Suppose that water is pouring into a swimming pool in the shape of a right circular cylinder at a constant rate of 3 cubic feet per minute. If the pool has radius 6 feet and height 10 feet, what is the rate of change of the height of the water in the pool when the depth of the water in the pool is 6 feet?
Well, that's a tall order! Let's dive in and compute the solution. We're given that the pool has a radius of 6 feet and a constant rate of water pouring in at 3 cubic feet per minute.
To find the rate of change of the height of the water, we need to differentiate the volume of the cylinder with respect to the height. The volume of a right circular cylinder is given by the formula V = πr^2h, where V is the volume, r is the radius, and h is the height.
Differentiating both sides of the formula with respect to h, we get dV/dh = 2πrh. This means that for every unit increase in height, the volume changes by 2πrh cubic feet.
Now, when the depth of the water is 6 feet, the height of the water in the cylinder is 10 - 6 = 4 feet.
So, plugging in the values into our formula, we get dV/dh = 2π(6)(4) = 48π cubic feet per unit increase in height.
Therefore, the rate of change of the height of the water in the pool when the depth is 6 feet is 48π cubic feet per minute.
Hope that didn't make your head swim!
To find the rate of change of the height of the water in the pool, we can use related rates. Let's call the height of the water in the pool h, and the time t.
Given:
- The rate of water pouring into the pool = 3 cubic feet per minute.
- The radius of the pool = 6 feet.
- The height of the pool = 10 feet.
We want to find the rate of change of the height, dh/dt, when the depth of the water is 6 feet.
We can start by finding the equation that relates the volume of water in the pool with the height:
Volume of a cylinder (V) = πr^2h
Differentiating the equation with respect to time (t), we get:
dV/dt = π * d(r^2h)/dt
Since we have r and h, we can write r^2 as (6^2):
dV/dt = π * d(6^2 * h)/dt
The rate of water pouring into the pool is given as 3 cubic feet per minute, so:
dV/dt = 3
Substituting the values, we get:
3 = π * d(36h)/dt
Now, we need to find the value of d(36h)/dt when h = 6.
Differentiating h with respect to t, we get:
d(36h)/dt = 36 * dh/dt
So the equation becomes:
3 = π * 36 * dh/dt
Simplifying, we find:
dh/dt = 3/(π*36)
Now, substitute the value of π as approximately 3.14:
dh/dt = 3 / (3.14 * 36)
Simplifying further, we get:
dh/dt ≈ 0.0265 feet per minute
Therefore, the rate of change of the height of the water in the pool when the depth of the water in the pool is 6 feet is approximately 0.0265 feet per minute.
To find the rate of change of the height of the water in the pool, we need to use the formula for the volume of a cylinder, which is given by V = πr^2h, where V is the volume, r is the radius, and h is the height.
Given that water is pouring into the pool at a constant rate of 3 cubic feet per minute, we can set up an equation for the rate of change of the volume with respect to time:
dV/dt = 3
We also need to find the rate of change of the height of the water with respect to time, which is what we're looking for. Let's call it dh/dt.
Now, we can differentiate the equation for the volume with respect to time using the chain rule:
dV/dt = d/dt(πr^2h)
= 2πrh(dh/dt)
Since the radius of the pool is given as 6 feet, and we are looking for the rate of change of the height when the depth of the water is 6 feet, we can substitute these values into the equation:
3 = 2π(6)(6)(dh/dt)
Simplifying the equation, we get:
3 = 72π(dh/dt)
To find the rate of change of the height of the water, we can divide both sides of the equation by 72π:
3/(72π) = (dh/dt)
Simplifying further, we have:
dh/dt = 1/(24π)
Therefore, the rate of change of the height of the water in the pool when the depth of the water is 6 feet is 1/(24π) feet per minute, or approximately 0.013 feet per minute.
at a time of t minutes, let the height of the water be h ft
V = π(6^2)(h) = 36π h
dV/dt = 36π dh/dt
subbing in our values
3 = 36π dh/dt
dh/dt = 1/(12π) ft/min
Did you notice that the rate was constant (it did not depend on what h was) ?