Question

Let \((a, b)\) be an arbitrary point on the graph of \(y=1 / x, x>0\). Prove that the area of the triangle formed by the tangent line through \((a, b)\) and the coordinate axes is 2.

Short answer

The area of the triangle formed by the tangent line to the curve through (a, b) and the coordinate axes is 2.

Step-by-step solution

Find the derivative of the function

The derivative of \(y = 1/x\) is \(y' = -1/x^2\). At point \((a, b)\), the slope of the tangent line is \(-1/a^2\). Thus, the equation of the tangent line can be expressed in slope-point format as \( y - b = -1/a^2(x-a) \).

Find the x and y intercepts of the tangent line

Setting y to 0 in our tangent line equation to find the x-intercept gives us \(a - ab = x\), which simplifies to \(x = a(1-b)\). For the y-intercept, we set x = 0 in the equation, and we get \(y = b + 1/a^2 * a\), which simplifies to \(y = b + a\). However, from the graph's equation, we have \(b = 1/a\). Substituting this into the y-intercept equation gives us \(y = 1/a + a = (1 + a^2) / a\). Simplifying this with \(a=1/b\) we obtain \(y = b + 1/b\).

Calculate the triangle area

We can use the concept that the area, A, of a triangle is given by the formula A = 0.5 * base * height. Here, the base is the x-intercept and the height is the y-intercept. Substituting these values in, we get \(A = 0.5 * a(1-b) * (b + 1/b)\). Substituting \(b=1/a\) simplifies the equation to \(A = 0.5 * a * (1-1/a) * (1/a + a)\). Further simplification yields \(A = 0.5 * (1 + a^2 - 1)\), thus \(A= 2\).

Key concepts

Tangent Line Calculation

Understanding how to calculate a tangent line to a function is critical in calculus. A tangent line is a straight line that just touches a curve at a given point without crossing it. To calculate the equation of a tangent line to a curve at a given point \( (a, b) \), you first need to find the derivative of the function at that point. This derivative represents the slope of the tangent line.

Once you have the slope \( m = f'(a) \), you can apply the point-slope form of the equation of a line, which is \( y - b = m(x - a) \). This equation uses the known point on the function \( (a, b) \) and the slope \( m \) to define the line.

Next, to find specific points such as the x and y intercepts of the tangent line, you set \( y \) to 0 to find the x-intercept, and \( x \) to 0 to find the y-intercept. These intercepts reveal where the tangent touches the axes, which is essential when finding the area of the corresponding triangle formed.

Derivative of a Function

The derivative of a function represents the rate of change of the function with respect to its independent variable. This concept is a foundational stone in calculus and is applied in various fields, from physics to economics.

To find the derivative of a simple function like \( y = 1/x \) at an arbitrary point \( (a, b) \), you can apply the power rule. The derivative, denoted as \( y' \) or \( f'(x) \) is then used to determine the behavior and characteristics of the function at that point.

Derivative Calculation Example

For \( y = 1/x \) the derivative is \( y' = -1/x^2 \). This negative sign indicates that the function is decreasing at that point. At \( (a, b) \) specifically, the slope of the tangent line (the derivative at \( a \) will be \( -1/a^2 \) which provides essential information about the tangent line and is used to find the equation of the tangent as previously described.

Area of a Triangle

Triangles are basic geometric shapes, and understanding how to calculate their area is fundamental. The area of a triangle is generally given by \( A = 0.5 * \text{base} * \text{height} \), where the base and height are perpendicular to each other.

When a tangent line creates a triangle with the coordinate axes, the intercepts found from the tangent line equation serve as the base and height. The x-intercept represents the base of the triangle formed along the x-axis, while the y-intercept represents the height along the y-axis.

Application of Area Formula

In our calculus proof exercise, by inputting the x and y intercepts as the base and height into the area formula, we validate that the area of our specific triangle is indeed 2. This process not only demonstrates geometric principles in a calculus context but also provides insight into how the abstract concepts of calculus have tangible geometric implications.