Problem 76
Question
Suspension bridge cables hang in parabolas The suspension bridge cable shown here supports a uniform load of \(w\) pounds per horizontal foot. It can be shown that if \(H\) is the horizontal tension of the cable at the origin, then the curve of the cable satisfies the equation $$\frac{d y}{d x}=\frac{w}{H} x$$ Show that the cable hangs in a parabola by solving this differential equation subject to the initial condition that \(y=0\) when \(x=0\) . (GRAPH NOT COPY)
Step-by-Step Solution
Verified Answer
The cable hangs in the shape of a parabola described by \(y = \frac{w}{2H}x^2\).
1Step 1: Understand the Differential Equation
The problem gives us a differential equation \(\frac{dy}{dx} = \frac{w}{H}x\). We need to solve this differential equation to find the function \(y(x)\). The solution will show the shape of the cable, which is expected to be a parabola.
2Step 2: Integrate the Differential Equation
Integrate both sides of the equation \(\frac{dy}{dx} = \frac{w}{H}x\) with respect to \(x\). \[\int \frac{dy}{dx} \, dx = \int \frac{w}{H} x \, dx\]This results in:\[y = \frac{w}{2H}x^2 + C\]where \(C\) is the constant of integration.
3Step 3: Apply the Initial Condition
Use the initial condition given in the problem: \(y = 0\) when \(x = 0\). Substitute these values into the equation:\[0 = \frac{w}{2H}(0)^2 + C\]This shows that \(C = 0\).
4Step 4: Write the Final Solution
Substitute \(C = 0\) back into the equation from Step 2. The final equation is:\[y = \frac{w}{2H}x^2\]This equation is a parabola, showing that the cable hangs in a parabolic shape.
Key Concepts
Differential EquationsSuspension BridgesIntegration Techniques
Differential Equations
Differential equations are powerful tools in mathematics that relate a function to its derivatives. This means they describe the rate of change of a quantity and often appear in problems involving motion, growth, or any dynamic systems.
In the context of suspension bridges, we're dealing with a differential equation that models the cable's shape. Specifically, the equation \( \frac{dy}{dx} = \frac{w}{H}x \) expresses how the slope of the cable changes along its length. Solving this equation reveals the geometry of the cable, which in this case, forms a parabola.
Understanding differential equations involves knowing how to manipulate functions and derivatives to find unknowns like shape or trajectory. It's an essential skill for engineers, physicists, and mathematicians who model real-world phenomena.
In the context of suspension bridges, we're dealing with a differential equation that models the cable's shape. Specifically, the equation \( \frac{dy}{dx} = \frac{w}{H}x \) expresses how the slope of the cable changes along its length. Solving this equation reveals the geometry of the cable, which in this case, forms a parabola.
Understanding differential equations involves knowing how to manipulate functions and derivatives to find unknowns like shape or trajectory. It's an essential skill for engineers, physicists, and mathematicians who model real-world phenomena.
Suspension Bridges
Suspension bridges are marvels of engineering, known for their distinct shape and efficiency in spanning large distances. A key component of these bridges is the cable, which hangs between two towers and supports the bridge deck.
The curve formed by the cable under uniform load is of particular interest. This curve is not just any curve; it is a parabola, as shown through mathematical modeling. The uniform load per horizontal foot \(w\) and the horizontal tension \(H\) play significant roles in determining this shape.
The parabolic nature of suspension bridge cables offers several advantages:
The curve formed by the cable under uniform load is of particular interest. This curve is not just any curve; it is a parabola, as shown through mathematical modeling. The uniform load per horizontal foot \(w\) and the horizontal tension \(H\) play significant roles in determining this shape.
The parabolic nature of suspension bridge cables offers several advantages:
- It distributes weight evenly, providing stability.
- It allows materials to be used efficiently, reducing costs.
- It enhances the aesthetic appeal of the bridge.
Integration Techniques
Integration is the process of finding the antiderivative of a function. In simpler terms, it's about finding the function's area under its curve or, as in our case, determining the shape of a curve based on its rate of change.
Here, the differential equation \( \frac{dy}{dx} = \frac{w}{H}x \) required integration to find the cable's equation \( y = \frac{w}{2H}x^2 + C \). This involves:
Here, the differential equation \( \frac{dy}{dx} = \frac{w}{H}x \) required integration to find the cable's equation \( y = \frac{w}{2H}x^2 + C \). This involves:
- Recognizing the form of the equation, which is linear in \(x\).
- Integrating \( \frac{w}{H}x \) to get \( \frac{w}{2H}x^2 \), illustrating the technique of basic polynomial integration.
- Applying initial conditions to solve for constants, in this case, setting \( C = 0 \) when \( x = 0 \).
Other exercises in this chapter
Problem 73
Sketch the regions in the \(x y\) -plane whose coordinates satisfy the inequalities or pairs of inequalities in Exercises \(69-74 .\) $$ 4 y^{2}-x^{2} \geq 4 $$
View solution Problem 75
Archimedes 'formula for the volume of a parabolic solid The region enclosed by the parabola \(y=\left(4 h / b^{2}\right) x^{2}\) and the line \(y=h\) is revolve
View solution Problem 77
Find an equation for the circle through the points \((1,0),(0,1),\) and \((2,2) .\)
View solution Problem 78
Find an equation for the circle through the points \((2,3),(3,2),\) and \((-4,3) .\)
View solution