A water tank has a shape of an inverted circular cone with base radius 3 m and height of 5 m. If the water is being pumped into the tank at a rate of
2 m^3 /min, find the rate at which the water level is rising when the water is 3 m deep.
let the height of the water be h m
let the radius of the water be r m
by ratios : h/r = 5/3
5r = 3h
r = 3h/5
Volume = (1/3)π r^2 h
= (1/3)π(9h^2/25)h
= (3π/25)h^3
d(Volume)/dt = (9π/25)h^2 dh/dt
2 = (9π/25)(9)dh/dt
dh/dt = 50/(81π) m/s
Oh boy, solving problems with cones and pumping water! Don't worry, I'm here to help - with a sprinkle of humor, of course.
First, let's find the volume of the cone. The formula for the volume of a cone is V = (1/3) * π * r^2 * h, where r is the radius and h is the height.
Given that the base radius is 3 m and the height is 5 m, we can substitute these values into the formula to find the volume of the cone. V = (1/3) * π * 3^2 * 5 = 15π m^3 (volume when the water is 5 m deep).
Now, let's find the rate at which the water level is rising. We know that the water is being pumped into the tank at a rate of 2 m^3/min. So, when the water is 3 m deep, the volume of the cone will be 3/5 * 15π = 9π m^3.
Differentiating the volume of the cone with respect to time, we get dV/dt = π * r^2 * dh/dt, where dV/dt represents the rate of change of volume and dh/dt represents the rate at which the height is changing.
Substituting the given values, 2 = π * 3^2 * dh/dt (when the water is 3 m deep and we're looking for dh/dt).
Simplifying further, dh/dt = 2 / (π * 9) = 2/9π m/min.
So, the rate at which the water level is rising when the water is 3 m deep is approximately 0.0707 m/min.
Hope this helps, and remember, water problems can be tricky! Just keep swimming... I mean, calculating!
To find the rate at which the water level is rising, we need to use related rates.
Let's denote the radius of the water level in the tank as r and the height of the water level as h.
First, let's establish a relationship between the radius and the height. The water tank has the shape of an inverted circular cone, so we can use similar triangles to determine that the ratio of the radius to the height is constant:
(r / 3) = (h / 5)
Next, we can differentiate this equation with respect to time t, noting that both r and h are functions of time:
(d/dt)(r / 3) = (d/dt)(h / 5)
To find the rate at which the water level is rising, we need to find (dh/dt), which represents the rate of change of the height with respect to time t.
Simplifying and rearranging the equation, we get:
(1/3)(dr/dt) = (1/5)(dh/dt)
Now, let's substitute the given information into the equation. We're given that the water is being pumped into the tank at a rate of 2 m^3/min, which means (dV/dt) = 2. The volume of the cone at any height h can be calculated as:
V = (1/3)πr^2h
To find (dh/dt) when the water is 3 m deep, we need to find the rate at which the height is changing at that particular depth, which corresponds to (h = 3).
V = (1/3)πr^2(3) – substituting h = 3 into the volume equation
(dV/dt) = (1/3)π(2r)(dh/dt) – differentiating the volume equation with respect to time t
2 = (2/3)πr(dh/dt) – plugging in (dV/dt) = 2 and (dr/dt) = 0, since the radius of the water level remains constant
Now, we can solve for (dh/dt):
2 = (2/3)πr(dh/dt)
(dh/dt) = (3/2πr)(2)
(dh/dt) = (3/πr)
Now, plug in the given radius of the water tank (3 m) to find the rate at which the water level is rising when the water is 3 m deep:
(dh/dt) = (3/π(3))
(dh/dt) = 1/π
Therefore, the rate at which the water level is rising when the water is 3 m deep is 1/π m/min.
To find the rate at which the water level is rising, we can use the related rates formula:
V = (1/3) * π * r^2 * h
Where V is the volume of the water in the tank, π is a constant (approximately 3.14159), r is the base radius of the tank, and h is the height of the water.
We are given that the water is being pumped into the tank at a rate of 2 m^3/min. This means that the rate of change of volume with respect to time, dV/dt, is 2 m^3/min.
Now, we need to find the rate at which the water level is rising, dh/dt, when the water is 3 m deep.
Using the related rates formula, we can differentiate both sides of the equation with respect to time:
dV/dt = (1/3) * π * (2 * r * dr/dt * h + r^2 * dh/dt)
We know that the base radius, r, is constant at 3 m. So, dr/dt = 0.
Substituting the given values into the formula:
2 = (1/3) * π * (2 * 3 * 0 * 3 + 3^2 * dh/dt)
Simplifying:
2 = (1/3) * π * (0 + 9 * dh/dt)
2 = (1/3) * π * 9 * dh/dt
Multiplying both sides by 3/(π * 9):
2 * 3/(π * 9) = dh/dt
Simplifying:
dh/dt = 2/(3π)
Therefore, the rate at which the water level is rising when the water is 3 m deep is approximately 0.212 m/min.