Problem 67 - Calculus Single Variable

Question

For what values of \(y_{0}\) is the distance between the points of intersection of \(y=y_{0}\) with \(y=2 x+1\) and \(y=x+2\) equal to \(1,000,000 ?\)

Step-by-Step Solution

Answer

The values of \(y_0\) are \(-2,000,002\) and \(999,998\).

Step 1: Identify Points of Intersection

First, we need to find the points where the line \(y = y_0\) intersects the lines \(y = 2x + 1\) and \(y = x + 2\). For \(y = 2x + 1\), set \(y_0 = 2x + 1\), which simplifies to \(x = \frac{y_0 - 1}{2}\). For \(y = x + 2\), set \(y_0 = x + 2\), which simplifies to \(x = y_0 - 2\). So, the points are \(\left(\frac{y_0 - 1}{2}, y_0\right)\) and \((y_0 - 2, y_0)\).

Step 2: Calculate Distance Between Points

Use the distance formula to find the distance between the points \(\left(\frac{y_0 - 1}{2}, y_0\right)\) and \((y_0 - 2, y_0)\). The distance formula is \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\). In this case, \(x_1 = \frac{y_0 - 1}{2}\), \(x_2 = y_0 - 2\), \(y_1 = y_2 = y_0\). This simplifies to \(d = \left|y_0 - 2 - \frac{y_0 - 1}{2}\right| = \left| -\frac{y_0 + 2}{2} \right| \).

Step 3: Set Distance Equal to 1,000,000

According to the problem, the distance should be 1,000,000: \[ \left| -\frac{y_0 + 2}{2} \right| = 1,000,000 \]. This leads to two equations: \(-\frac{y_0 + 2}{2} = 1,000,000\) or \(-\frac{y_0 + 2}{2} = -1,000,000\).

Step 4: Solve Equations

Solve \(-\frac{y_0 + 2}{2} = 1,000,000\) by multiplying both sides by -2: \(-y_0 - 2 = 2,000,000\), hence \(y_0 = -2,000,002\). Solve \(-\frac{y_0 + 2}{2} = -1,000,000\) similarly: \(y_0 = 999,998\).

Step 5: Verify Solutions

We found two solutions: \(y_0 = -2,000,002\) and \(y_0 = 999,998\). Substituting these back into the distance formula confirms both give a distance of 1,000,000, satisfying the problem's condition.

Key Concepts

Intersection Points

To solve problems involving intersection points, we first need to understand how to find where two equations meet. This means finding a point that satisfies both equations. In our example, we're trying to find points where the horizontal line equation \(y = y_0\) intersects with two given lines: \(y = 2x + 1\) and \(y = x + 2\).

• For the line \(y = 2x + 1\), setting \(y_0 = 2x + 1\) allows us to solve for \(x\). The solution is \(x = \frac{y_0 - 1}{2}\).
• For the line \(y = x + 2\), setting \(y_0 = x + 2\) gives us \(x = y_0 - 2\).

Thus, these expressions are used to find the coordinates of the intersection points: \(\left(\frac{y_0 - 1}{2}, y_0\right)\) and \((y_0 - 2, y_0)\). Intersection points are key to understanding how different functions relate spatially.

Distance Formula

The distance formula is an essential tool in calculating the distance between two points in a plane. It is derived from the Pythagorean theorem and allows you to find the length of the line segment that connects two points.For two points, \((x_1, y_1)\) and \((x_2, y_2)\), the distance \(d\) is given by:\[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]In our specific problem, checking the distance between the points of intersection is vital. The \(y\)-coordinates are the same for both points, simplifying our distance calculation.

• From point \(\left(\frac{y_0 - 1}{2}, y_0\right)\) to \((y_0 - 2, y_0)\).
• Distance formula simplifies to: \(d = \left|y_0 - 2 - \frac{y_0 - 1}{2}\right| = \left| -\frac{y_0 + 2}{2} \right|\).

Understanding how this tool is applied helps in many areas of problem solving and graphing.

Linear Equations

Linear equations represent straight lines when graphed on a coordinate plane. They are usually expressed in the form \(y = mx + b\), where \(m\) is the slope and \(b\) is the y-intercept. In our problem, the given linear equations, \(y = 2x + 1\) and \(y = x + 2\), describe two lines with distinct slopes and intercepts.

• The equation \(y = 2x + 1\) has a slope of 2, meaning the line rises steeply as \(x\) increases.
• The equation \(y = x + 2\) has a slope of 1, indicating a less steep incline than the first line.

These formulas are straightforward yet powerful, allowing you to understand the orientation and position of lines on a graph. Knowing how to manipulate and understand these equations aids in determining where lines intersect and how they relate.

Problem Solving Steps

Breaking down a complex problem into smaller, manageable steps is crucial for effective problem solving in calculus and beyond. In this particular exercise, you start by finding intersection points, then move through a series of logical steps to find the desired outcome.1. **Identify Points of Intersection:** Determine where the graphs cross each other by equating their formulas.2. **Apply the Distance Formula:** Calculate the exact distance between these points using the established mathematics.3. **Equate for Specific Condition:** Here, set the calculated distance to 1,000,000 to find the appropriate \(y_0\) value.4. **Solve the Equations:** Work through the algebra required to isolate and solve for \(y_0\).5. **Verify Solutions:** Check your result by plugging back into the formula to ensure the conditions are met.Following these structured steps helps in not only solving the problem but also understanding the process and logic behind each step.