Answer:
The height is increasing at the rate of 0.42 ft/min
Step-by-step explanation:
Given that:
The diameter = height
and
diameter d = 2 r
2 r = h
r = [tex]\frac{h}{2}[/tex]
The volume of the cone formed = v
v ⇒[tex]\frac{dv}{dt}[/tex] = 40 ft³/ min
[tex]v = \frac{1}{3} \pi r^2h[/tex]
[tex]v = \frac{1}{3} \pi (\frac{h}{2})^2 h[/tex]
[tex]v = \frac{1}{3} \pi \frac{h^3}{4}[/tex]
[tex]v = \frac{\pi}{12}h^3[/tex]
[tex]\frac{dv}{dt}= \frac{\pi}{12}3h^2\frac{dh}{dt}[/tex]
[tex]\frac{dv}{dt}= \frac{\pi}{4}h^2\frac{dh}{dt}[/tex]
[tex]\frac{dh}{dt}= \frac{4}{\pi h^2}\frac{dv}{dt}[/tex]
at h = 11 ft high
[tex]\frac{dh}{dt}= \frac{4}{\pi (11)^2}* (40)[/tex]
[tex]\frac{dh}{dt}= 0.42 \ ft/min[/tex]
Hence, the height of the pile was increasing at the rate of 0.42 ft/min
