The base of a cone-shaped tank is a circle of radius 5 feet, and the vertex of the cone is 12 feet above the base. The tank is being filled at a rate of 3 cubic feet per minute. Find the rate of change of the depth of water in the tank when then depth is 7 feet.
Well, let's dive into this problem! We know that the tank is shaped like a cone with a radius of 5 feet and a height of 12 feet. Now, we want to find the rate of change of the depth of water in the tank when the depth is 7 feet.
To begin, let's first find the volume of the cone-shaped tank. The formula for the volume of a cone is V = (1/3) * π * r^2 * h, where V represents volume, r is the radius of the base, and h is the height of the cone.
Substituting in the given values, we have V = (1/3) * π * 5^2 * 12.
Now, let's find the rate of change of the volume with respect to time. Given that the tank is being filled at a rate of 3 cubic feet per minute, we can write dV/dt = 3.
Next, we need to find the rate of change of the depth of water, which we'll call dh/dt, when the depth is 7 feet.
Now, the volume of a cone can also be expressed as V = (1/3) * π * r^2 * H, where H is the height of the water in the tank.
Since the depth of the water is 7 feet, the height of the water in the tank is equal to the height of the cone minus the depth of the water, which is 12 - 7 = 5 feet.
So, now we can rewrite the volume as V = (1/3) * π * 5^2 * 5.
Differentiating both sides of the equation with respect to time, we get dV/dt = (1/3) * π * 5^2 * dH/dt.
Since dV/dt is given as 3 cubic feet per minute, we can substitute those values in.
3 = (1/3) * π * 5^2 * dH/dt.
Now, we just need to solve for dH/dt, which represents the rate of change of the depth of water.
dH/dt = (3 * 3) / ((1/3) * π * 5^2).
Calculating this, we find dH/dt ≈ 0.114 feet per minute.
So, the rate of change of the depth of water in the tank when the depth is 7 feet is approximately 0.114 feet per minute.
I hope that made a splash in terms of explanation!
To find the rate of change of the depth of water in the tank, we need to calculate the rate at which the volume of water in the tank is changing with respect to time.
The volume V of a cone can be calculated using the formula:
V = (1/3)πr²h
Where r is the radius of the base, h is the height or depth of the cone, and π is a constant approximately equal to 3.14159.
In this case, the radius of the base is 5 feet, and the height of the cone (distance from the base to the vertex) is 12 feet.
So, the volume of the cone can be written as:
V = (1/3)π(5²)h
Now let's differentiate both sides of the equation with respect to time (t) using the chain rule:
dV/dt = (1/3)π(5²)(dh/dt)
Here, dV/dt represents the rate of change of the volume of water in the tank with respect to time (which is given as 3 cubic feet per minute).
We need to find the rate of change of the depth of water in the tank (dh/dt) when the depth is 7 feet.
Plugging in the values into the equation, we have:
3 = (1/3)π(5²)(dh/dt)
Now, let's solve for dh/dt:
dh/dt = (3 * 3) / [(1/3)π(5²)]
Calculating the value, we get:
dh/dt ≈ 0.0365 ft/min
Therefore, the rate of change of the depth of water in the tank when the depth is 7 feet is approximately 0.0365 feet per minute.
To find the rate of change of the depth of water in the tank, we can use related rates. Let's assign some variables to the given quantities:
- r: radius of the circular base of the tank (5 feet)
- h: depth of water in the tank at a given time
- V: volume of water in the tank at a given time (related to h)
- dV/dt: rate of change of the volume of water in the tank with respect to time (3 cubic feet per minute)
- dh/dt: rate of change of the depth of water in the tank with respect to time (what we need to find)
We can start by expressing the volume of a cone in terms of its radius and height. The volume of a cone is given by the formula:
V = (1/3) * π * r^2 * h
We can differentiate this equation with respect to time (t) using implicit differentiation, since both V and r are changing:
dV/dt = (1/3) * π * 2r * dr/dt * h + (1/3) * π * r^2 * dh/dt
Now we have an equation relating dV/dt, dr/dt, r, dh/dt, and h. Since we need to find dh/dt when h = 7 feet, we can substitute the given values into the equation and solve for dh/dt.
Given:
- r = 5 feet
- dV/dt = 3 cubic feet per minute
- h = 7 feet
By substituting these values into the equation, we get:
3 = (1/3) * π * 2(5) * dr/dt * 7 + (1/3) * π * (5^2) * dh/dt
Simplifying the equation, we get:
3 = (10/3) * π * dr/dt * 7 + (25/3) * π * dh/dt
Next, isolate dh/dt by rearranging the equation:
(25/3) * π * dh/dt = 3 - (10/3) * π * dr/dt * 7
Finally, solve for dh/dt by dividing both sides of the equation by (25/3) * π:
dh/dt = (3 - (10/3) * π * dr/dt * 7) / [(25/3) * π]
Now you can substitute the value of dr/dt given in the problem statement and evaluate the expression to find the rate of change of the depth of water in the tank when the depth is 7 feet.
when the water is at depth x, the radius of the surface of the water is 5/12 (12-x)
so, the volume of the air space is 1/3 pi r^2 h
= 1/3 pi (5/12 (12-x))^2 (12-x)
= 25pi/432 (12-x)^3
the volume of water is thus the tank volume less the air space:
v = pi/3 * 25^2 * 12 - 25pi/432 (12-x)^3
= 100 pi - 25pi/432 (12-x)^3
dv/dt = 25/144 pi (12-x)^2 dx/dt
3 = 25/144pi * 25 dx/dt
dx/dt = 432/(625pi) = 0.22 ft/min