Water at the rate of 10 cm3/ min is pouring into the leaky cistern whose shape is a cone 16 cm. deep and 8 cm. in diameter at the top. At the time the water is 12 cm. deep, the water level is observed to be rising 1/3 cm/min. how fast is the water leaking away?
diameter /depth = 8/16 = 1/2
so when deprh h = 12 diameter of water surface = 6 and r = 3 cm
area of water surface = pi r^2 = 9 pi cm^2
so pi r^2 dh/dt = 9 pi (1/3) cm^3/min = 3 pi cm^3/min
At a time of t min, let the height of the water level be h cm, and the radius
of its surface be r cm
Vol of water = (1/3)π r^2 h
but by simple ratios:
r/h = 4/16 = 1/4
h = 4r or r = h/4
since we need dh/dt, let's keep the h's
V = (1/3)π (h^2/16)(h)
= (1/48)π h^3
dV/dt = (1/16) π h^2 dh/dt
let the leakage be x cm^3/min
so dV/dt = (10-x) cm^3/min
when h = 12
V = (1/48)π (1728) = 36π cm^3
"At the time the water is 12 cm. deep, the water level is observed to be rising 1/3 cm/min"
----> dV/dt = (1/16) π h^2 dh/dt
10-x = (1/16)π(144)(1/3)
10-x = 3π
x = 10-3π = appr .575
So the leak is appr .575 cm^3/min
check my arithmetic, I did not write this out first
Well, it seems like we have a leaky situation here! Let's dive into the problem.
To find out how fast the water is leaking away, we need to determine the rate at which the volume of water is changing with respect to time. This can be calculated using the formula for the volume of a cone:
Volume = (1/3) * π * r² * h,
where r is the radius of the cone and h is its height.
Given that the diameter at the top is 8 cm, we can find the radius by dividing it by 2: r = 8 cm / 2 = 4 cm.
The water is pouring in at a rate of 10 cm³/min, but the water level is rising at a different rate. So, let's focus on the change in height over time.
When the water level is 12 cm deep, it is observed to be rising at a rate of 1/3 cm/min. This means dh/dt = 1/3 cm/min.
Next, we can differentiate the volume formula with respect to time to find an expression for dV/dt:
dV/dt = (1/3) * π * [(2r * dr/dt * h) + (r² * dh/dt)],
where dr/dt represents the rate at which the radius is changing with respect to time. Since the radius remains constant, dr/dt = 0.
Now we have all the values we need to calculate dV/dt:
dV/dt = (1/3) * π * (0 + (4 cm)² * (1/3 cm/min)),
dV/dt = (1/3) * π * (16 cm²/3 min),
dV/dt ≈ 16.96 cm³/min.
So, it seems like the leaky cistern is losing water at a rate of approximately 16.96 cm³/min. Watch out for those water-coated clowns!
To find out how fast the water is leaking away, we need to determine the rate at which the volume of water is changing with respect to time.
First, let's find the volume of the cone at any given time using the formula for the volume of a cone:
V = (1/3) * π * r^2 * h
where V is the volume, π is the mathematical constant pi, r is the radius of the cone, and h is the height of the cone.
Since the diameter at the top of the cone is 8 cm, the radius (r) is half of that, which is 4 cm. We can use this radius because the shape of the cone remains constant.
Now, let's differentiate the volume equation with respect to time to find the rate of change of volume:
dV/dt = (1/3) * π * (2r) * dh/dt
where dV/dt is the rate of change of volume with respect to time, and dh/dt is the rate at which the height of the water is changing.
Given that the rate at which the water level is rising is 1/3 cm/min, we have:
dh/dt = 1/3 cm/min
Substituting the values into the equation, we have:
dV/dt = (1/3) * π * (2 * 4) * (1/3) = (8/3) * π * (1/3) cm^3/min
Now, let's calculate the value of dV/dt:
dV/dt = (8/3) * (π/3) cm^3/min
dV/dt ≈ 2.82 cm^3/min
Therefore, the water is leaking away at a rate of approximately 2.82 cm^3/min.
To find the rate at which the water is leaking away, we need to calculate the rate of change of the volume of water in the cistern with respect to time.
First, let's calculate the volume of the cistern when the water level is 12 cm deep. Since the cistern is in the shape of a cone, we can use the formula for the volume of a cone:
V = (1/3) * π * r^2 * h
where V is the volume, π is the mathematical constant approximately equal to 3.14159, r is the radius of the circular top of the cone, and h is the height of the cone.
Given that the diameter at the top is 8 cm, we can find the radius as half of the diameter: r = 8 cm / 2 = 4 cm.
Substituting the radius and the height (h = 12 cm) into the volume formula, we get:
V = (1/3) * 3.14159 * (4 cm)^2 * 12 cm
= 201.06176 cm^3
Now, let's find the rate at which the volume is changing with respect to time. We are told that water is pouring into the cistern at a rate of 10 cm^3/min and the water level is rising at a rate of 1/3 cm/min when the water is 12 cm deep.
In other words, the rate of change of the volume with respect to time (dV/dt) is equal to the rate of water pouring in (10 cm^3/min) minus the rate of water rising (1/3 cm/min), since water is leaking away.
So, dV/dt = 10 cm^3/min - (1/3) cm/min
Simplifying this equation, we get:
dV/dt = 10 cm^3/min - 1/3 cm/min
= 29/3 cm^3/min
Therefore, the water is leaking away from the cistern at a rate of approximately 9.67 cm^3/min (rounded to two decimal places).