Respuesta :
Water is leaking out of an inverted conical tank at a rate of 10,500 cm3/min at the same time that water is being pumped into the tank at a constant rate. The tank has height 6 m and the diameter at the top is 4 m. If the water level is rising at a rate of 20 cm/min when the height of the water is 2 m, find the rate at which water is being pumped into the tank
Answer:
The rate at which water is being pumped into the tank is 289,752 [tex]cm^3/min[/tex]
Solution:
According to question,
There is an inverted conical tank, through which water is leaking at a rate of 10,500 cm3/min at the same time that water is being pumped into the tank at a constant rate
The dimension of tank are:
Diameter = 4cm
Radius(r) = [tex]\frac{diameter}{2}[/tex] = 2cm
Height = 6cm
Clearly we can see that height is 3 times radius so, we can write
h = 3r OR r = h/3   ……………………. (1)
The volume of cone "V" is given as:
[tex]\text { volume of cone }(\mathrm{V})=\frac{1}{3} \pi r^{2} h[/tex] Â -------- (2)
From (1) and (2)
[tex]\text { Volume of cone(V) }=\frac{1}{3} \pi\left(\frac{h}{3}\right)^{2} h[/tex]
[tex]\mathrm{V}=\frac{\pi h^{3}}{27}[/tex] Â -------- (3)
Now we calculate the derivate:- Â
[tex]\frac{d V}{d t}=\frac{3 \pi h^{2}}{27} \frac{d h}{d t}[/tex]
[tex]\frac{d V}{d t}=\frac{\pi h^{2}}{9} \frac{d h}{d t}[/tex] Â --------- (4)
According to question, when height is 2m = 200cm, the water level is rising at a rate of 20 cm/min
[tex]\frac{d h}{d t}=20 \mathrm{cm} / \mathrm{min}[/tex]
On putting above values in equation(4) and solving we get
[tex]\frac{d V}{d t}=279252[/tex]
Hence, the rate at which water is being pumped is 289,752 [tex]cm^3/min[/tex] which is the sum of water volume increasing at rate of 279,252 and 10,500 leaking out
