a water tank in the shape of a circular cone with its top down, with a base radius of 5cm and height of 10cm. pour water into it at rate of 9pi cm^3/min. calculate the rate of change of height of water two minutes after the start of pouring water.

Answer 1

The volume of the water tank is V = (1/3)πr^2h = (1/3)π(5cm)^2(10cm) = 250π cm^3.

The rate of change of height of water two minutes after the start of pouring water is equal to the rate of change of volume of water in the tank.

Rate of change of volume of water in the tank = rate of change of water poured in - rate of change of water out

Rate of change of water poured in = 9π cm^3/min

Rate of change of water out = (1/3)πr^2dh/dt = (1/3)π(5cm)^2dh/dt

dh/dt = (9π - (1/3)π(5cm)^2dh/dt)/(5cm)^2

dh/dt = (9π - (1/3)π(25 cm^2)dh/dt)/(25 cm^2)

dh/dt = (9π - (25/3)dh/dt)/25

dh/dt = (225π - 25dh/dt)/25

25dh/dt + 225π = 25

dh/dt = (25 - 225π)/25

dh/dt = -9π cm/min

Therefore, the rate of change of height of water two minutes after the start of pouring water is -9π cm/min.

Answer 2

WOW Mr. Bot, just Wow!!, How can the water level be falling when we are pouring water in ?????
This type of problem is clearly too tough for you, better let a human do the right way.

at some time of t minutes, let the height of water be h cm and
let the radius of the water level be r cm
assuming a sketch was made,
r/h = 5/10 = 1/2 ----> r = h/2

V = (1/3)pi r^2 h = (1/3)pi (h^2/4)h = 1/12 pi h^3
dV/dt = 1/4 pi h^2 dh/dt

when t = 2, V = 18pi
1/12 pi h^3 = 18 pi
h^3 = 216
h = 6

9 pi = (1/4 pi (36)dh/dt
dh/dt = 1

at 2 minutes the water level is rising at 1 cm/min

Answer 3

impatient much?
Once is enough, thank you.
Repeated postings will not get faster or better responses.