Problem 58 - Calculus - AP Edition

Question

Show that the segment cut off by the coordinate axes from any tangent line to the graph of \(y=1 / x\) is bisected by the point of tangency.

Step-by-Step Solution

Answer

The segment is bisected by the point of tangency \((a, \frac{1}{a})\).

Step 1: Find the Derivative

Start by finding the derivative of the function \(y = \frac{1}{x}\). The derivative, \(y'\), can be found using the power rule. Rewriting \(y = x^{-1}\), we have \(y' = -x^{-2} = -\frac{1}{x^2}\). This derivative represents the slope of the tangent line at any point on the graph.

Step 2: Equation of the Tangent Line

To find the equation of the tangent line at a point \((a, \frac{1}{a})\), use the point-slope form of a line equation: \(y - y_1 = m(x - x_1)\). The slope \(m\) is the derivative evaluated at \(x = a\), which is \(-\frac{1}{a^2}\). Hence, the equation becomes: \(y - \frac{1}{a} = -\frac{1}{a^2}(x - a)\). Simplifying gives the line equation as \(y = -\frac{1}{a^2}x + \frac{2}{a}\).

Step 3: Determine Where the Tangent Line Cuts the Axes

Set \(y = 0\) to find the x-intercept. Solving \(0 = -\frac{1}{a^2}x + \frac{2}{a}\), we find the x-intercept \(x = 2a\). Next, set \(x = 0\) to find the y-intercept. Solving \(y = \frac{2}{a}\), the y-intercept is \(y = \frac{2}{a}\). Therefore, the x and y-intercepts are \((2a, 0)\) and \(\left(0, \frac{2}{a}\right)\) respectively.

Step 4: Midpoint of the Segment Between Intercepts

Calculate the midpoint of the segment between the intercepts. The midpoint formula \( \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) \) gives the midpoint between \((2a, 0)\) and \(\left(0, \frac{2}{a}\right)\) as \(\left(\frac{2a + 0}{2}, \frac{0 + \frac{2}{a}}{2}\right) = (a, \frac{1}{a})\).

Step 5: Compare with the Point of Tangency

The point of tangency was initially given as \((a, \frac{1}{a})\). This matches the midpoint of the segment calculated in the previous step, confirming that the point of tangency bisects the segment cut off by the axes.

Key Concepts

Derivative

In calculus, one of the key concepts is the derivative, which measures how a function changes as its input changes. For the function given as \(y = \frac{1}{x}\), the derivative can be found by rewriting it in power form: \(y = x^{-1}\). This allows us to use the power rule, easy for calculating derivatives. Using the power rule, which is \(\frac{d}{dx}(x^n) = nx^{n-1}\), the derivative of \(x^{-1}\) becomes \(-x^{-2}\). In standard form, this is \(-\frac{1}{x^2}\). The derivative here represents the slope of the tangent line at any point on the curve. It's vital to understand that this slope indicates how steeply the curve rises or falls at that specific point.

• This derivative urges us to consider the concept of instantaneous rate of change.
• It helps in understanding the behavior of the curve.
• Different derivatives at different points lead to different tangent lines.

Tangent Line

The tangent line is like the shadow of the curve, just touching it without crossing at a particular point, depicting the curve's immediate direction at that point. Finding the equation of a tangent line is straightforward using the point-slope form: \(y - y_1 = m(x - x_1)\), where \(m\) is the slope from the derivative.In the exercise's context, the slope \(m\) is evaluated at point \(a\), so \(-\frac{1}{a^2}\) becomes the slope. The tangent point on the graph of \(y = \frac{1}{x}\) is \((a, \frac{1}{a})\). Substitute and simplify the expression: \( y - \frac{1}{a} = -\frac{1}{a^2}(x - a)\), resulting in a linear equation: \(y = -\frac{1}{a^2}x + \frac{2}{a}\).

• The tangent line's slope aligns with the behavior of the curve at that point.
• Understanding tangent lines is crucial for approximating values and predicting trends.

Coordinate Axes

The coordinate axes consist of two perpendicular lines on a graph: the x-axis and y-axis, forming a plane to visualize equations. In a typical two-dimensional space:

• The x-axis runs horizontally.
• The y-axis runs vertically.
• The intersection of these axes is the origin, \((0,0)\).

Understanding how lines interact with these axes gives us essential insight into mathematical relationships. For instance, when analyzing tangent lines' intersections with axes, we're essentially identifying intercepts that plot the segment the line cuts on the coordinate space. These relations help visualize and solve multi-step problems efficiently.

X-intercept

The x-intercept is where a line or curve crosses the x-axis on a graph, showing the point where the equation's output \(y\) is zero. This is computed by setting the equation equal at \(y = 0\) and solving for \(x\).In the exercise, the tangent line equation \(y = -\frac{1}{a^2}x + \frac{2}{a}\) is used. Set \(y = 0\), solve for \(x\): \(0 = -\frac{1}{a^2}x + \frac{2}{a}\), which simplifies to \(x = 2a\).

• X-intercepts are vital in understanding where a function changes behaviors.
• The x-intercept is a foundational concept in graphing techniques.
• Helps determine the reach or "spread" of graphs along the x-axis.

Y-intercept

Similarly, the y-intercept is where the graph intersects the y-axis, marking the point where \(x = 0\). This illustrates how high or low the line first appears on the vertical scale when reading the graph. Using the same tangent line equation: \(y = -\frac{1}{a^2}x + \frac{2}{a}\), setting \(x = 0\), we quickly find the y-intercept as \( \frac{2}{a}\).

• Y-intercepts are crucial for initial value problems in real-world applications.
• Useful for identifying vertical shifts in functions.
• Aids in sketching and understanding the overall shape of graphs.