Answer:
The volume is increasing at a rate of 16286 in³/min.
Step-by-step explanation:
a) The volume of a cone is given by:
[tex] V = \frac{1}{3}h\pi r^{2} [/tex]
Where:
r: is the radius
h: is the height
The rate of change of the volume can be calculated by using the chain rule:
[tex] \frac{dV}{dt} = \frac{\pi}{3}[\frac{dh}{dt}r^{2} + h\frac{d(r^{2})}{dt}] [/tex]
Since h = 4/3 r we have:
[tex] \frac{dV}{dt} = \frac{\pi}{3}[\frac{d(\frac{4r}{3})}{dt}r^{2} + \frac{4r}{3}\frac{d(r^{2})}{dt}] [/tex]
[tex] \frac{dV}{dt} = \frac{4\pi}{9}[\frac{dr}{dt}r^{2} + r\frac{d(r^{2})}{dt}] [/tex]
[tex] \frac{dV}{dt} = \frac{4\pi}{9}[\frac{dr}{dt}r^{2} + 2r^{2}\frac{dr}{dt}] [/tex] (1)
With:
[tex]\frac{dr}{dt}[/tex] = 3 in/min
[tex]r = 3 ft*\frac{12 in}{1 ft} = 36 in[/tex]
And by entering the above values into equation (1) we have:
[tex] \frac{dV}{dt} = \frac{4\pi}{9}[(3 in/min)*(36 in)^{2} + 2*(36 in)^{2}*3 in/min] [/tex]
[tex]\frac{dV}{dt} = 16286 in^{3}/min[/tex]
Therefore, the volume is increasing at a rate of 16286 in³/min.
I hope it helps you!
