1. A company that produces landscaping materials is dumping sand into a conical pile. The sand is being poured at a rate of 5 ft????3/sec; the physical proper????es of the sand, in conjunc- ????on with gravity, ensure that the cone’s height is roughly 2/3 the length of the diameter of the circular base. How fast is the cone rising when it has a height of 30 feet?

\[ height = \frac{2}{3} diameter = \frac{4}{3}radius; \ radius = \frac {3}{4}height \\ V = \frac{\pi r^2 h}{3} = \frac{3 \pi h^3}{16}; \ h = \sqrt[3]{\frac {16V}{3 \pi}} \\ \frac {dV}{dt} = 5 ft^3/sec = \frac {9 \pi h^2}{16} \frac {dh}{dt} \\ \frac {dh}{dt} = \frac {5*16}{9 \pi h^2} = \frac {80}{9 \pi h^2} \\ h = 30, \frac {dh}{dt} = \frac {80}{8100 \pi} = \\ sand \ pile \ height \ is \ increasing \ 0.003 \ ft \ per \ second \]

Let’s double check our calculation by calculating the difference in heights one second apart, with the first height at 30 ft.

rate = 5
height = 30
radius = height*3/4
volume = pi*radius^2*height/3

volume1 = volume + 5
height1 = (16*volume1/(3*pi))^(1/3)
diff = height1 - height
print(paste0("The increase in height after 1 second, at initial height = 30 ft: ",
             round(diff, 4), "ft"))
## [1] "The increase in height after 1 second, at initial height = 30 ft: 0.0031ft"