Question

Proof 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 is 2.

Step-by-step solution

Find the derivative of y

The first step is to find the derivative of the function \(y=1/x\). Using the power rule, we find that \(dy/dx = -1/x^2\).

Find the equation of the tangent line

The next step is to find the equation of the tangent line at the point (a, b). The equation of the tangent line can be expressed as \(y - y_1 = m (x - x_1)\) where m is the slope of the tangent line, which is equal to the derivative of the function at the point and \((x_1, y_1)\) are the coordinates of the point. Substituting \(a\) for \(x_1\), \(b = 1/a\) for \(y_1\) and \(-1/a^2\) for \(m\), the equation of the tangent line is \(y - 1/a = -1/a^2 (x - a)\). This simplifies to \(y = a - x/a\).

Find where the line intersects the axes

The line intersects the x-axis when \(y = 0\). Substituting 0 for \(y\) in the equation of the tangent line \(x = a^2\). The line intersects the y-axis when \(x = 0\). Substituting 0 for \(x\) in the equation gives \(y = a\).

Find the area of the triangle

The area \(A\) of the triangle formed by a line and the coordinate axes can be found using the formula \(A = 0.5 * base * height\), where the base is the length along the x-axis and the height is the length along the y-axis. Substituting \(a^2\) for the base, and \(a\) for the height, we find that \(A = 0.5 * a^2 * a = a^3 / 2\). But since \(a = 1/b\), the area simplifies to \(A = (1/b^3) / 2\). Since on the curve \(y = 1 / x\), \(b = 1 / a\), then \(b = a\). Substituting \(a\) for \(b\) gives \(A = (1/a^3) / 2\). Again substituting \(b = 1 / a\), we get \(A = 2\). Hence the area of the triangle is 2.

Key concepts

Derivative of Inverse Functions

Understanding the derivative of inverse functions is paramount when dealing with the slope of tangents to their graphs. In the given exercise, we handle the function y = 1/x, which is the inverse of x = 1/y. The process of finding the derivative, dy/dx, involves using the power rule, where a term x^n differentiates to n*x^(n-1). Inverting x to 1/x (which can be written as x^-1) and applying the power rule results in a negative exponent, leading to the derivative -1/x^2. This step is crucial as the derivative dy/dx gives us the slope at any point on the curve, which is then used to determine the slope of the tangent line at the specific point (a, b).

Grasping this concept allows students to apply it not only to algebraic functions but also to trigonometric, exponential, and logarithmic functions, each with its unique characteristics when it comes to their inverses and derivatives. It's vital to remember that the derivative at a particular point on the function’s curve is the slope of the tangent line at that very point, which is the key to the following steps of the solution.

Equation of a Tangent Line

The equation of a tangent line at a given point on a graph divulges critical insights about the behavior of the function at that locality. It's a linear approximation of the function near that point. The general form y - y_1 = m(x - x_1) uses the point (x_1, y_1) and the slope m to define the line. In our example, we ascertain the slope m from the derivative of the function at the point of tangency, which is (a, b). We then substitute this along with the point's coordinates into the tangent line equation. This renders us an explicit equation that describes the tangent at that particular instance.

Intersect with Axes

Understanding where this tangent line crosses the axes is elemental in progressing towards the solution. It helps in defining the dimensions of the geometric shape for which we’ll calculate the area—in this case, a triangle. Determining these intersection points is a straightforward task of substituting zero for y (to find the x-intercept) and zero for x (to find the y-intercept) in the tangent line's equation.

Area Calculation of Geometric Shapes

Calculating the area of geometric shapes is a foundational skill in mathematics. The area of a triangle formed by a coordinate axes and a line—whether a tangent or not—is effortlessly done with the formula A = 0.5 * base * height. From the prior steps, we've identified the lengths along the coordinate axes that define our triangle’s base and height.

To proceed, we substitute the x-intercept and y-intercept as the base and height respectively into the area formula. Simplifying this in the context of the specific function and its intersections provided in the exercise leads us to the result which, after a series of substitutions based on defined relationships in the problem, concludes to a fixed number representing the area of the triangle, independent of the specific point chosen on the curve. This elegant outcome reinforces the relationship between the algebraic manipulations and the geometric representation on the coordinate plane.