A conical cistern is 10 ft. across the top and 12 ft. deep. If water is poured into the cistern at the rate of 1 cubic foot per second, how fast is the surface rising when the water is 8 ft. deep?
Well, I must say, this is one deep question!
To find the rate at which the surface of the water is rising, we need to use some tricks of geometry.
First, we can use the formula for the volume of a cone:
V = (1/3)πr²h
Where V is the volume, r is the radius (half of the diameter), and h is the height of the cone. In this case, the radius (r) is 5 ft and the height (h) is 12 ft.
Now that we have the volume, we can differentiate it to get the rate of change with respect to time (t), which is the rate at which water is poured in:
dV/dt = (1/3)πr²(dh/dt)
Since we're given that water is poured in at a rate of 1 cubic foot per second, we can substitute that value in:
1 = (1/3)π(5²)(dh/dt)
Simplifying, we get:
dh/dt = 3/(25π)
Now, when the water is 8 ft deep, we can substitute that value into the equation and calculate the rate at that specific depth:
dh/dt = 3/(25π) ≈ 0.038 ft/s (rounded to three decimal places)
So, the surface of the water is rising at a rate of approximately 0.038 feet per second when the water is 8 feet deep. Hope that answers your deep question!
To find the rate at which the surface is rising, we need to use similar triangles and the formula for the volume of a cone.
Let's denote the radius of the surface of the water as r and the depth of the water as h.
Given that the cistern is conical, we have the following similar triangles:
r / h = 10 / 12
We can rearrange this to:
r = (10 / 12) * h
The formula for the volume of a cone is:
V = (1/3) * π * r^2 * h
We are told that water is poured into the cistern at a rate of 1 cubic foot per second, so:
dV / dt = 1
where dV / dt is the rate at which the volume is increasing with respect to time.
Now, let's find an expression for dV / dt:
V = (1/3) * π * [(10 / 12) * h]^2 * h
V = (1/3) * π * (100 / 144) * h^3
Taking the derivative of both sides with respect to time:
dV / dt = (1/3) * π * (100 / 144) * 3 * h^2 * dh / dt
Simplifying:
dV / dt = (π / 48) * h^2 * dh / dt
We need to find dh / dt, the rate at which the depth of the water is changing when it is 8 ft deep. So, we set h = 8 ft:
dh / dt = (48 / π) * (dV / dt) / (h^2)
Substituting the values we know:
dh / dt = (48 / π) * 1 / (8^2)
dh / dt = (48 / π) * 1 / 64
Simplifying further:
dh / dt ≈ 0.2387 ft/s
Therefore, when the water is 8 ft deep, the surface is rising at a rate of approximately 0.2387 ft/s.
To find the rate at which the surface of the water is rising, we need to use related rates.
Let's assign the variables:
- V: Volume of water in the cistern (cubic feet)
- r: Radius of the circular water surface in the cistern (feet)
- h: Depth of the water in the cistern (feet)
- A: Area of the circular water surface in the cistern (square feet)
Given that the cistern is conical, we need to use the formula for the volume of a cone to relate the variables:
V = (1/3) * π * r^2 * h
First, we need to express all the variables in terms of h. Since we know the diameter of the cistern (10 ft), we can find the radius (r) using the formula r = D/2:
r = 10/2 = 5 ft
The area of the circular surface (A) can be calculated using the formula for the area of a circle:
A = π * r^2
Now let's differentiate V with respect to t (time) to find the rate of change of volume with respect to time:
dV/dt = (1/3) * π * (2r * dr/dt * h + r^2 * dh/dt)
Since dr/dt represents the rate at which the radius is changing, and we're not given this information, we need to find a way to express dr/dt in terms of h.
Using similar triangles, we can relate the radius to the depth of the water:
r/h = (10/2)/(12) = 5/12
Now we can solve for r:
r = (5/12) * h
To find dr/dt, we differentiate this equation with respect to t:
dr/dt = (5/12) * dh/dt
Substituting this expression into the equation for dV/dt, we get:
dV/dt = (1/3) * π * (2 * (5/12) * h * (5/12) * dh/dt * h + (5/12)^2 * dh/dt)
Simplifying further:
dV/dt = (1/3) * π * (25/144) * (2h^2 * dh/dt + h^2)
We know that the rate at which water is being poured into the cistern is 1 cubic foot per second, so dV/dt = 1 cubic foot per second.
Now we can solve for dh/dt, which represents the rate at which the water level is rising:
1 = (1/3) * π * (25/144) * (2h^2 * dh/dt + h^2)
Simplifying:
2h^2 * dh/dt + h^2 = 144/25
2h^2 * dh/dt = 144/25 - h^2
dh/dt = (144/25 - h^2) / (2h^2)
Now we can substitute h = 8 ft into the equation to find the rate at which the surface is rising when the water is 8 ft deep:
dh/dt = (144/25 - (8)^2) / (2(8)^2)
dh/dt = (144/25 - 64) / 128
dh/dt = 80/320
dh/dt = 1/4 ft/s
Therefore, the surface of the water is rising at a rate of 1/4 ft/s when the water is 8 ft deep.
let the water level have a radius of r ft
let the height of the water be h ft
by ratio:
r/h = 5/12
12r = 5h ---> r = 5h/12
V = (1/3)π r^2 h
= (1/3)π(25h^2/144)(h)
= (25/432)π h^3
dV/dt = (25/144)π h^2 dh/dt
plug in our given stuff
1 = (25/144)π(64) dh/dt
dh/dt = 144/(25(64)π) = 9/(100π)
check my arithmetic