Problem 20 - Calculus Early Transcendentals: Pearson New International Edition

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 \(V=\frac{1}{3} \pi h \cdot\left(a^{2}+a b+b^{2}\right)\).

Step-by-Step Solution

Answer

The water level rises at approximately 0.84 cm/min when the depth is 30 cm.

Step 1: Understand the Problem

We need to find the rate at which the water level is rising when the water depth is 30 cm. The tank is a frustum of a cone, and water is being added at a rate of 2 liters per minute.

Step 2: Convert Units

Since 1 liter = 1000 cubic centimeters, the water is being pumped in at a rate of 2000 cubic centimeters per minute. We will use this conversion when calculating how fast the water level is rising.

Step 3: Volume Formula for Frustum

The volume of a frustum of a cone is given by the formula:\[ V = \frac{1}{3} \pi h (a^2 + ab + b^2) \]where \( h \) is the height, \( a \) and \( b \) are the lower and upper radii, respectively. In this problem, these are the lower and upper radii of the frustum: 20 cm and 40 cm, and the total height of the frustum: 80 cm.

Step 4: Find Relationship Between h and r

Since similar triangles are formed by the proportions, the radius of the water at depth \( h_w = b_0 (1 - \frac{h_w}{h}) + b \frac{h_w}{h} \). When \( h_w = 30 \) cm, we need the corresponding radius \( r_w \). Solve the proportional change due to the depth: \[ r = 20 + (40-20) \frac{30}{80} = 27.5 \text{ cm} \]

Step 5: Calculate Volume for Depth of Water

Substitute \( r_w = 27.5 \text{ cm}, h_w = 30 \text{ cm} \) in volume formula for the frustum of section of water:\[ V_w = \frac{1}{3} \pi 30 ((27.5)^2 + 27.5 \times 27.5 + (27.5)^2) \]Simplify and calculate this volume depiction of water section at 30cm.

Step 6: Determine Rate of Change of Volume

The rate of change of volume of water, \( \frac{dV_w}{dt} \), is given by 2000 cm³/min from the pump.

Step 7: Relate Rate of Change of Height with Volume

We use the formula for the derivative of the volume with respect to height:\[ \frac{dV_w}{dh_w} = \frac{d}{dh_w}\left(\frac{1}{3} \pi h_w (27.5^2 + 27.5 \times 27.5 + 27.5^2)\right) \]\( V_w = \frac{1}{3} \pi h_w \times 3 \times (27.5)^2 \)\[ \frac{dV_w}{dh_w} = \pi \times (27.5)^2 \]

Step 8: Solve for Rate of Rise in Water Level

Now, use \( \frac{dV_w}{dt} = \pi \times (27.5)^2 \times \frac{dh_w}{dt} = 2000 \)\[ \frac{dh_w}{dt} = \frac{2000}{\pi \times (27.5)^2} \]Calculate \( \frac{dh_w}{dt} \) to find the rate at which the water level is rising when the depth is 30 cm.

Key Concepts

Rate of Change

In differential calculus, the concept of "rate of change" is crucial for understanding how one quantity varies in relation to another. In the given problem, we are primarily concerned with how fast the water level is rising over time. To find this rate, we need to consider the amount of water being pumped into the conical frustum per minute.

• The given rate of water flow is 2000 cubic centimeters per minute, which represents the change in volume over time, expressed as \( \frac{dV}{dt} \).
• To find the rate at which the water level is rising, indicated by \( \frac{dh}{dt} \), we must relate this volume change to the change in height.

By calculating \( \frac{dh}{dt} \), we discover how quickly the water rises in the tank, illustrating the practical application of differential calculus in solving real-world problems.

Volume of a Frustum

The frustum is a type of geometric shape derived from a cone, cut horizontally across to form two parallel circular ends of different sizes. Calculating the volume of a frustum is an essential step to determine how the water fills up this space.
The formula for the volume \( V \) of a frustum of a cone is:\[V = \frac{1}{3} \pi h (a^2 + ab + b^2),\]where:

• \( h \) is the vertical height of the frustum,
• \( a \) is the radius of the smaller base, and
• \( b \) is the radius of the larger base.

This formula allows us to compute the volume of the water within the given depth when these dimensions are placed into the equation. Understanding the structure of the frustum and how its geometry affects volume calculation is key in efficiently solving the problem.

Related Rates

Related rates problems involve finding how one variable changes in relation to another. They are a common application of derivatives in calculus.
In this problem, related rates are at play in calculating how the water level (height) changes as more water is added.

• We start with the known rate of change of the volume, \( \frac{dV}{dt} = 2000 \) cm³/min.
• We then need to find \( \frac{dh}{dt} \), the rate at which the water level rises. This involves using the chain rule to relate the changes in volume and height.

By understanding related rates, students can solve problems involving changes over time, which can be directly applied to fields like engineering and physics.

Geometry Applications

Using geometry to solve real-world problems offers tangible insights into how abstract mathematical concepts can be applied practically. In our exercise, we see several geometry applications at work:

• Shape of the Tank: Understanding the shape of the frustum helps in deriving the formula for the volume, allowing us to work out the exact amount of space available for the water to fill.
• Similar Triangles: The equivalence of triangles formed by water levels and tank dimensions allows us to derive different radii of the part-filled tanks at specific depths.

These applications emphasize the intersection between geometry and calculus – using geometric intuition to set up equations that calculus can solve for rate changes, showcasing the power of mathematical synergy.