Answer:
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](https://tex.z-dn.net/?f=%5Cfrac%7Bh%7D%7Br%7D%20%3D500%2F5000)
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 ![\frac{dr}{dt}](https://tex.z-dn.net/?f=%5Cfrac%7Bdr%7D%7Bdt%7D)
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![\frac{dr}{dt}](https://tex.z-dn.net/?f=%5Cfrac%7Bdr%7D%7Bdt%7D)
40 x
= 1/10 π (3000)² ![\frac{dr}{dt}](https://tex.z-dn.net/?f=%5Cfrac%7Bdr%7D%7Bdt%7D)
= 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