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pochemuha
3 years ago
15

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
Mathematics
1 answer:
solong [7]3 years ago
3 0

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!                                    

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