Problem 69 - Calculus Single Variable

Question

Suppose that \(c\) is a positive constant. Let \(L_{c}\) be the line that is tangent to the graph of \(f(x)=1 / x\) at \(P=(c, 1 / c)\). Show that the area of the triangle formed by \(L_{c}\) and the positive axes is independent of \(c\). Compute that area.

Step-by-Step Solution

Answer

The area of the triangle is 2 and it is independent of the constant \(c\).

Step 1: Find the Derivative

To find the equation of the tangent line, we need the slope at point \(P=(c, 1/c)\). First, find the derivative of \(f(x) = \frac{1}{x}\). The derivative \(f'(x) = -\frac{1}{x^2}\).

Step 2: Compute the Slope of the Tangent Line

Evaluate the derivative at \(x = c\) to find the slope of the tangent line. Thus, the slope is \(f'(c) = -\frac{1}{c^2}\).

Step 3: Write the Equation of the Tangent Line

Using the point-slope form of a line equation, \(y - y_1 = m(x - x_1)\), substitute \(m = -\frac{1}{c^2}\), \(x_1 = c\), and \(y_1 = \frac{1}{c}\). The equation becomes \(y - \frac{1}{c} = -\frac{1}{c^2}(x - c)\).

Step 4: Simplify the Tangent Line Equation

Rearrange to find the line equation in slope-intercept form: \[ y = -\frac{1}{c^2}x + \frac{2}{c} \].

Step 5: Determine x-intercept and y-intercept

Set \(y = 0\) to find the \(x\)-intercept: \(0 = -\frac{1}{c^2}x + \frac{2}{c}\) gives \(x = 2c\). Set \(x = 0\) to find the \(y\)-intercept: \(y = \frac{2}{c}\).

Step 6: Calculate the Area of the Triangle

The triangle formed by the line \(L_c\) and the axes has vertices at \((0, 0)\), \((2c, 0)\), and \((0, \frac{2}{c})\). The area of a triangle is \(\frac{1}{2} \times \text{base} \times \text{height}\). Here, the base is \(2c\) and the height is \(\frac{2}{c}\), so the area is \[ \frac{1}{2} \times 2c \times \frac{2}{c} = 2 \].

Step 7: Conclude with the Independence of c

The area calculation shows that the value \(c\) cancels out, confirming the area is independent of \(c\). Thus, the area of the triangle is always \(2\), regardless of \(c\).

Key Concepts

Derivative

In calculus, derivatives play a key role in understanding how a function changes at any given point. For the function \(f(x) = \frac{1}{x}\), the derivative helps us find the slope of the tangent line at a specific point. The process involves calculating the rate of change of the function, which is found by differentiating \(f(x)\). In this case, the derivative is \(f'(x) = -\frac{1}{x^2}\). This tells us how steep or flat the function is at any point \(x\).
Obtaining the correct derivative is crucial, as it allows us to proceed with finding the tangent line's equation. This understanding provides the foundation for further analysis, such as discovering how the tangent line interacts with the graph and other components like triangle areas.
Having this derivative, we pinpoint the slope of the tangent line at our point of interest, which significantly eases following calculations.

Tangent Line

The tangent line is an essential concept when analyzing the behavior of functions graphically. It represents the line that touches the graph of the function at only one point without crossing it. For the function \(f(x) = \frac{1}{x}\), the tangent line at point \(P = (c, \frac{1}{c})\) is derived using the slope obtained from the derivative. With the slope \(-\frac{1}{c^2}\) from the derivative, we utilize the point-slope form of the line equation:
\[ y - \frac{1}{c} = -\frac{1}{c^2}(x - c) \]
Rearranging this equation gives us the slope-intercept form:
\[ y = -\frac{1}{c^2}x + \frac{2}{c} \]
This line precisely reflects how the function behaves around the point \(c\). Understanding this line helps illustrate how local changes in the function graph are mirrored by the tangent's slope and position.

Triangle Area

Calculating the area of the triangle formed by the tangent line and the axes involves geometric principles combined with calculus insights. The vertices of the triangle are \((0, 0)\), \((2c, 0)\), and \((0, \frac{2}{c})\). The base of this triangle resides on the x-axis from \((0,0)\) to \((2c,0)\), making it \(2c\), and its height stretches up the y-axis to \((0, \frac{2}{c})\).
The formula for the area of a triangle is \(A = \frac{1}{2} \times \text{base} \times \text{height}\). Substituting the base and height values, we find:
\[ A = \frac{1}{2} \times 2c \times \frac{2}{c} = 2 \]
Notice how the constant \(c\) cancels out, indicating the area of the triangle is always 2, irrespective of \(c\). This elegant result showcases the interconnectedness of geometry and calculus in providing deep insights into function behavior and geometric shape properties.

Function Graph

The graph of a function like \(f(x) = \frac{1}{x}\) serves as a visual roadmap for its behavior and relationships with other elements like tangents or geometric figures. This hyperbola reflects the inverse relationship between \(x\) and \(f(x)\), showing that as \(x\) increases, \(f(x)\) decreases and vice versa.
This insight helps when considering the tangent lines and how they interact with the axes to form triangles. The tangent lines provide snapshots of the function's slope at particular points, and their intersections with the axes can form geometric shapes, like the triangles we've explored.
Such functions and their graphs provide foundational understanding in many calculus problems, helping to visually and analytically bridge function behavior with algebraic and geometric concepts.