Problem 58 - Calculus Early Transcendentals

Question

Show that the segment of the tangent line to the graph of \(y=1 / x\) that is cut off by the coordinate axes is bisected by the point of tangency.

Step-by-Step Solution

Answer

The segment is bisected by the point of tangency at \((a, \frac{1}{a})\).

Step 1: Find the Derivative

Start by differentiating the function \( y = \frac{1}{x} \) to find the slope of the tangent line. The derivative is \( y' = -\frac{1}{x^2} \).

Step 2: Determine the Point of Tangency

Assume the point of tangency is \((a, \frac{1}{a})\). This point lies on the curve \( y = \frac{1}{x} \).

Step 3: Write the Equation of the Tangent Line

Using the point-slope form of a line, the equation of the tangent line at \((a, \frac{1}{a})\) is given by \( y - \frac{1}{a} = -\frac{1}{a^2}(x - a) \), simplifying to \( y = -\frac{1}{a^2}x + \frac{2}{a} \).

Step 4: Find the Intercepts of the Tangent Line

To find the x-intercept, set \( y = 0 \): \( 0 = -\frac{1}{a^2}x + \frac{2}{a} \), leading to \( x = 2a \). To find the y-intercept, set \( x = 0 \): \( y = -\frac{1}{a^2}(0) + \frac{2}{a} = \frac{2}{a} \).

Step 5: Analyze the Segment and the Point of Tangency

The segment from the x-intercept (\(2a, 0\)) to the y-intercept (\(0, \frac{2}{a}\)) has its midpoint at \( \left( \frac{2a + 0}{2}, \frac{0 + \frac{2}{a}}{2} \right) = (a, \frac{1}{a}) \), which matches the point of tangency.

Key Concepts

Tangent Line

A tangent line is a straight line that touches a curve at one specific point without crossing it. This point is known as the "point of tangency." For the curve described by the function \( y = \frac{1}{x} \), the tangent line at any given point provides the slope and direction at that exact point.

• The tangent line can be visualized as the closest linear approximation of the curve at any given point.
• It is essential for understanding the behavior of the function near the point of contact.

To write the equation of a tangent line, we must know both its slope and a point it passes through. In this problem, the point is the point of tangency \(a, \frac{1}{a}\). The slope is determined by the derivative of the function at this point.

Derivative

The derivative of a function measures how the function value changes as its input changes, which is crucial for finding the slope of the tangent line. For the function \( y = \frac{1}{x} \), the derivative can be found using basic differentiation rules.

• The derivative is calculated as \( y' = -\frac{1}{x^2} \).
• This derivative provides the slope of the tangent line at any point \( x = a \), indicating whether the curve is rising or falling at that point.

The steeper the slope, the more rapid the change in the curve's value. A positive slope indicates an increasing function, while a negative slope shows a decreasing function. Here, the negative slope at any point means the graph is decreasing as \( x \) increases.

Coordinate Axes Intersection

When analyzing the tangent line on a graph, it's important to find where this line intersects the x-axis and the y-axis. This information helps us understand the full segment of the tangent that lies within the first quadrant.

• The x-intercept is found by setting \( y = 0 \) in the tangent line equation, yielding the point \((2a, 0)\).
• The y-intercept results from setting \( x = 0 \), giving us \((0, \frac{2}{a})\).

These intercepts are crucial for constructing the segment of the tangent line, contributing to a deeper understanding of how the line interacts with the coordinate axes.

Point of Tangency

The point of tangency is where the tangent line touches the curve precisely. In this problem, the point of tangency for the curve \( y = \frac{1}{x} \) is given by \((a, \frac{1}{a})\).

• This point lies exactly on the curve, establishing the accuracy of the tangent line at that spot.
• Moreover, it acts as a midpoint for the tangent line's segment between the axes, ensuring the line equally bisects the segment.

These features collectively ensure that any characteristics of the curve at \( (a, \frac{1}{a}) \) are reflected evenly on either side, highlighting the symmetrical nature of the tangent segment.