Problem 35 - Technical Mathematics with Calculus

Question

A parabolic curve is to be used at a dip in a highway. The road dips \(32.0 \mathrm{m}\) in a horizontal distance of \(125 \mathrm{m}\) and then rises to its previous height in another \(125 \mathrm{m}\) Write the equation of the curve of the roadway, taking the origin at the bottom of the dip and the \(y\) axis vertical.

Step-by-Step Solution

Answer

The equation of the parabolic curve is \(y = \frac{-32}{125^2}x^2\).

Step 1 - Understand the Vertex of the Parabola

Firstly, we identify that the origin is at the bottom of the dip which is the vertex of our parabola. The vertex form of a parabola is given by the equation: \(y = a(x - h)^2 + k\), where \((h,k)\) are the coordinates of the vertex. Since the vertex is at the bottom of the dip and we're choosing the origin (0,0) as the vertex, \(h=0\) and \(k=0\). So the equation simplifies to \(y = ax^2\).

Step 2 - Use Symmetry to Find a Point on the Parabola

The road dips 32.0 m vertically and then rises back to the original height at 125 m horizontally from the vertex, on both left and right sides. So, we have the point \((125, -32)\) on the parabola.

Step 3 - Substitute the Known Point to Find the Coefficient a

We plug the point \((125, -32)\) into the simplified parabola equation to find the coefficient \(a\). Substituting 125 for \(x\) and -32 for \(y\): \(-32 = a(125)^2\). Solving for \(a\) gives us \(a = \frac{-32}{125^2}\).

Step 4 - Write the Final Equation of the Parabola

Having determined \(a\), we plug it back into the simplified parabola's equation \(y = ax^2\). This gives us the final equation of the parabola as \(y = \frac{-32}{125^2}x^2\).

Key Concepts

Vertex Form of a Parabola

Understanding the vertex form of a parabola is essential when dealing with quadratic functions in geometry and algebra. The vertex form is expressed as
\(y = a(x - h)^2 + k\),
where \((h, k)\) represents the vertex, the highest or lowest point of the parabola, and \(a\) determines the direction and width of the parabola.

Finding the Vertex

In our exercise, identifying the vertex is simple because it's given at the bottom of the dip, which is at the origin, thus \(h = 0\) and \(k = 0\). The advantage of having a vertex at the origin is that our equation simplifies significantly to \(y = ax^2\), eliminating the need to calculate for \(h\) and \(k\).

Determining 'a'

The coefficient \(a\) affects the parabola's 'openness'. A larger absolute value of \(a\) makes the parabola narrower, while a smaller absolute value makes it wider. Additionally, if \(a\) is positive, the parabola opens upwards; if negative, it opens downwards. This is critical for our roadway, where we expect the parabola to open upwards after the dip.

Symmetry in Parabolas

Parabolas are symmetrical, meaning one half is a mirror image of the other. The axis of symmetry passes through the vertex and splits the parabola into two congruent halves.

Applying Symmetry

For our highway parabolic curve problem, symmetry is key to understanding the curve's behavior. Knowing the road returns to the original height at the same horizontal distance on both sides allows us to deduce an additional point on the parabola: \((125, -32)\), located 125 m horizontally from the vertex and 32 m below the initial height.
This understanding not only provides us with a necessary point to solve for \(a\), but also confirms the mirrored nature of the parabola, helping us visualize and verify the accuracy of our curve.
Note: The axis of symmetry in this case is the y-axis, since the vertex is at the origin.

Solving Quadratic Equations

Quadratic equations form the backbone of many geometric and algebraic problems, particularly in defining the shape of parabolas. A standard quadratic equation looks like \(ax^2 + bx + c = 0\), with the solutions of the variable \(x\) being the x-intercepts of the parabola.

Setting up Our Equation

In the exercise, we focus on solving for the coefficient \(a\) to derive our parabola's equation. After substituting the known point \((125, -32)\) into the simplified vertex form, we obtain \(a = \frac{-32}{125^2}\)), which when solved gives us the specific 'stretch' factor of our parabola.

Completing the Square

In cases where the vertex isn't at the origin, completing the square is a method used to transform a standard quadratic equation into vertex form. This step 'creates' a perfect square trinomial allowing us to express the equation in a way that readily identifies the vertex and axis of symmetry of the parabola. However, for the given exercise, since our vertex is at the origin, completing the square isn't necessary.
With these concepts in mind, students can better understand not just how to solve the problem but also the geometric properties and applications of quadratic equations shaping the world around us.