Problem 58 - Calculus Early Transcendentals

Question

Water-level changes A conical tank with radius \(0.50 \mathrm{m}\) and height \(2.00 \mathrm{m}\) is filled with water (see figure). Water is released from the tank, and the water level drops by \(0.05 \mathrm{m}\) (from \(2.00 \mathrm{m}\) to \(1.95 \mathrm{m}\) ). Approximate the change in the volume of water in the tank. (Hint: When the water level drops, both the radius and height of the cone of water change.)

Step-by-Step Solution

Answer

Answer: To find the change in the volume of water in the conical tank, follow the steps mentioned above to first find the initial and final volumes of water in the tank and then calculate the difference between them. Step 1: The initial volume of water in the tank is: \(V_1 = \frac{1}{3}\pi (0.50\,\text{m})^2(2.00\,\text{m})\) Step 2: The change in radius when the height drops by 0.05 m is: \(r_2 = \frac{0.50\,\text{m} \times 1.95\,\text{m}}{2.00\,\text{m}}\) Step 3: The final volume of water in the tank is: \(V_2 = \frac{1}{3}\pi r_2^2 (1.95\,\text{m})\) Step 4: The change in volume is: \(\Delta V = V_1 - V_2\) Step 5: Substitute the expressions from Steps 1, 2, and 3 and simplify the expression for \(\Delta V\). By following these steps, you can find the change in the volume of water in the conical tank as the water level drops by 0.05 m.

Step 1: Find the initial volume of water in the tank

To find the initial volume of water, we can use the formula for the volume of a cone, which is given by \(V = \frac{1}{3}\pi r^2h\). The initial radius and height of the water in the tank are \(0.50\,\text{m}\) and \(2.00\,\text{m}\), respectively. Plugging these values into the formula, we get: \(V_1 = \frac{1}{3}\pi (0.50\,\text{m})^2(2.00\,\text{m})\)

Step 2: Find the change in radius

The ratio of the radius to the height of the tank remains constant when the water level drops. We can set up a proportion using the given dimensions and solve for the new radius, \(r_2\): \(\frac{0.50\,\text{m}}{2.00\,\text{m}} = \frac{r_2}{1.95\,\text{m}}\). Solving for \(r_2\), we get: \(r_2 = \frac{0.50\,\text{m} \times 1.95\,\text{m}}{2.00\,\text{m}}\)

Step 3: Find the final volume of water in the tank

Using the new radius and height, we can find the final volume of water in the tank using the cone volume formula: \(V_2 = \frac{1}{3}\pi r_2^2 (1.95\,\text{m})\)

Step 4: Calculate the change in volume

To find the change in volume, we can subtract the final volume of water \(V_2\) from the initial volume \(V_1\): \(\Delta V = V_1 - V_2\)

Step 5: Substitute and simplify

Now, simply substitute the expressions for \(V_1\), \(r_2\), and \(V_2\) found in Steps 1, 2, and 3, and simplify the expression for \(\Delta V\) to find the change in the volume of water in the tank. By following these steps, the change in the volume of water in the conical tank can be found.

Key Concepts

Water-level changes

Understanding water-level changes in a conical tank involves observing how the dimensions of the water column shift as water is added or removed. In this particular scenario, the water level in the tank decreases by 0.05 meters, moving from 2.00 meters to 1.95 meters. When the water level drops, it's not just the height that changes—its radius changes too. This simultaneously affects the volume since the water still forms a cone shape within the tank.

By understanding these changes, students can better appreciate how small adjustments in height may lead to significant changes in the overall volume of the tank. This serves as a real-world application of geometric principles in fluid dynamics, where simple measurements directly relate to complex calculations. Keep in mind that variables are interconnected, and changes in one dimension necessitate corresponding adjustments in others for precise volume assessments.

Conical tank

A conical tank plays a crucial role in this exercise as it acts as the vessel containing the water. The tank has a fixed radius of 0.50 meters and a height of 2.00 meters, which creates a perfect cone shape. Cones are characterized by their circular base and smoothly tapering sides that meet at a point called the apex.

When dealing with conical tanks, it's essential to remember that all measurements relate to this conical shape. This shape's geometry is defined by the proportion of its height to its radius, determining the tank's overall capacity. Hence, any change in the water level, as described in the exercise, means recalibrating these dimensions to understand the remaining or reduced volume.

This knowledge of conical properties is vital not just in theoretical math problems but also in practical applications like designing and using tanks for water storage and other fluids.

Volume approximation

Approximating volume changes in a conical tank utilizes basic mathematical concepts combined with logical reasoning. The formula for the volume of a cone, \( V = \frac{1}{3} \pi r^2 h \), lies at the core of this calculation. Initially, you calculate the volume with the original height and radius, providing the baseline measurement.

When the water level drops, both the height and radius changes must be considered to calculate the new volume. By using the proportion between the original radius and height, you can determine the new radius corresponding to the altered height. Substituting these values back into the cone volume formula gives the revised volume.

Finally, by subtracting the new volume from the initial volume, you can find the approximate change in volume. This step-wise approach enhances students' understanding of how precise mathematical tools and observations allow approximations and real-world applications, bridging theoretical math with tangible problems.