Problem 51 - Algebra and Trigonometry

Question

A hill is inclined so that its “slope” is \(\frac{1}{2}\) , as shown in the figure. We introduce a coordinate system with the origin at the base of the hill and with the scales on the axes measured in meters. A rocket is fired from the base of the hill in such a way that its trajectory is the parabola \(y=-x^{2}+401 x\) . At what point does the rocket strike the hillside? How far is this point from the base of the hill (to the nearest \(\mathrm{cm} )\) ?

Step-by-Step Solution

Answer

The rocket strikes at (400.5, 200.25); the distance is approximately 447.21 meters from the base.

Step 1: Write the equation of the hill

The slope of the hill is given as \( \frac{1}{2} \). Thus, the equation of the hill, given it intersects the origin, is \( y = \frac{1}{2}x \).

Step 2: Set up the equation to find the intersection point

We need to find where the trajectory of the rocket, \( y = -x^{2} + 401x \), intersects with the hill, \( y = \frac{1}{2}x \). Set the equations equal to each other: \[ -x^{2} + 401x = \frac{1}{2}x \]

Step 3: Simplify the intersection equation

Rearrange and simplify the equation:\[ -x^{2} + 401x - \frac{1}{2} x = 0 \]Combine the terms:\[ -x^2 + 400.5x = 0 \]

Step 4: Factor the equation

Factor out \(x\) from the equation:\[ x(-x + 400.5) = 0 \]This gives us the solutions \( x = 0 \) and \( x = 400.5 \). Since \( x = 0 \) is the starting point, the intersection is at \( x = 400.5 \).

Step 5: Calculate the y-coordinate of the intersection point

Substitute \( x = 400.5 \) into the equation of the hill, \( y = \frac{1}{2}x \), to find the y-coordinate:\[ y = \frac{1}{2} \times 400.5 = 200.25 \]

Step 6: Determine the straight-line distance from the base

Use the distance formula to calculate the distance from the base at point \((0, 0)\) to \((400.5, 200.25)\):\[ \text{Distance} = \sqrt{(400.5 - 0)^2 + (200.25 - 0)^2} \]\[ \text{Distance} = \sqrt{400.5^2 + 200.25^2} \]Calculating this gives:\[ \text{Distance} \approx 447.21 \text{ meters} \]

Key Concepts

Slope of a Line

The slope of a line is a measure of how steep the line is. It's essentially the ratio of the vertical change to the horizontal change between two points on the line. If you think of driving up a hill, the slope tells you how steep that hill is. When a problem states the slope is \( \frac{1}{2} \), it means that for every 2 meters you move horizontally, the hill rises by 1 meter.
In mathematical terms, if a line passes through points \((x_1, y_1)\) and \((x_2, y_2)\), the slope \(m\) is calculated by \(m = \frac{y_2 - y_1}{x_2 - x_1}\).
In this exercise, the slope helps form the equation of the hill as \(y = \frac{1}{2}x\). A gentle slope results in a small angle with the horizontal axis, while a steep slope gives a larger angle.

Coordinate System

A coordinate system is used to locate points in a plane. It's like a map for positioning things in space. Typically, it's made of two axes, the x-axis (horizontal) and the y-axis (vertical), intersecting at a point called the origin.
In this exercise, the origin is placed at the base of the hill, simplifying calculations by setting the starting point for the rocket launch as \((0, 0)\).
This system provides a way to express the location of any point as coordinates \((x, y)\), where \(x\) is the horizontal distance from the origin and \(y\) is the vertical distance. Such a setup is very useful for plotting curves like the parabola in the rocket's trajectory.”As curves and lines are sketched on this plane using their respective equations, the coordinate system becomes the canvas for solving where and how these lines intersect.

Intersection Points

Intersection points are where two or more graphs meet. They represent solutions shared by each equation. In our example, the rocket's trajectory is described by a quadratic curve and the hill by a straight line.
To find where these intersect, we equate their equations: \(-x^2 + 401x = \frac{1}{2}x\), and solve for \(x\).
Solving becomes a matter of rearranging and simplifying the terms to find values of \(x\) that satisfy both equations. In this problem, the solutions found were \(x = 0\) and \(x = 400.5\). Since the intersection point occurs where \(x = 400.5\), calculations are verified.
Then, substitute \(x = 400.5\) back into either equation to find the corresponding \(y\)-coordinate. This final point \((400.5, 200.25)\) is where the rocket meets the hill.

Distance Formula

The distance formula calculates the straight-line distance between two points on a plane. It's based on the Pythagorean theorem and is expressed as \( \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \).
In this problem, we need to determine how far the intersection point \((400.5, 200.25)\) is from the base of the hill \((0, 0)\). Plugging these coordinates into the formula:

• The difference in the x-coordinates is \(400.5 - 0\), which is \(400.5\).
• The difference in the y-coordinates is \(200.25 - 0\), which is \(200.25\).

Using these, the distance calculation is: \(\sqrt{400.5^2 + 200.25^2} \approx 447.21\) meters.
This result gives a precise measure of the path the rocket takes from its launch to its destination on the hill.