Problem 59 - Calculus Early Transcendentals

Question

Show that the triangle that is formed by any tangent line to the graph of \(y=1 / x, x>0,\) and the coordinate axes has an area of 2 square units.

Step-by-Step Solution

Answer

The area is always 2 square units, regardless of the point of tangency.

Step 1: Find the Point of Tangency

Let the point of tangency be \((a, \frac{1}{a})\). We know this point lies on the curve \(y = \frac{1}{x}\).

Step 2: Determine the Slope of the Tangent Line

Differentiate \(y = \frac{1}{x}\) to obtain\[\frac{dy}{dx} = -\frac{1}{x^2}\].At \(x = a\), the slope (\(m\)) is \(-\frac{1}{a^2}\).

Step 3: Write the Equation of the Tangent Line

The equation of the tangent line at \((a, \frac{1}{a})\) can be expressed as:\[y - \frac{1}{a} = -\frac{1}{a^2}(x - a)\].Simplifying, we obtain:\[y = -\frac{1}{a^2}x + \frac{2}{a}\].

Step 4: Find the Intercepts

The x-intercept occurs when \(y = 0\):\[0 = -\frac{1}{a^2}x + \frac{2}{a}\].Solving, \(x = 2a\).The y-intercept occurs when \(x = 0\):\[y = \frac{2}{a}\].

Step 5: Calculate the Area of the Triangle

The base of the triangle is \(2a\) (x-intercept) and the height is \(\frac{2}{a}\) (y-intercept).The area \(A\) is given by:\[A = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2a \times \frac{2}{a} = 2\] square units.

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus that allows us to determine the rate at which a function changes at any given point.

• In simpler terms, differentiation helps us find the slope or steepness of the tangent line to a curve at a particular point.
• To differentiate a function means to compute its derivative, which can give us valuable information, like the slope of a tangent line.

In the exercise, we have a curve defined by the function \(y = \frac{1}{x}\), where \(x > 0\). To find the derivative, we apply differentiation rules to obtain:\[\frac{dy}{dx} = -\frac{1}{x^2}\]Here, \(\frac{dy}{dx}\) represents the derivative of \(y\) with respect to \(x\), which tells us the slope of the tangent line at any point on the curve. When evaluated at a specific \(x=a\), it gives \(-\frac{1}{a^2}\). This is the slope of the tangent line at the point \((a, \frac{1}{a})\) on the graph.

Area of a Triangle

The area of a triangle can be easily calculated if you know the base and the height.

• These dimensions are typically perpendicular to each other, making it straightforward to use the area formula for triangles.
• The formula to find the area \(A\) is: \[ A = \frac{1}{2} \times \text{base} \times \text{height} \]

In the context of this exercise, the triangle's base and height are derived from the x-intercept and y-intercept of the tangent line.

• The base of the triangle is the segment along the x-axis from the origin to the x-intercept \(2a\).
• The height is the y-intercept, \(\frac{2}{a}\), which is the segment along the y-axis from the origin to where the tangent line intersects.

Therefore, the area of the triangle can be calculated as:\[A = \frac{1}{2} \times 2a \times \frac{2}{a} = 2 \text{ square units}\]

Equation of a Line

The equation of a line in the slope-intercept form is crucial for understanding how lines behave on a graph.

• The general form of this equation is \(y = mx + c\), where \(m\) represents the slope and \(c\) the y-intercept.
• The slope \(m\) tells us how steep the line is, and in which direction it is slanting.

For the tangent line in this problem, starting from the point \((a, \frac{1}{a})\), and with a slope \(-\frac{1}{a^2}\), the equation is:\[y - \frac{1}{a} = -\frac{1}{a^2}(x - a)\]Simplifying this gives us:\[y = -\frac{1}{a^2}x + \frac{2}{a}\]

• Here, \(-\frac{1}{a^2}\) is the slope, which indicates the tangent's direction and steepness.
• The y-intercept \(\frac{2}{a}\) is the point where the tangent crosses the y-axis.

This is how the equation helps in forming and visualizing the line on a graph.