Question

Water is pumped at a uniform rate of 2 liters \((1\) liter \(=1000\) cubic centimeters \()\) per minute into a tank shaped like a frustum of a right circular cone. The tank has altitude 80 centimeters and lower and upper radii of 20 and 40 centimeters, respectively (Figure 11 ). How fast is the water level rising when the depth of the water is 30 centimeters? Note: The volume, \(V\), of a frustum of a right circular cone of altitude \(h\) and lower and upper radii \(a\) and \(b\) is \(V=\frac{1}{3} \pi h \cdot\left(a^{2}+a b+b^{2}\right)\).

Short answer

The water level is rising at approximately 1.14 cm/min.

Step-by-step solution

Understand the Problem

We need to determine the rate at which the water level is rising inside a frustum of a cone. We are given the rates of flow and dimensions of the frustum. This involves differentiating the formula for the volume of the frustum with respect to time.

Find the Volume Function in Relation to Water Level Height

Given the volume of a frustum as \( V = \frac{1}{3} \pi h \left(a^2 + ab + b^2\right) \), we need to express the function in terms of the water level, \( h = x \), where the height of water is increasing. Here, \( a = 20 \text{ cm} \), \( b = 40 \text{ cm} \), and \( x \) is the current height of the water. The top radius is variable, depending on x, governed by similar triangles.

Use Similar Triangles to Find Variable Radius

Since the radius varies as the water level rises, use the similar triangles relationship: \( \frac{R}{40} = \frac{x}{80} \). Therefore, the radius at any point, \( R \), when the water level is \( x \), is \( R = \frac{x}{2} \).

Formulate Volume Expression with Depth x

The volume V as a function of \( x \) is given by substituting for \( R \): \[ V = \frac{1}{3} \pi x \left(20^2 + 20\cdot\frac{x}{2} + \left(\frac{x}{2}\right)^2\right) = \frac{1}{3} \pi x \left(400 + 10x + \frac{x^2}{4}\right) \] simplifying gives: \[ V = \frac{1}{3} \pi \left(400x + 10x^2 + \frac{x^3}{4}\right) \].

Differentiate Volume Function with Respect to Time

Compute \( \frac{dV}{dt} \) given that the tank is filled at a rate of 2000 cm³/min. First calculate \( \frac{dV}{dx} = \frac{1}{3} \pi \left(400 + 20x + \frac{3x^2}{4}\right) \) and use the formula \( \frac{dV}{dt} = \frac{dV}{dx} \cdot \frac{dx}{dt} \). Thus, \( 2000 = \frac{1}{3} \pi (400 + 20 \times 30 + \frac{3 \times 30^2}{4}) \cdot \frac{dx}{dt} \).

Solve for Rate of Rising Water Level

Substitute known values: \({dV}/{dt} = 2000 \text{ cm}^3/\text{min} \). Calculate the expression for \( \frac{dV}{dx} \): \( \frac{dV}{dx} \big|_{x=30} = \frac{1}{3} \pi (400 + 600 + 675) = \frac{1}{3} \pi \times 1675 \). Solving \( 2000 = \frac{1}{3}\pi \times 1675 \cdot \frac{dx}{dt} \), isolate \( \frac{dx}{dt} \approx 1.14 \text{ cm/min} \).

Key concepts

Volume of a Frustum

The volume of a frustum of a right circular cone plays a crucial role in solving problems involving water tanks, like this one. A frustum is formed from a cone by slicing it parallel to its base, resulting in a shape with two parallel circular ends of different sizes. The volume formula for a frustum is given by:

• \[ V = \frac{1}{3} \pi h (a^2 + ab + b^2) \]

where \( h \) is the altitude or height of the frustum, and \( a \) and \( b \) are the radii of the lower and upper bases, respectively.
In our water tank problem, we substitute the known values to find the volume in terms of the water level height. When the water reaches a depth of 30 cm, the volume as a function will help us understand how the water's volume changes as the water level rises.
Understanding this formula allows us to link the volume at any given height to the related rates problem.

Similar Triangles

When the water level increases, the frustum's radius changes, and this relationship is uncovered using similar triangles. Similar triangles have equivalent corresponding angles and proportional sides. In the context of our frustum:

• The top and bottom surfaces of the frustum form two similar triangles with each other.
• Using the properties of similar triangles, we can write:
  • \[ \frac{R}{40} = \frac{x}{80} \]
• This equation simplifies to find the radius \( R \) at any water level height \( x \):
  • \[ R = \frac{x}{2} \]

We use this relationship of changing radius to adjust the volume formula, as it affects how we express the volume dependent on the height of the water.
This step is essential because it sets up the foundation for how the geometric principles align with the algebraic manipulation needed for our calculus operations.

Differentiation in Calculus

Differentiation is a key tool in calculus that aids in determining the rate at which quantities change. To solve the problem of how fast the water level rises, we need to differentiate the volume of the frustum with respect to the water level height, x.
Given:

• Start with the function for volume:
  • \[ V = \frac{1}{3} \pi x \left(400 + 10x + \frac{x^2}{4}\right) \]
• The differentiation process requires finding \( \frac{dV}{dx} \), the derivative of V with respect to x.

The formula becomes a tool to connect physical changes in the volume of water (as the water level rises) with temporal changes.
This lets us later equate this to the flow rate, forming the foundation for our rate equation. Differentiation transforms our static volume expression into a dynamic rate of change representation.

Rate of Change

In related rates problems like this one, the goal is to connect how one quantity's rate of change affects another. Given that water is pumped into the frustum at a steady rate of 2000 cm³/min, we aim to find how quickly the water level height \( x \), rises.
The relation:

• \[ \frac{dV}{dt} = \frac{dV}{dx} \cdot \frac{dx}{dt} \]

allows us to find \( \frac{dx}{dt} \), the rate at which the water level rises, by using the known rate of \( \frac{dV}{dt} \).

• Calculate \( \frac{dV}{dx} \) and substitute into the equation along with \( \frac{dV}{dt} \).
• Solving gives us exactly how fast the water is rising at any moment, which, when x is 30, results in a rate of approximately 1.14 cm/min.

The chain rule in calculus neatly ties together how a change in one rate can affect another, making this link clear and approachable. This comprehension underpins the solution to similar problems you'll encounter.