Problem 59 - Calculus - AP Edition

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 area of 2 square units,

Step-by-Step Solution

Answer

The area of the triangle is 2 square units.

Step 1: Find the derivative of the function

The given function is \( y = \frac{1}{x} \). To find the slope of the tangent line at any point \((x_0, y_0)\) on the curve, we first need the derivative of the function. The derivative \( y' \) of \( y = \frac{1}{x} \) is \( y' = -\frac{1}{x^2} \).

Step 2: Write the equation of the tangent line

Using the point-slope form of a line, the equation of the tangent line at any point \((x_0, \frac{1}{x_0})\) is: \( y - \frac{1}{x_0} = -\frac{1}{x_0^2}(x - x_0) \). This simplifies to: \( y = -\frac{1}{x_0^2}x + \frac{2}{x_0} \).

Step 3: Find intercepts of the tangent line

To find the x-intercept, set \( y = 0 \) in the tangent equation: \( 0 = -\frac{1}{x_0^2}x + \frac{2}{x_0} \). Solving for \( x \), we get \( x = 2x_0 \). For the y-intercept, set \( x = 0 \) in the tangent equation: \( y = \frac{2}{x_0} \).

Step 4: Calculate the area of the triangle

The vertices of the triangle are \((2x_0, 0)\), \((0, \frac{2}{x_0})\), and the origin \((0, 0)\). The base of the triangle is \( 2x_0 \) and the height is \( \frac{2}{x_0} \). Therefore, the area \( A \) of the triangle is given by \( A = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2x_0 \times \frac{2}{x_0} = 2 \).

Step 5: Conclusion

Thus, the area of the triangle formed by the tangent line and the coordinate axes is 2 square units, independent of the point of tangency \((x_0, \frac{1}{x_0})\).

Key Concepts

Derivatives

The derivative of a function measures how the function changes as its input changes. This is essential for understanding the behavior of curves and their tangents. When we look at the graph of a function like \( y = \frac{1}{x} \), the derivative helps us determine the slope of the tangent line at any given point on the curve.

• The "slope" is simply a measure of how steep the curve is at any point.
• For \( y = \frac{1}{x} \), the derivative is \( y' = -\frac{1}{x^2} \).

This derivative tells us that, as \( x \) increases, the slope becomes less negative, and conversely, as \( x \) decreases, the slope becomes more negative.Understanding derivatives is crucial because they give us the equation of the tangent. Knowing the equation of the tangent is key to forming triangles with other lines, such as the coordinate axes.

Coordinate Geometry

Coordinate geometry involves plotting points, lines, and curves on a coordinate plane, allowing us to analyze and solve geometric problems algebraically.In our exercise, once we have the derivative, it helps us form a tangent to the curve \( y = \frac{1}{x} \). The equation of the tangent line is derived using the point-slope formula, which is fundamental to coordinate geometry:

• Point-slope form is given by \( y - y_1 = m(x - x_1) \).
• For any point \((x_0, y_0)\) on our curve, \( m \) is \(-\frac{1}{x_0^2}\).
• The tangent line equation simplifies to \( y = -\frac{1}{x_0^2}x + \frac{2}{x_0} \).

Using this form, we can find where this line intersects the coordinate axes. These intercepts are crucial as they form the base and height of the triangle in our area calculation.

Areas of Triangles

The area of a triangle is an invaluable concept in geometry, calculated as \( \frac{1}{2} \times \text{base} \times \text{height} \). By understanding the positions of the triangle's vertices, we can easily compute its area.For our specific problem concerning the tangent line:

• The x-intercept is \((2x_0, 0)\).
• The y-intercept is \((0, \frac{2}{x_0})\).
• The origin \((0, 0)\) is also a vertex.

The triangle thus formed has its base along the x-axis from \(0\) to \(2x_0\) and its height equal to \( \frac{2}{x_0} \). The vertices form a right-angle triangle, simplifying our calculation:\[ A = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2x_0 \times \frac{2}{x_0} = 2 \]This straightforward calculation shows that, no matter where the tangent touches the curve, the area of the resulting triangle remains constant at 2 square units.