Problem 67 - Intermediate Algebra

Question

Archery. See the illustration below. An arrow shot from the base of a hill follows the parabolic path \(y=-\frac{1}{6} x^{2}+2 x,\) with distances measured in meters. The inclined hill has a slope of \(\frac{1}{3}\) and can therefore be modeled by the equation \(y=\frac{1}{3} x\). Find the coordinates of the point of impact of the arrow and then its distance from the archer. (GRAPH CANNOT COPY)

Step-by-Step Solution

Answer

The impact point is \(\left(\frac{10}{3}, \frac{10}{9}\right)\); the distance from the archer is \(\frac{10\sqrt{10}}{9}\) meters.

Step 1: Equate the equations of the paths

Set the equation of the arrow's path equal to the equation of the hill's slope to find the point of impact: \[-\frac{1}{6}x^2 + 2x = \frac{1}{3}x.\] This will give us the x-coordinate where the arrow hits the hill.

Step 2: Rearrange and simplify the equation

Subtract \(\frac{1}{3}x\) from both sides of the equation to form a quadratic equation: \[-\frac{1}{6}x^2 + \frac{5}{3}x = 0.\] Factor out the common factor \(x\): \[x \left(-\frac{1}{6}x + \frac{5}{3}\right) = 0.\]

Step 3: Solve the factored equation

From the factored equation \(x \left(-\frac{1}{6}x + \frac{5}{3}\right) = 0\), we obtain two solutions: 1. \(x = 0\) (the starting point of the arrow),2. \(-\frac{1}{6}x + \frac{5}{3} = 0.\)Solve the second part: \(x = 0\) is trivial, and for the second solution, rearrange to find \(x = \frac{10}{3}\) meters.

Step 4: Find the y-coordinate of the impact point

Substitute \(x = \frac{10}{3}\) into the hill's equation to get the y-coordinate:\[y = \frac{1}{3} \left(\frac{10}{3}\right) = \frac{10}{9}.\] The point of impact is \(\left(\frac{10}{3}, \frac{10}{9}\right).\)

Step 5: Calculate the distance from the archer

Use the distance formula from the origin (the archer's location) to the point of impact:\[d = \sqrt{\left(\frac{10}{3}\right)^2 + \left(\frac{10}{9}\right)^2}.\] Simplify:\[d = \sqrt{\frac{100}{9} + \frac{100}{81}} = \sqrt{\frac{900}{81} + \frac{100}{81}} = \sqrt{\frac{1000}{81}} = \frac{10\sqrt{10}}{9}\text{ meters}.\]

Key Concepts

Quadratic Equation

A quadratic equation is a fundamental concept in algebra that represents a polynomial of degree 2. It typically has the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \). Quadratic equations can be solved using various methods:

• Factoring: Express the quadratic equation as a product of two binomial expressions.
• Completing the Square: Convert the quadratic into a perfect square trinomial.
• Quadratic Formula: Use the formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) to find the roots.

Quadratic equations often model parabolic motion, like an arrow's path in our exercise. The arrow's path equation is \( y = -\frac{1}{6}x^2 + 2x \), which shows its parabolic trajectory. Understanding the shape and position of a parabola can help predict where the arrow will hit corresponding features like a hill.
The process of finding where the arrow hits the hill involves setting this quadratic equation equal to the linear equation of the hill. This allows us to solve for \( x \), giving us the exact point where the parabolic path intersects with another path.

Slope-Intercept Form

The slope-intercept form is a straightforward way to represent linear equations. The equation is written as \( y = mx + b \), where \( m \) represents the slope and \( b \) the y-intercept. The slope \( m \) indicates the inclination or steepness of the line, and the intercept \( b \) marks where the line crosses the y-axis.
In our archery exercise, the slope-intercept form equation of the hill is \( y = \frac{1}{3}x \). Here, the slope \( m = \frac{1}{3} \), meaning for every unit increase in \( x \), \( y \) increases by \( \frac{1}{3} \). This gives the hill a gentle incline.

• The slope-intercept form is beneficial for easily graphing a line since it provides clear information about its slope and position.
• It allows us to rapidly find the y-coordinate for any x-value.

By using the slope-intercept form, we equated the hill's line with the arrow's path to find their intersection. The point of intersection reveals the coordinates where the arrow impacts the hill, seamlessly combining the quadratic curve with a linear slope.

Distance Formula

The distance formula is used to calculate the straight-line distance between two points in a plane. For points \( (x_1, y_1) \) and \( (x_2, y_2) \), the formula is written as:\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]This powerful tool helps determine how far one point is from another.
In our exercise, we applied the distance formula to find how far the point where the arrow hits the hill is from the archer's starting position (origin). The impact coordinates are \( \left(\frac{10}{3}, \frac{10}{9}\right) \), and the archer is at the origin \((0, 0)\). The formula then becomes:\[ d = \sqrt{\left(\frac{10}{3}\right)^2 + \left(\frac{10}{9}\right)^2} \]

• This calculation gives a precise distance, ensuring accurate results.
• Using the distance formula is essential for solving problems involving vertical, horizontal, or diagonal measurements.

Understanding this formula ensures that the solutions are grounded in proper geometric reasoning, as seen when calculating the arrow's traveled distance on its parabolic path.