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4. The profile of the cables of a suspension bridge may be modeled by a parabola. The centr span of the Golden Gate Bridge is 1280 meters long and 152 meters high. The parabola \( y \) \( 0.00037 x^{2} \) gives a good fit to the shape of the cables, where \( |x| \leq 640 \), and \( x \) and \( y \) are measured in meters. Find the definite integral that gives the length of the cables that stretch between the tops of the two towers (set-up but do not solve).

Ask by Carlson Whittaker. in the United States
Jan 25,2025

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The length of the cables can be found using the integral: \[ \int_{-640}^{640} \sqrt{1 + (0.00074x)^2} \, dx \]

Solución

To find the length of the cables modeled by the parabola \( y = 0.00037x^2 \) between the two towers of the Golden Gate Bridge, we'll use the formula for the arc length of a curve defined by a function \( y = f(x) \). **Arc Length Formula:** \[ L = \int_{a}^{b} \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \] **Step-by-Step Setup:** 1. **Identify the Function and Its Derivative:** \[ y = 0.00037x^2 \] \[ \frac{dy}{dx} = 2 \times 0.00037x = 0.00074x \] 2. **Square the Derivative:** \[ \left( \frac{dy}{dx} \right)^2 = (0.00074x)^2 = 0.0005476x^2 \] 3. **Set Up the Integral Limits:** The span is from \( x = -640 \) meters to \( x = 640 \) meters. 4. **Write the Definite Integral for Arc Length:** \[ L = \int_{-640}^{640} \sqrt{1 + 0.0005476x^2} \, dx \] **Final Integral Expression:** \[ \boxed{ \int_{-640}^{640} \sqrt{1 + \left(0.00074\,x\,\right)^2} \; dx } \] This integral represents the total length of the suspension bridge cables between the two tower tops. To find the numerical value of the length, you would evaluate this definite integral.

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To find the length of the cables stretching between the tops of the two towers of the Golden Gate Bridge, we need to set up the definite integral for the arc length of the parabola described by the function \( y = 0.00037x^2 \). The formula for the arc length \( L \) of a curve from \( a \) to \( b \) is given by: \[ L = \int_{a}^{b} \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \] First, we need to find \( \frac{dy}{dx} \) for our parabola: \[ \frac{dy}{dx} = 0.00074x \] Now substitute this into the arc length formula. If we are considering the length of the cables from the left tower at \( x = -640 \) meters to the right tower at \( x = 640 \) meters, our definite integral will be: \[ L = \int_{-640}^{640} \sqrt{1 + (0.00074x)^2} \, dx \] This integral represents the total length of the cables stretching between the tops of the two towers.

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