Water is flowing at the rate of 40 m cubed divided by min from a shallow concrete conical reservoir (vertex down) of base radius 50 m and height 5 m. a. How fast (centimeters per minute) is the water level falling when the water is 3 m deep? b. How fast is the radius of the water's surface changing then? Answer in centimeters per minute.

a)1.414 cm/min

b)14.14 cm/min

Step-by-step explanation:

Volume of a cone can be determined by the formula i.e

V= 1/3 πr²h

where, 'h'is height and 'r' is radius

Now, when height and radius of  shallow concrete conical reservoir is,

r= 50m = 5000cm

h= 5m = 500cm

As height and radius are proportional, we can write

h= ⇒ r= 10h

As the water is flowing and filling up reservoir at the rate of 40 m³/min

So,

40 x cm³/min

a) in this part, we have to find and the height given is:

height 'h'= 3m=> 300m

Therefore,

V= 1/3 πr²h => 1/3 π (10h)² h

V= 100 π h³/ 3

π 3h²

40 x =   π300²

=  40 x  / (100 x 300²)π

= 1.414 cm/min

b) in this part, we need to find how  fast is the radius of the water's surface changing i.e

We have h=300 cm . therefore, r= 10h => 10(300)

r= 3000

V= 1/3 πr²h =>1/3 πr²

V= 1/30 πr³

Applying Derivation both side w.r.t to 't'

π 3r²

40 x = 1/10 π (3000)²

= 40 x x 10 / π (3000)²

= 14.14 cm/min

Thus,  the radius of the water's surface is changing at the rate of 14.14 cm/min

b i think

step-by-step explanation:

not really sure

6e14 is the answer.