A barn is 100 ft long and 40 feet wide. A cross section of the roof is the inverted catenary
y=31-10(e^x/20+e^-x/20). Find the number of square feet of roofing of the barn.
I assume roofing of the barn is length of catenary * wide of the barn:
\(\begin{array}{|rcll|} \hline f(x) = y &=& 31-10\cdot \left( e^{\frac{x}{20}} + e^{-\frac{x}{20}} \right) \quad & |\quad \left( e^{\frac{x}{20}} + e^{-\frac{x}{20}} \right) = 2\cdot \cosh(\frac{x}{20}) \\ &=& 31-10\cdot \left[ 2\cdot \cosh(\frac{x}{20}) \right] \\ f(x) &=& 31-20\cdot \cosh(\frac{x}{20}) \\\\ f'(x) &=& -20\cdot \sinh(\frac{x}{20}) \cdot \frac{1}{20} \\ f'(x) &=& - \sinh(\frac{x}{20}) \\\\ [f'(x)]^2 &=& \sinh^2(\frac{x}{20}) \\ \hline \end{array} \)
The graph of the function f(x):
The length L of the catenary is:
\(\begin{array}{|rcll|} \hline L &=& \int \limits_{-50}^{50} \sqrt{1+[f'(x)]^2}\ dx \\ &=& \int \limits_{-50}^{50} \sqrt{1+\sinh^2(\frac{x}{20})}\ dx \quad & |\quad 1+\sinh^2(\frac{x}{20}) = \cosh^2(\frac{x}{20}) \\ &=& \int \limits_{-50}^{50} \sqrt{\cosh^2(\frac{x}{20})}\ dx \\ &=& \int \limits_{-50}^{50} \cosh(\frac{x}{20})\ dx \\ &=& [~ 20\cdot \sinh(\frac{x}{20}) ~]_{-50}^{50} \\ &=& 20\cdot [~ \sinh(\frac{x}{20}) ~]_{-50}^{50} \\ &=& 20\cdot [~ \sinh(\frac{50}{20})-\sinh(\frac{-50}{20}) ~]\\ &=& 20\cdot [~ \sinh(2.5)-\sinh(-2.5) ~]\\ &=& 20\cdot [~ \sinh(2.5)+\sinh(2.5) ~]\\ &=& 20\cdot 2\cdot \sinh(2.5) \\ &=& 40\cdot \sinh(2.5) \\ &=& 40\cdot 6.0502044810397875 \\ &=&\mathbf{ 242.0081792415915\ feet} \\ \hline \end{array}\)
The number of square feet of roofing of the barn is:
\(\begin{array}{|rcll|} \hline && 242.0081792415915\ feet \times 40 feet \\ &=& \mathbf{9680.32716964\ square\ feet} \\ \hline \end{array} \)
