Answer:
[tex]\Rightarrow \frac{dh}{dt}\approx -31.690 [/tex]
Explanation:
The volume of a cone, V = 1/3πr^2h
Where r is the radius, and h is the height of the cone.
Given:
Volume is constant at 56 cubic feet.
Radius of cone increases at a constant rate of 8 feet per second.
The radius of the cone is 3 feet.
When radius is 3 feet
[tex]\Rightarrow h= \frac{3V}{\pi r^2}[/tex]
[tex]\Rightarrow h= \frac{3\times 56}{\pi \times 3^2} [/tex]
[tex]\Rightarrow h= \frac{3\times 56}{\pi \times 9} [/tex]
[tex]\Rightarrow h= \frac{ 56}{3\pi} [/tex]
Where h is in feets.
[tex]\frac{dV}{dt}=\frac{d (\frac{1}{3}\pi r^2h) }{dt} [/tex]
[tex]\Rightarrow \frac{dV}{dt}=(\frac{1}{3}\pi \frac{dr^2}{dt}h)+(\frac{1}{3}\pi r^2 \frac{dh}{dt}) [/tex]
[tex]\Rightarrow \frac{dV}{dt}=(\frac{1}{3}\pi \frac{dr^2}{dr}.\frac{dr}{dt}h)+(\frac{1}{3}\pi r^2 \frac{dh}{dt}) [/tex]
[tex]\Rightarrow \frac{dV}{dt}=(\frac{1}{3} \pi 2r\frac{dr}{dt}h)+(\frac{1}{3}\pi r^2 \frac{dh}{dt}) [/tex]
Since the volume is constant so [tex] \frac{dV}{dt}=0 [/tex]
The rate of change of radius is [tex]\frac{dr}{dt}=8 [/tex]
[tex]\Rightarrow 0=(\frac{1}{3} \pi (2\times 3)(8)(\frac{56}{3\pi}))+(\frac{1}{3}\pi (3^2) \frac{dh}{dt}) [/tex]
[tex]\Rightarrow 0=(\frac{56\times 16}{3})+(\frac{1}{3}\pi (9) \frac{dh}{dt}) [/tex]
[tex]\Rightarrow 0=(\frac{56\times 16}{3})+(3\pi \frac{dh}{dt}) [/tex]
[tex]\Rightarrow 3\pi \frac{dh}{dt}=-\frac{56\times 16}{3} [/tex]
[tex]\Rightarrow \frac{dh}{dt}=-\frac{56\times 16}{3\times 3 \times \pi} [/tex]
[tex]\Rightarrow \frac{dh}{dt}=-\frac{56\times 16}{9\pi} [/tex]
[tex]\Rightarrow \frac{dh}{dt}=-31.689518 [/tex]
[tex]\Rightarrow \frac{dh}{dt}\approx -31.690 [/tex]
The minus indicates that the rate of change of height is decreasing.
