Problem 348 - Calculus Volume 1

Question

For the following exercises, consider the points on the given graphs. Use a calculator to graph the functions. Where is the line \(y=5-2 x\) closest to point \((1,1) ?\)

Step-by-Step Solution

Answer

The point on the line closest to \((1,1)\) is \((1.8, 1.4)\).

Step 1: Understanding the Problem

We need to find the point on the line \(y = 5 - 2x\) that is closest to the point \((1,1)\) by minimizing the distance between a point \((x, 5 - 2x)\) on the line and the point \((1, 1)\).

Step 2: Distance Formula Setup

The distance between the point \((x, 5 - 2x)\) on the line and \((1, 1)\) is given by \(d = \sqrt{(x - 1)^2 + (5 - 2x - 1)^2}\).

Step 3: Simplifying the Distance Expression

Simplify the distance formula: \(d = \sqrt{(x - 1)^2 + (4 - 2x)^2}\). This reduces to \(d = \sqrt{(x-1)^2 + (4-2x)^2}\).

Step 4: Squaring the Distance

For simplicity, minimize the squared distance: \(d^2 = (x - 1)^2 + (4 - 2x)^2\).

Step 5: Expand the Squared Distance

Expand \(d^2 = (x - 1)^2 + (4 - 2x)^2\) to get \(d^2 = x^2 - 2x + 1 + 16 - 16x + 4x^2\).

Step 6: Combine Like Terms

Combine like terms: \(d^2 = 5x^2 - 18x + 17\). This is a quadratic equation in \(x\).

Step 7: Find Critical Points

Find the minimum by taking the derivative \(d^2 = 5x^2 - 18x + 17\), which gives \(d'(x) = 10x - 18\). Set \(d'(x) = 0\) to find \(x = \frac{18}{10} = 1.8\).

Step 8: Calculate Closest Point

Substitute \(x = 1.8\) back into the line equation \(y = 5 - 2x\) to find \(y = 5 - 2(1.8) = 1.4\). So the point is \((1.8, 1.4)\).

Step 9: Conclusion

Thus, the point on the line \(y = 5 - 2x\) closest to \((1, 1)\) is \((1.8, 1.4)\).

Key Concepts

Distance Minimization

When working on problems related to identifying the closest point on a curve or line to a specific point in the Cartesian plane, you are engaging in distance minimization. This concept is crucial in understanding how to analyze geometric relationships effectively. The task is to find a point on one entity (like a line) that is closest to another point.
Distance minimization typically involves parameterizing a line or curve and then applying the distance formula to this in order to create an expression for distance. This expression is often minimized by setting its derivative to zero. By turning the challenge into one of minimizing an expression mathematically, students can hone their problem-solving skills efficiently.

• Identify all the points involved.
• Formulate the distance between these points using the distance formula.
• Simplify when possible, and focus on minimizing the expression.

Quadratic Equations

Understanding quadratic equations is key when dealing with optimization problems like finding minimum distance. In the context of distance minimization, the squared distance often creates a quadratic equation. These equations take the form:
\[ ax^2 + bx + c = 0 \]Where \(a\), \(b\), and \(c\) are constants by collecting like terms from the distance expression.
Quadratic equations have specific properties, such as symmetry and a parabolic shape when graphed, that makes them predictable and solvable efficiently. Solving these either by factoring, completing the square, or using the quadratic formula helps find the roots corresponding to critical points in the problem.

• Gather coefficients by adding like terms from your simplified expression.
• Recognize the general structure of a quadratic for easier processing.
• Apply methods like the quadratic formula if needed.

Derivative Calculations

Derivative calculations are an essential tool used to find minimum or maximum points on graphs. By taking the derivative of a function, you determine its rate of change, which can help locate critical points, where there may be a transition in increase or decrease—essential for optimizing problems.
In this context, you start with your quadratic expression of the squared distance. Calculating its derivative involves using basic differentiation rules:
\[ ext{For } f(x) = ax^2 + bx + c, ext{ derivative is } f'(x) = 2ax + b \]Setting the derivative equal to zero allows you to solve for \(x\), pinpointing where the distance is minimized.

• Understand that the derivative finds slope, guiding you to levels where change stops.
• Always set your derivative equal to zero to find critical points.
• Check the second derivative, if necessary, to confirm minima or maxima.

Geometry of Lines

The geometry of lines is foundational in problems like determining the shortest distance from a point to a line. Lines in geometry can be expressed in various forms, such as the slope-intercept form, which is being used here:
\[ y = mx + b \]Where \(m\) is the slope and \(b\) is the y-intercept.
Understanding linear functions means knowing how changes in \(x\) affect \(y\), ultimately allowing you to characterize all points along that line. Such insights help in plotting accurate graphs, which assist in visual verifications of solutions. Calculating intersection points or understanding their relative positioning further aids in solving minimization problems effectively.

• Interpret key characteristics of the line, like slope and intercept.
• Visualize geometric positioning on the plane for greater clarity.
• Calculate specific points by substituting values into the line equation.