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.
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.
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
impatient much?
Once is enough, thank you.
Repeated postings will not get faster or better responses.
To solve this problem, we can use the related rates approach. We are given that water is pouring into the cone at a rate of 9π cm^3/min and we need to find the rate of change of the height of the water two minutes after the start.
Let's denote the radius of the water column as r and the height as h at any given time t. We are interested in finding dh/dt when t = 2 minutes.
First, let's express the volume of the cone using the given dimensions. We know that the volume of a cone is given by V = (1/3)πr^2h.
Given that the base radius of the cone is 5 cm and the height of the cone is 10 cm, we can substitute these values into the volume equation:
V = (1/3)π(5^2)(10) = (1/3)π(25)(10) = (1/3)π(250) = (250/3)π cm^3
Since water is pouring in at a rate of 9π cm^3/min, we can express the volume with respect to time as V = 9πt.
Now, let's differentiate both sides of this equation with respect to t:
dV/dt = d(9πt)/dt
The rate of change of volume with respect to time is given by:
dV/dt = 9π
Since the volume V is related to the height h, we can use the volume equation to express V in terms of h:
(250/3)π = (1/3)π(5^2)h
250 = 25h
h = 10
Now, let's differentiate both sides of this equation with respect to t:
dh/dt = d(10)/dt
dh/dt = 0
dh/dt = 0 cm/min
Therefore, the rate of change of the height of the water two minutes after the start of pouring water is 0 cm/min. This means that the water level remains constant after two minutes.