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
[tex]\frac{h}{r} =500/5000[/tex]
h= [tex]\frac{1}{10}r[/tex] ⇒ r= 10h
As the water is flowing and filling up reservoir at the rate of 40 m³/min
So,
[tex]\frac{dV}{dt} =[/tex] 40 x [tex]10^{6}[/tex] cm³/min
a) in this part, we have to find [tex]\frac{dh}{dt}[/tex] 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
[tex]\frac{dV}{dt} =[/tex] [tex]\frac{100}{3}[/tex]π 3h² [tex]\frac{dh}{dt}[/tex]
40 x [tex]10^{6}[/tex] = [tex]\frac{300}{3}[/tex] π300²[tex]\frac{dh}{dt}[/tex]
[tex]\frac{dh}{dt}[/tex] = 40 x [tex]10^{6}[/tex] / (100 x 300²)π
[tex]\frac{dh}{dt}[/tex] = 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 [tex]\frac{dr}{dt}[/tex]
We have h=300 cm . therefore, r= 10h => 10(300)
r= 3000
V= 1/3 πr²h =>1/3 πr² [tex]\frac{1}{10}r[/tex]
V= 1/30 πr³
Applying Derivation both side w.r.t to 't'
[tex]\frac{dV}{dt} = \frac{1}{30}[/tex]π 3r²[tex]\frac{dr}{dt}[/tex]
40 x [tex]10^{6}[/tex] = 1/10 π (3000)² [tex]\frac{dr}{dt}[/tex]
[tex]\frac{dr}{dt}[/tex]= 40 x [tex]10^{6}[/tex] x 10 / π (3000)²
[tex]\frac{dr}{dt}[/tex]= 14.14 cm/min
Thus, the radius of the water's surface is changing at the rate of 14.14 cm/min
