+0

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

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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.

Guest Mar 2, 2017

#2
+18956
+15

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:

$$\begin{array}{|rcll|} \hline && 242.0081792415915\ feet \times 40 feet \\ &=& \mathbf{9680.32716964\ square\ feet} \\ \hline \end{array}$$

heureka  Mar 3, 2017
edited by heureka  Mar 3, 2017
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#1
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Can't speak for others on this site, but show that you have made some sort of attempt and I will help you.

I'm  not willing to simply write out a solution.

Guest Mar 3, 2017
#2
+18956
+15

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:

$$\begin{array}{|rcll|} \hline && 242.0081792415915\ feet \times 40 feet \\ &=& \mathbf{9680.32716964\ square\ feet} \\ \hline \end{array}$$

heureka  Mar 3, 2017
edited by heureka  Mar 3, 2017
#3
+926
+5

Hello Herr Heureka,

It seems odd that a barn (or even an airplane hangar) would have such a high and massively large roof for a short length.

Wouldn’t you just integrate from -20 to 20 --the width of the barn? This would make the roof area 4701 Ft2 for a 40’ x 100’ barn.  This seems more aesthetic and consistent with common barn structures.

Gruß Ginger

GingerAle  Mar 3, 2017
#4
+91793
+5

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.

Here is the cross section of my barn, I painted it purple :)

consider the length of the arc of the roof on the 40 foot side.  Using Pythagoras    (this is really cool)

$$dL^2=dx^2+dy^2\\ (\frac{dL}{dx})^2=1+(\frac{dy}{dx})^2\\ \frac{dL}{dx}=\sqrt{1+(\frac{dy}{dx})^2}\\ L=\displaystyle\int_{-20}^{20} \sqrt{1+(\frac{dy}{dx})^2}\;dx\\ \text{So the area of the roof would be}\\ Area=100\displaystyle\int_{-20}^{20} \sqrt{1+(\frac{dy}{dx})^2}\;dx\qquad feet^2$$

so my starting point is now the same as Heurekas except if have got the catenary over the short side which makes more sense to me :)

Heureka has done the rest already :)))))

Melody  Mar 3, 2017

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