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so the sand pile looks more or less like the one in the image below.
let's check its circumference first. We know that dC/dt = 5π
[tex]C=2\pi r\implies \cfrac{dC}{dt}=2\pi \cfrac{dr}{dt}\implies 5\pi =2\pi \cfrac{dr}{dt}\implies \cfrac{5\pi }{2\pi }=\cfrac{dr}{dt}\implies \boxed{\cfrac{5}{2}=\cfrac{dr}{dt}}[/tex]
hmmm when the circumference C = 8π, what's the radius "r"?
[tex]8\pi =2\pi r\implies \cfrac{8\pi }{2\pi }=r\implies 4=r[/tex]
now let's take the derivative to the volume equation, and check what dV/dt is
[tex]V=\cfrac{r^3}{3}\implies V=\cfrac{1}{3}r^3\implies \cfrac{dV}{dt}=\stackrel{\textit{chain rule}}{\cfrac{3}{3}r^2\cdot \cfrac{dr}{dt}}\implies \cfrac{dV}{dt}=r^2\cdot \cfrac{5}{2}\implies \cfrac{dV}{dt}=\cfrac{5r^2}{2} \\\\\\ \stackrel{\textit{when }C=8\pi \textit{ we know that }r=4}{\cfrac{dV}{dt}=\left. \cfrac{5r^2}{2}\right|_{r=4}} \implies \cfrac{dV}{dt}=\cfrac{5(4)^2}{2}\implies \cfrac{dV}{dt}=40[/tex]
The rate of change of the volume is mathematically given as
dV/dt=40
Question Parameter(s):
The volume V of the pile is given by V = r^3/3,
at a constant rate of 5 pi feet per hour
Generally, the equation for the Circumference is mathematically given as
[tex]C=2\pi r\\\\\ \frac{5}{2}=\frac{dr}{dt}}[/tex]
Therefore
8π=2πr
r=8π/2π
r=4
In conclusion
[tex]V=\cfrac{r^3}{3}[/tex]
Hence
[tex]\frac{dV}{dt}=\cfrac{5r^2}{2} \\\\\\ \frac{dV}{dt}=\cfrac{5(4)^2}{2} \frac{dV}{dt}[/tex]
dV/dt=40
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