Water is poured into a conical tank 6m across the top and 8m deep at the rate of 10m/min. How fast is the water level rising when the water in the tank is 5m deep?
Answer should be 0.9054m/min, 6m across means diameter therefore r=3m,
Well, well, well, it seems we have a conical tank craving some water! Let's see what we can do with this situation.
First things first, we need to figure out the volume of the tank. We know it's a cone, so the volume can be calculated using the formula V = (1/3) * π * r² * h, where r is the radius of the top and h is the height of the water.
Given that the tank is 6m across the top, the radius is 3m. We also know that water is being poured in at a rate of 10m/min and we want to know how fast the water level is rising when it's 5m deep.
Now, let's differentiate the volume V with respect to time t to find the rate at which the water level is rising.
dV/dt = (1/3) * π * (2r) * dr/dt * h
Here, dr/dt represents the rate at which the radius is changing, and we want to find dh/dt, the rate at which the height is changing.
Since we know the dimensions of the tank, we can use similar triangles to find that the height h is (4/3) times the radius r. So, dh/dt = (4/3) * dr/dt.
Now we have all the pieces of the puzzle. We know that dr/dt = 10m/min (as given), and we want to find dh/dt when h = 5m.
dh/dt = (4/3) * dr/dt = (4/3) * 10m/min ≈ 13.33m/min
Voila! The water level is rising at approximately 13.33 meters per minute when the water in the tank is 5 meters deep. Just make sure not to drown in all that high-speed rising water!
To solve this problem, we can use related rates. The key idea is to relate the rates at which the water level and the volume of water in the tank are changing.
Let's denote the radius of the conical tank as "r" and the depth of the water as "h." In this case, we know that the radius is constant at 6m and the rate at which the water is poured into the tank is 10m/min. We need to find the rate at which the water level is rising when the water in the tank is 5m deep (dh/dt).
First, we need to establish the relationship between the radius, height, and volume of the conical tank.
The formula for the volume of a cone is given by V = (1/3) * pi * r^2 * h.
Since the radius is constant at 6m, we can simplify the formula to V = (1/3) * pi * 6^2 * h.
Differentiating the formula with respect to time (t), we get:
dV/dt = (1/3) * pi * 6^2 * dh/dt.
Now, we need to solve for dh/dt.
Since the rate of change of volume (dV/dt) is given as 10m/min, and we want to find the rate of change of height (dh/dt) when h = 5m, we can substitute these values into the equation:
10 = (1/3) * pi * 6^2 * dh/dt.
We can rearrange the equation to solve for dh/dt:
dh/dt = 10 * 3 / (pi * 36).
Evaluating this expression, we get:
dh/dt ≈ 0.265 m/min.
Therefore, the water level is rising at a rate of approximately 0.265 m/min when the water in the tank is 5m deep.
why is it the answer become 0.905 when you are going to solve it in calculator it will give a results of 0.045
radius of water level --- r
height of water level --- h
by ratio:
r/h = 6/8
8r= 6h
r = 3h/5
V = (1/3)πr^2 h
= (1/3)π(9h^2/25)(h)
=3/25)πh^3
dV/dt = (9/25)π h^2 dh/dt
given: when h = 5 , dV/dt = 10
10 = (9/25)π (25) dh/dt
dh/dt = 10/(9π)
check my arithmetic