Math Problem Statement

The question concerns modeling and calculating the properties of the cable curve of the Golden Gate Bridge using a quadratic function.

Solution

The question concerns modeling and calculating the properties of the cable curve of the Golden Gate Bridge using a quadratic function. Let's break this into parts:


Given:

  1. Function form: f(x)=ax2+bx+cf(x) = ax^2 + bx + c (quadratic polynomial for the cable's shape).

  2. Key Points:

    • Anchor points at (0,220)(0, 220) and (1280,220)(1280, 220) (cable starts and ends 220 m above the water).
    • The lowest point of the cable is at (640,80)(640, 80), where f(640)=80f(640) = 80.
  3. Cable length formula:
    The arc length of f(x)f(x) from x1x_1 to x2x_2 is given by: Arc Length=x1x21+(f(x))2dx\text{Arc Length} = \int_{x_1}^{x_2} \sqrt{1 + \left( f'(x) \right)^2} \, dx


(a) Find the equation for f(x)f(x):

The equation f(x)=ax2+bx+cf(x) = ax^2 + bx + c has three unknowns (a,b,ca, b, c). We use the given points to form equations:

  1. At (0,220)(0, 220): f(0)=220    c=220f(0) = 220 \implies c = 220

  2. At (1280,220)(1280, 220): f(1280)=220    a(1280)2+b(1280)+220=220f(1280) = 220 \implies a(1280)^2 + b(1280) + 220 = 220 Simplifies to: 1638400a + 1280b = 0 \tag{1}

  3. At (640,80)(640, 80): f(640)=80    a(640)2+b(640)+220=80f(640) = 80 \implies a(640)^2 + b(640) + 220 = 80 Simplifies to: 409600a + 640b = -140 \tag{2}


Solve for aa and bb:

From (1)(1):
b = -1280a \tag{3}

Substitute b=1280ab = -1280a into (2)(2): 409600a+640(1280a)=140409600a + 640(-1280a) = -140 409600a819200a=140409600a - 819200a = -140 409600a=140    a=140409600=720480-409600a = -140 \implies a = \frac{-140}{-409600} = \frac{7}{20480}

Substitute a=720480a = \frac{7}{20480} into (3)(3): b=1280720480=896020480=716b = -1280 \cdot \frac{7}{20480} = -\frac{8960}{20480} = -\frac{7}{16}

Thus: f(x)=720480x2716x+220f(x) = \frac{7}{20480}x^2 - \frac{7}{16}x + 220


(b) Calculate the cable length:

The arc length is: L=012801+(f(x))2dxL = \int_0^{1280} \sqrt{1 + \left( f'(x) \right)^2} \, dx

  1. Derivative f(x)f'(x): f(x)=2720480x716=710240x716f'(x) = 2 \cdot \frac{7}{20480}x - \frac{7}{16} = \frac{7}{10240}x - \frac{7}{16}

  2. Simplify 1+(f(x))21 + (f'(x))^2: 1+(f(x))2=1+(710240x716)21 + \left( f'(x) \right)^2 = 1 + \left( \frac{7}{10240}x - \frac{7}{16} \right)^2

This integral can now be evaluated numerically to find the cable length.


Would you like a detailed calculation of the integral or any clarification?

Related Questions:

  1. How does a quadratic polynomial model a parabolic cable shape accurately?
  2. What assumptions are made in calculating arc length of a cable?
  3. How do you derive f(x)f'(x) and compute arc length step by step?
  4. How does the placement of anchor points affect the shape of f(x)f(x)?
  5. What is the impact of numerical errors in approximating cable length?

Tip: Always double-check units when working with physical models to ensure consistency across calculations!

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Math Problem Analysis

Mathematical Concepts

Quadratic Functions
Arc Length
Differentiation
Calculus

Formulas

f(x) = ax^2 + bx + c
Arc Length = ∫(x1 to x2) √(1 + (f'(x))^2) dx
f'(x) = 2 * (7/20480)x - 7/16

Theorems

Arc Length Theorem
Differentiation of Quadratic Functions

Suitable Grade Level

Grades 11-12