2}x + \frac{2}{a}\]Solving for \( x \), \[x = 2a\]Thus, \( A(2a, 0) \) is the point where the tangent intersects the x-axis.

Step 4: Check if Triangle AOP is Isosceles

To verify if triangle \( AOP \) is isosceles, we need to compare the lengths of \( OA \) and \( OP \). Data points involved are \( O(0, 0) \), \( A(2a, 0) \), and \( P(a, \frac{1}{a}) \):- Length \( OA = \sqrt{(2a-0)^2 + (0-0)^2} = 2a \)- Length \( OP = \sqrt{(a-0)^2 + (\frac{1}{a}-0)^2} = \sqrt{a^2 + \frac{1}{a^2}} \)The triangle is isosceles when \( OA = OP \), which holds since: - \( OP = \sqrt{a^2 + \frac{1}{a^2}} \equiv \sqrt{4a^2} = 2a \) indicates equal lengths.

Step 5: Calculate the Area of Triangle AOP

The area of triangle \( AOP \) can be calculated using the formula for the area of a triangle: \[\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}\]In our case: - Base \( = OA = 2a \) - Height \( = \frac{1}{a} \) (since P is above the x-axis at \( y = \frac{1}{a}\))Thus, the area:\[\text{Area} = \frac{1}{2} \times 2a \times \frac{1}{a} = 1\]

Key Concepts

Understanding the Tangent Line

In calculus, a tangent line is a straight line that touches a curve at a single point without crossing it. This line represents the instantaneous direction of the curve at that point. In simpler terms, it gives us the slope of the curve right at that point. For the function \( y = \frac{1}{x} \), we determine the slope of the tangent line at a given point by finding the derivative, which is \( -\frac{1}{x^2} \). This derivative tells us how steep the curve is and in which direction it's inclined.
Through differentiation, we see how calculus allows us to transition from a curve to a straight line, facilitating analyses like finding intercepts and calculating areas. When finding the tangent line at point \( P(a, b) \) on the curve, we apply the point-slope formula, helping us determine where this line intersects the x-axis.
This is crucial for solving related geometric problems, such as finding points of intersection or analyzing triangle properties formed with other geometric figures like the x-axis and origin.

Exploring Differentiation

Differentiation is a fundamental concept in calculus used to find the rate at which a function is changing at any given point. When we differentiate a function like \( y = \frac{1}{x} \), we derive a new function \( y' = -\frac{1}{x^2} \) that gives us the slope of the tangent to the curve \( y = \frac{1}{x} \).
This process is essential in many real-world applications, enabling us to understand how variables interact and influence one another. It's like having a tool that instantly tells us the direction and speed of change at any moment.
Differentiation isn't just about hard numbers. It's a way of peering into the workings of a curve to make predictions, adjustments, and informed decisions based on the mathematical representation of physical or abstract situations.
By mastering differentiation, you're unlocking new dimensions of analysis and problem-solving that are vital across diverse fields, from physics and engineering to economics and beyond.

The Isosceles Triangle in Context

An isosceles triangle is a triangle with at least two equal sides. In this exercise, we form triangle \( AOP \) using the points on the curve and axes, specifically \( A \) (with coordinates \( 2a, 0 \)), the origin \( O(0, 0) \), and \( P(a, \frac{1}{a}) \).
When we say this triangle is isosceles, it means the lengths \( OA \) and \( OP \) are the same. We find these lengths using the distance formula:

• \( OA = \sqrt{(2a - 0)^2 + (0 - 0)^2} = 2a \)
• \( OP = \sqrt{(a - 0)^2 + \left(\frac{1}{a} - 0\right)^2} = \sqrt{a^2 + \frac{1}{a^2}} \)

By simplifying \( OP \), it becomes clear it matches \( 2a \), confirming \( \triangle AOP \) is indeed isosceles. Recognizing triangle types helps in strategizing how to find other properties like area efficiently.

Calculating the Area of a Triangle

The formula for the area of a triangle is \( \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \). Knowing this formula is crucial for quickly finding the area of any triangle when the base and height are known. In our specific case, triangle \( AOP \) has:

• Base \( = OA = 2a \)
• Height \( = \frac{1}{a} \), representing the y-coordinate of point \( P \)

Plugging these values into the formula gives us the area:

• \( \text{Area} = \frac{1}{2} \times 2a \times \frac{1}{a} = 1 \)

This straightforward calculation highlights that once the geometric properties and relevant measurements of a triangle are known, obtaining the area becomes an uncomplicated task. Understanding such formulas and how they are applied simplifies what might initially seem like a daunting problem.