Problem 28 - University Calculus: Early Transcendentals

Question

A draining conical reservoir Water is flowing at the rate of \(50 \mathrm{m}^{3} / \mathrm{min}\) from a shallow concrete conical reservoir (vertex down) of base radius \(45 \mathrm{m}\) and height \(6 \mathrm{m} .\) a. How fast (centimeters per minute) is the water level falling when the water is 5 m deep? b. How fast is the radius of the water's surface changing then? Answer in centimeters per minute.

Step-by-Step Solution

Answer

a) Water level falls at about -1.13 cm/min. b) Water surface radius changes at about -8.475 cm/min.

Step 1: Write the Formula for the Volume of a Cone

The volume \( V \) of a cone is given by the formula: \[V = \frac{1}{3} \pi r^2 h\]where \( r \) is the radius of the base and \( h \) is the height.

Step 2: Relate the Variables to the Dimensions of the Cone

In the given problem, the radius and height vary but maintain the same ratio as the full cone. The full cone has a base radius of 45m and height of 6m, giving a ratio of \( \frac{r}{h} = \frac{45}{6} = 7.5 \). This implies \( r = 7.5h \).

Step 3: Substitute Radius in the Volume Formula

Substitute \( r = 7.5h \) into the volume formula:\[V = \frac{1}{3} \pi (7.5h)^2 h = \frac{1}{3} \pi \times 56.25 h^3 = 18.75 \pi h^3\]

Step 4: Differentiate the Volume with Respect to Time

We need to differentiate the volume with respect to time:\[\frac{dV}{dt} = 18.75 \pi \cdot 3h^2 \cdot \frac{dh}{dt} = 56.25 \pi h^2 \cdot \frac{dh}{dt}\]We know the change in volume over time is \( \frac{dV}{dt} = -50 \) m³/min.

Step 5: Solve for the Rate of Change of Height

Set up the equation using \( \frac{dV}{dt} = -50 \) to solve for \( \frac{dh}{dt} \):\[-50 = 56.25 \pi (5)^2 \frac{dh}{dt}\]Thus, solving for \( \frac{dh}{dt} \):\[\frac{dh}{dt} = \frac{-50}{56.25 \pi \times 25} = \frac{-2}{56.25 \pi} = \frac{-8}{225\pi} \approx -0.0113 \text{ m/min}\]Convert to cm/min: \( \approx -1.13 \text{ cm/min}\).

Step 6: Differentiate the Relationship Between Radius and Height

Using \( r = 7.5h \), differentiate with respect to time:\[\frac{dr}{dt} = 7.5 \cdot \frac{dh}{dt}\]Substitute \( \frac{dh}{dt} = -0.0113 \) m/min:\[\frac{dr}{dt} = 7.5 \times -0.0113 = -0.08475 \text{ m/min}\]Convert to cm/min: \( \approx -8.475 \text{ cm/min}\).

Key Concepts

Conical Reservoir

A conical reservoir is essentially a large cone-shaped tank that holds fluids like water. These structures are quite common in fields like hydrology and engineering, as they efficiently store and manage liquid volumes. Understanding how the dimensions of this reservoir change with time involves studying the dynamics of its drainage process. In this exercise, we're looking at a conical reservoir with a base radius of 45 meters and a height of 6 meters.
This cone has its vertex facing downward, and as water drains out, both the height of the water level and the radius of the water's surface decrease. The challenge lies in linking these dimensions, as the relationship between the radius and the height of the water is crucial. Because the cone maintains its shape as it drains, the ratio of the radius to the height remains constant, which is practical information when setting up equations to find how fast each dimension changes as the water drains.
Understanding a conical reservoir helps visualize how water flow dynamics apply in real-world scenarios, contributing to diverse applications such as water supply management and flood control.

Volume of a Cone

The volume of a cone is a fundamental concept in geometry and is used extensively in calculus when dealing with problems involving varying quantities. The formula for the volume of a cone is expressed as:
\[V = \frac{1}{3} \pi r^2 h\]
Where:

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

This equation helps us understand how changes in the height or radius affect the overall volume. In the context of our problem, as the water drains from the conical reservoir, both \( r \) and \( h \) are functions of time. However, they are not independent since the cone retains its shape, meaning that the ratio of the radius to the height remains constant at \( 7.5 \).
By substituting and simplifying the variables into the volume formula, we reduce the dimensions we need to track. This simplification is vital for differentiating with respect to time since it reflects how the various dimensions affect the rate at which the volume changes. Understanding the volume's dependency on both the radius and height is crucial when calculating how fast these parameters change over time.

Differentiation with Respect to Time

Differentiation with respect to time, often referred to as finding related rates, is a powerful tool in calculus used to understand how different variables change in relation to each other over time. This technique is especially useful in scenarios like our conical reservoir, where variables such as height and radius are interdependent and evolve as the water flows out.
Using the chain rule, we differentiate the volume of the cone with respect to time. We first express the volume in terms of one variable—height, in this case—by using the relationship \( r = 7.5h \). The differentiated formula with respect to time is given as:
\[\frac{dV}{dt} = 18.75 \pi h^2 \frac{dh}{dt}\]
Given \( \frac{dV}{dt} = -50 \) m³/min, we can solve for \( \frac{dh}{dt} \), which represents how quickly the water level is falling. Alongside, by differentiating the radius-height relationship, we obtain \( \frac{dr}{dt} \), indicating how fast the surface radius shrinks.
Hence, by applying related rates, we gain insight into the speed of change within the reservoir's dimensions, illustrating the dynamic nature of systems undergoing various transformations over time.