![y=-0.1480x^3+1.353x^2-4.068x+5.651](https://tex.z-dn.net/?f=y%3D-0.1480x%5E3%2B1.353x%5E2-4.068x%2B5.651)
![\implies y'=-0.444x^2+2.706x-4.068](https://tex.z-dn.net/?f=%5Cimplies%20y%27%3D-0.444x%5E2%2B2.706x-4.068)
The arc length is given by the integral
![\displaystyle\int_{0.2}^{5.25}\sqrt{1+(y')^2}\,\mathrm dx](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint_%7B0.2%7D%5E%7B5.25%7D%5Csqrt%7B1%2B%28y%27%29%5E2%7D%5C%2C%5Cmathrm%20dx)
which has a value of about
![7.6683](https://tex.z-dn.net/?f=7.6683)
.
A=8ft^2
A=l•w
8=l•w
l=8/w
P=2(l+w)
12=2(8/w+w)
12=2(8/w+w/1)
12=2(8/w+w^2/w)
12=16w^2/w
12=16w
w=12/16
w=3/4=0.75ft
l=8/0.75=10.67ft
The solution would be like this for this specific problem:
Volume of a cylinder = pi * r^2 * h
Volume of a cone = 1/3 * pi * r^2 * h
Total Height = 47
Height of the cone = 12
Height of the cylinder = 35
If the top half is filled with sand, then:
volume (sand) = pi * 4^2 * 36
volume (cone) = 1/3 * pi * 4^2 * 12
Total volume = 1960.353816 cubic millimeters
353816 / (10 * pi) = 62.4 seconds.
It will take 62.4 seconds until all of the sand has dripped to the bottom of the hourglass.