Mathematics: 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.

Water is flowing at the rate of 40 m cubed divided

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23.01.2022

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

$\frac{h}{r} =500/5000$

h= $\frac{1}{10}r$ ⇒ r= 10h

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

So,

$\frac{dV}{dt} =$ 40 x $10^{6}$ cm³/min

a) in this part, we have to find $\frac{dh}{dt}$ 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

$\frac{dV}{dt} =$ $\frac{100}{3}$ π 3h² $\frac{dh}{dt}$

40 x $10^{6}$ = $\frac{300}{3}$ π300² $\frac{dh}{dt}$

$\frac{dh}{dt}$ = 40 x $10^{6}$ / (100 x 300²)π

$\frac{dh}{dt}$ = 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 $\frac{dr}{dt}$

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

r= 3000

V= 1/3 πr²h =>1/3 πr² $\frac{1}{10}r$

V= 1/30 πr³

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

$\frac{dV}{dt} = \frac{1}{30}$ π 3r² $\frac{dr}{dt}$

40 x $10^{6}$ = 1/10 π (3000)² $\frac{dr}{dt}$

$\frac{dr}{dt}$ = 40 x $10^{6}$ x 10 / π (3000)²

$\frac{dr}{dt}$ = 14.14 cm/min

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

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