Question

An inverted cylindrical cone, \(20 \mathrm{ft}\) deep and \(10 \mathrm{ft}\) across at the top, is being filled with water at a rate of \(10 \mathrm{ft}^{3} / \mathrm{min}\). At what rate is the water rising in the tank when the depth of the water is: (a) 1 foot? (b) 10 feet? (c) 19 feet? How long will the tank take to fill when starting at empty?

Short answer

(a) 50.93 ft/min, (b) 0.509 ft/min, (c) 0.141 ft/min. Time to fill: ~104.72 min.

Step-by-step solution

Understand the Problem

We have an inverted cone with height 20 ft and diameter 10 ft. This translates to a radius of 5 ft at the top. We're given a filling rate of 10 ft³/min and need to determine how fast the water depth (height) changes at specific depths.

Determine the Relationship

Volume of a cone is given by the formula \( V = \frac{1}{3} \pi r^2 h \), where \( r \) is the radius and \( h \) is the height. In our inverted cone, radius and height are related due to similarity with the full cone's dimensions. Therefore, \( \frac{r}{h} = \frac{5}{20} = \frac{1}{4} \), giving \( r = \frac{h}{4} \).

Relate Volume to Height

Substitute \( r = \frac{h}{4} \) into the volume formula:\[ V = \frac{1}{3} \pi \left(\frac{h}{4}\right)^2 h = \frac{1}{3} \pi \frac{h^3}{16} = \frac{\pi h^3}{48} \]. This expresses volume as a function of water height.

Differentiate the Volume Equation

To find how the water height \( h \) changes with time, differentiate \( V = \frac{\pi h^3}{48} \) with respect to time \( t \):\[ \frac{dV}{dt} = \frac{\pi}{48} \times 3h^2 \frac{dh}{dt} = \frac{\pi h^2}{16} \frac{dh}{dt} \]. Given \( \frac{dV}{dt} = 10 \), solve for \( \frac{dh}{dt} \):\[ 10 = \frac{\pi h^2}{16} \frac{dh}{dt} \].

Solve for dh/dt at h = 1

Substitute \( h = 1 \) ft into the differentiated equation:\[ 10 = \frac{\pi (1)^2}{16} \frac{dh}{dt} \right) \rightarrow 10 = \frac{\pi}{16} \frac{dh}{dt} \]. Solve for \( \frac{dh}{dt} \):\[ \frac{dh}{dt} = \frac{160}{\pi} \approx 50.93 \text{ ft/min} \].

Solve for dh/dt at h = 10

Substitute \( h = 10 \) ft into the equation:\[ 10 = \frac{\pi (10)^2}{16} \frac{dh}{dt} \right) \rightarrow 10 = \frac{100\pi}{16} \frac{dh}{dt} \]. Solve for \( \frac{dh}{dt} \):\[ \frac{dh}{dt} = \frac{1.6}{\pi} \approx 0.509 \text{ ft/min} \].

Solve for dh/dt at h = 19

Substitute \( h = 19 \) ft into the equation:\[ 10 = \frac{\pi (19)^2}{16} \frac{dh}{dt} \right) \rightarrow 10 = \frac{361\pi}{16} \frac{dh}{dt} \]. Solve for \( \frac{dh}{dt} \):\[ \frac{dh}{dt} = \frac{160}{361\pi} \approx 0.141 \text{ ft/min} \].

Time to Fill the Tank

To find the time to fill the tank, calculate the total volume: For \( h = 20 \), \( r = 5 \), the total volume \( V = \frac{1}{3} \pi (5)^2 (20) = \frac{1}{3} \pi (100) \). This simplifies to \( rac{1000\pi}{3} \) ft³. Given the volume fills at 10 ft³/min:\[ t = \frac{1000\pi / 3}{10} = \frac{100\pi}{3} \approx 104.72 \text{ minutes} \].

Key concepts

Inverted Cone

An inverted cone is exactly what it sounds like—an upside-down cone, similar to a funnel shape. This geometric structure is seen in the water tank problem at hand. The tank is 20 feet deep, with a diameter of 10 feet at the top.
The diameter tells us the breadth of the circle at the wider end of the cone. To find the radius, which is half of the diameter, we divide the diameter by 2, giving us a radius of 5 feet. Understanding how the radius changes as the depth of water changes in the cone is crucial, especially when solving problems involving water levels in such containers.
In mathematics, inverted cones provide an interesting context for applying formulas related to volume, and exploring the relationship between height, radius, and volume using differential calculus.

Differential Calculus

Differential calculus is a field of mathematics focused on how functions change, which is especially handy when dealing with problems involving rates. In this exercise, differential calculus helps us understand how quickly the depth of water (\( h \)) is rising in the inverted cone as water flows in.
To address this, we use derivatives, which provide a means of expressing how a function changes over time. For instance, \( rac{dV}{dt} \) represents the rate at which water is filling the cone, known to be 10 cubic feet per minute.
We also derive another equation \( rac{dh}{dt} \), representing how fast the water's height is changing, using the relationship between volume and height. The core of solving related rates problems is to differentiate a function with respect to time, allowing us to relate various changing quantities within the problem.

Volume of a Cone

The volume of a cone can be calculated using the formula: \( V = \frac{1}{3} \pi r^2 h \), where \( V \) is the volume, \( r \) is the radius, and \( h \) is the height. This formula is central for determining how much water the inverted cone can hold.
In our specific water filling problem, because the cone is inverted and the rate of the radius to height is fixed (\( \frac{1}{4} \)), it simplifies the calculation. We substitute \( r = \frac{h}{4} \), which leads to the function \( V = \frac{\pi h^3}{48} \).
By substituting for \( h \) and differentiating, we can find how the volume changes as the water level changes, helping solve how quickly the water level rises depending on its current depth.

Water Filling Problem

The water filling problem exemplifies how real-life scenarios can be tackled using mathematical models. Here, the tank is an inverted cone, and the question involves understanding how fast the depth of the water (\( h \)) changes over time as the tank fills at a rate of 10 cubic feet per minute.
By using the relationship of volume and differentiating it with respect to time, we can establish how the water level rises. Substituting specific values for depths (such as 1, 10, and 19 feet) into the derived rate function allows us to see exactly how quickly the height is increasing at these points.

• At 1 foot, it rises quite quickly.
• At 10 feet, the rate lowers.
• Close to full (19 feet), the rate decreases even further.

Finally, to figure out how long it takes to completely fill the tank, we calculate the total volume capacity and divide it by the rate of filling, arriving at around 104.72 minutes.