Problem 27 - Calculus Volume 1

Question

Consider a right cone that is leaking water. The dimensions of the conical tank are a height of \(16 \mathrm{ft}\) and a radius of \(5 \mathrm{ft}\). If the water level is decreasing at a rate of 3 in./min when the depth of the water is \(8 \mathrm{ft}\), determine the rate at which water is leaking out of the cone.

Step-by-Step Solution

Answer

Water is leaking out at a rate of \(5\pi \text{ ft}^3/\text{min}\).

Step 1: Understand the Problem

We are given a right cone leaking water with a height of 16 ft and a radius of 5 ft. The water level decreases at a rate of 3 in./min when the current water depth is 8 ft. We need to determine the rate at which water is leaking out of the cone.

Step 2: Convert Units

The rate at which the water level decreases, given as 3 in./min, should be converted to feet per minute for consistency with the given measurements. There are 12 inches in a foot, so \[ \text{rate} = \frac{3}{12} \text{ ft/min} = 0.25 \text{ ft/min} \]

Step 3: Relate the Variables

Let \(h\) be the height of water and \(r\) the radius when the water height is \(h\). The formula for the volume \(V\) of a cone is \[ V = \frac{1}{3} \pi r^2 h \]We need to express \(r\) in terms of \(h\) using similar triangles. The full height is 16 ft and the full radius is 5 ft, so the ratio is \[ \frac{r}{h} = \frac{5}{16} \Rightarrow r = \frac{5}{16} h \]

Step 4: Differentiate the Volume Equation

Substitute \(r = \frac{5}{16} h\) into the volume formula \[ V = \frac{1}{3} \pi \left(\frac{5}{16} h\right)^2 h = \frac{1}{3} \pi \times \frac{25}{256} h^3 \]This simplifies to \[ V = \frac{25\pi}{768} h^3 \]Differentiate with respect to time \(t\):\[ \frac{dV}{dt} = \frac{25\pi}{768} \times 3h^2 \frac{dh}{dt} \]

Step 5: Substitute Given Values and Solve

We are given the water level \(h = 8\) ft and \(\frac{dh}{dt} = -0.25\) ft/min. Substitute these into the differentiated equation:\[ \frac{dV}{dt} = \frac{25\pi}{768} \times 3 \times 8^2 \times (-0.25) \]Calculate:\[ \frac{dV}{dt} = \frac{25\pi}{768} \times 3 \times 64 \times (-0.25) = -5\pi \text{ ft}^3/\text{min} \]

Step 6: Interpret the Result

The negative sign indicates that the volume of water is decreasing. Therefore, the rate at which water is leaking out of the cone is \(5\pi \text{ ft}^3/\text{min} \).

Key Concepts

Volume of a Cone

The volume of a cone is an essential concept in geometry and calculus, especially when dealing with related rates problems. The formula to find the volume of a cone is given by: \[ V = \frac{1}{3} \pi r^2 h \]where:

• \(V\) is the volume of the cone,
• \(r\) is the radius of the cone's base, and
• \(h\) is the height of the cone.

It's important to note that the formula incorporates both the area of the circular base, \(\pi r^2\), and the height \(h\).
This formula helps in calculating how the volume changes as either the height or the radius changes. Understanding how these variables are connected allows us to determine rates of change, which is essential for solving related rates problems like the one given in the exercise.

Similar Triangles

Similar triangles play a crucial role when we need to relate different dimensions of geometric shapes. In the given exercise, the full cone and the water section form two similar triangles.
These triangles have:

• the same shape but may differ in size,
• equal corresponding angles, and
• proportional sides.

For the cone problem, the ratio of the radius to the height in the full cone is the same as the ratio in the smaller, water-filled section. Therefore, if the full height is 16 ft and the full radius is 5 ft, the relationship can be expressed as:\[ \frac{r}{h} = \frac{5}{16} \Rightarrow r = \frac{5}{16} h \]This relationship allows us to express one variable in terms of another, greatly simplifying the problem-solving process. By using similar triangles, complex 3D shapes can be transformed into simpler 2D problems, making it easier to track how one measurement affects another.

Differentiation

Differentiation is a fundamental tool in calculus used to determine the rate of change of one variable with respect to another. In the context of related rates, differentiation helps us analyze how changing one quantity affects another over time.
For the cone exercise, after substituting the radius \(r\) in terms of the height \(h\) into the volume formula, we differentiate the volume \(V\) with respect to time \(t\):\[ \frac{dV}{dt} = \frac{25\pi}{768} \times 3h^2 \frac{dh}{dt} \]Here,

• \(\frac{dV}{dt}\) represents the rate at which the volume is changing, and
• \(\frac{dh}{dt}\) denotes the rate at which the height is changing.

Differentiation provides a way to connect these rates. It helps in the practical interpretation of how fast a particular feature of a physical system, such as the volume of water in a cone, is changing. When combined with the given rates in the exercise, the differentiation process helps solve for unknown rates such as the rate at which water is leaking from the cone.

Unit Conversion

Unit conversion is critical for consistency and accuracy in calculations. When solving the cone problem, we deal with a height decreasing at a rate of 3 inches per minute, while other measurements are in feet.
To maintain uniformity:

• Convert inches to feet using the conversion factor: 12 inches equals 1 foot.

Thus,\[ \frac{3\text{ in}}{12} = 0.25 \text{ ft/min} \]
Using consistent units simplifies problem-solving and reduces errors.Neglecting the step of unit conversion can lead to incorrect solutions and misunderstandings.

• Always verify that all measurements are in compatible units before starting calculations.

This ensures that when we calculate and interpret rates of change, like those involved in the cone problem, we do so accurately.