Question

1. Sand falling from a chute forms a conical pile whose altitude is always equal to 4/3 the radius of the base. (a) How fast is the volume increasing when the radius of the base is 3 ft and is increasing at a rate of 3 in/min

Answer 1

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:

$V = \frac{1}{3}h\pi r^{2}$

Where:    

r: is the radius

h: is the height

The rate of change of the volume can be calculated by using the chain rule:

$\frac{dV}{dt} = \frac{\pi}{3}[\frac{dh}{dt}r^{2} + h\frac{d(r^{2})}{dt}]$

Since h = 4/3 r we have:

$\frac{dV}{dt} = \frac{\pi}{3}[\frac{d(\frac{4r}{3})}{dt}r^{2} + \frac{4r}{3}\frac{d(r^{2})}{dt}]$

$\frac{dV}{dt} = \frac{4\pi}{9}[\frac{dr}{dt}r^{2} + r\frac{d(r^{2})}{dt}]$        

$\frac{dV}{dt} = \frac{4\pi}{9}[\frac{dr}{dt}r^{2} + 2r^{2}\frac{dr}{dt}]$    (1)    

With:

$\frac{dr}{dt}$ = 3 in/min

$r = 3 ft*\frac{12 in}{1 ft} = 36 in$

And by entering the above values into equation (1) we have:

$\frac{dV}{dt} = \frac{4\pi}{9}[(3 in/min)*(36 in)^{2} + 2*(36 in)^{2}*3 in/min]$

$\frac{dV}{dt} = 16286 in^{3}/min$

Therefore, the volume is increasing at a rate of 16286 in³/min.

I hope it helps you!