Problem 348 - Calculus Volume 1

Question

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 line is closest to (1.8, 1.4).

Step 1: Understanding the Problem

We need to find the point on the line represented by the equation \(y = 5 - 2x\) that is closest to the point (1,1). This involves determining the minimum distance between the line and the point.

Step 2: Distance Formula Setup

The distance \(d\) between a point \((x_1, y_1)\) and another point on the line \((x, y)\) is given by \(d = \sqrt{(x - x_1)^2 + (y - y_1)^2}\). In our problem, \((x_1, y_1) = (1,1)\) and \(y = 5 - 2x\). Substitute \(y\) as \(5 - 2x\): \[ d = \sqrt{(x - 1)^2 + (5 - 2x - 1)^2} = \sqrt{(x -1)^2 + (4 - 2x)^2}. \]

Step 3: Simplify the Distance Expression

Simplify \((4 - 2x)^2\): \[ (4 - 2x)^2 = 16 - 16x + 4x^2. \]Thus, \[ d^2 = (x - 1)^2 + (16 - 16x + 4x^2). \]

Step 4: Complete the Expanded Form

Expand \((x - 1)^2\): \[ (x - 1)^2 = x^2 - 2x + 1. \]Substitute back into the equation:\[ d^2 = x^2 - 2x + 1 + 16 - 16x + 4x^2. \]Combine terms to get: \[ d^2 = 5x^2 - 18x + 17. \]

Step 5: Find the Minimum Distance

To find the \(x\) where \(d^2\) is minimized, take the derivative and set it to zero.\[ \frac{d}{dx}(5x^2 - 18x + 17) = 10x - 18. \]Set the derivative equal to zero:\[ 10x - 18 = 0 \]Solve for \(x\): \[ x = \frac{18}{10} = 1.8. \]

Step 6: Calculate the Corresponding Y-Value

Use \(x = 1.8\) in the line equation to find \(y\):\[ y = 5 - 2(1.8) = 5 - 3.6 = 1.4. \]

Step 7: Conclusion

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

Key Concepts

Line Equation

A line equation is a mathematical expression that describes all the points lying on a straight line. In our exercise, the line equation is given by \( y = 5 - 2x \). This equation defines a linear relationship between \( x \) and \( y \), where for any value of \( x \), we can find the corresponding \( y \) value on the line. The slope of this line is \(-2\), which indicates that for every increase of 1 in \( x \), \( y \) decreases by 2.
The y-intercept is at \( y = 5 \), where the line crosses the y-axis. To visualize the line, you could plot at least two points using the equation and draw a line through them. For example:

• When \( x = 0 \), \( y = 5 - 2(0) = 5 \).
• When \( x = 1 \), \( y = 5 - 2(1) = 3 \).

Connecting these points will create the complete line on a graph.

Distance Formula

The distance formula is used to find the distance between two points in a coordinate plane. This formula helps us establish the exact length of the line segment between given points. For our exercise, it is crucial because we want to calculate the distance from a point \((1,1)\) to any point on the line \( y = 5 - 2x \).
The distance between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \). Here, \((x_2, y_2)\) will be replaced by the points on our line \((x, 5 - 2x)\).
For our specific problem, the formula adjusts to:

• \( d = \sqrt{(x - 1)^2 + (5 - 2x - 1)^2} = \sqrt{(x -1)^2 + (4 - 2x)^2} \).

Simplifying this formula leads to our expression for \( d^2 \), which is much easier to differentiate when we look for the minimum distance.

Derivatives

Derivatives are a fundamental concept in calculus that help us find the rate at which a quantity changes. In the context of optimization problems like ours, derivatives help us determine the minimum or maximum values of a function.
In our exercise, we first derived the expression for \( d^2 \) in terms of \( x \):

• \( d^2 = 5x^2 - 18x + 17 \).

To find where this distance is minimized, we calculate the derivative of \( d^2 \) with respect to \( x \), not \( d \) itself, to avoid dealing with square roots. The derivative is:

• \( \frac{d}{dx}(5x^2 - 18x + 17) = 10x - 18 \).

Setting the derivative equal to zero gives us the point where the function doesn't increase or decrease, indicating a minimum or maximum value. Solving \( 10x - 18 = 0 \) provides the critical point \( x = 1.8 \). This is where the distance from our point to the line is shortest.

Minimum Distance

Finding the minimum distance between a point and a line is the ultimate goal of this exercise. After deriving the equation and its squared distance expression, and solving for the point where this distance is minimized, we identified \( x = 1.8 \) as the critical point. This means at \( x = 1.8 \), the perpendicular distance from the point \((1,1)\) to the line \( y = 5 - 2x \) is at its shortest.
Next, we use \( x = 1.8 \) to find the corresponding y-value on the line:

• \( y = 5 - 2(1.8) = 1.4 \).

So, the coordinates of the point on the line nearest to (1,1) are \((1.8, 1.4)\). This results in the minimum distance between the point and the line, solving our optimization problem. Understanding the relationship between the line's equation and distance calculations leads us to this optimized solution.