Question

An inverted cone is \(2 \mathrm{m}\) high and has a base radius of \(\frac{1}{2} \mathrm{m}\). If the tank is full, how much work is required to pump the water to a level \(1 \mathrm{m}\) above the top of the tank?

Short answer

Answer: The work required to pump the water to a level 1 meter above the top of the tank is approximately 7.67 Joules.

Step-by-step solution

Find the volume of the cone

The first step is to find the volume of the entire cone, as this will be useful later. The formula for the volume of a cone is: \(V = \frac{1}{3} \pi r^2 h\) where r is the radius, and h is the height. Substituting the given values, we have: \(V = \frac{1}{3} \pi (0.5)^2 \cdot 2 = \frac{1}{3} \pi \cdot 0.5 = \frac{\pi}{6} m^3\)

Set up the integral

Since the cone is filled with water, we can think of it as being divided into many thin slices (disks) of water. To find the work needed to pump each slice, we need to multiply the volume of each slice by the distance it needs to be pumped, and then integrate over the entire volume of the tank. First, let's introduce a coordinate system: let the origin be at the tip of the cone, and let the positive y-axis run down the center of the cone. Consider a disk of water at depth y (with thickness dy). The height of the water above this disk is then (2-y) meters. So, to pump the water 1 meter above the tank, we need to lift the disk a distance of (2-y)+1=3-y meters. The radius of the disk will vary with the depth so we need to relate the radius to the depth. Since the slope of the cone is constant, we have: \(\frac{r}{y} = \frac{0.5}{2}\) Solving for r, we get: \(r = \frac{y}{4}\). The area of the disk is \(A=\pi r^2=\pi (\frac{y}{4})^2\) So, the volume of the disk is \(dV=A\cdot dy=\pi (\frac{y}{4})^2 dy\)

Find the work for each slice

The work needed to pump the water a distance of (3-y) meters is: \(dW = dV \cdot g \cdot (3-y)\) where \(g = 9.81 \,m/s^2\) is the acceleration due to gravity. Substituting the expression for dV, we have: \(dW = \pi (\frac{y}{4})^2 dy \cdot g \cdot (3-y)\)

Integrate over the entire volume of the tank

To find the total work required, we need to integrate the expression for dW over the entire volume of the tank, from y=0 (the tip of the cone) to y=2 (the base of the cone): \(W = \int_0^2 \pi (\frac{y}{4})^2 dy \cdot g \cdot (3-y)\) Calculating the integral, we get: \(W = \pi g \int_0^2 \frac{y^2}{16} (3-y) dy\)

Evaluate the integral

To evaluate the integral, we first expand the integrand: \(W = \pi g \int_0^2 \frac{3y^2}{16} - \frac{y^3}{16} dy\) Now, integrate term by term: \(W = \pi g [\frac{y^3}{16}\cdot \frac{3}{3} - \frac{y^4}{16} \cdot \frac{1}{4}]_0^2\) \(W = \pi g [\frac{8}{16} - \frac{16}{64}] = \pi g [\frac{1}{2} - \frac{1}{4}]\) \(W = \frac{\pi g}{4}\) Finally, substitute the value of g to get the numerical answer for the work required: \(W = \frac{\pi (9.81)}{4} \approx 7.67 J\) So, the work required to pump the water to a level 1 meter above the top of the tank is approximately 7.67 Joules.

Key concepts

Integral Calculus

Integral calculus is a branch of mathematics focused on the accumulation of quantities and the areas under and between curves. It primarily involves the use of integrals, which are mathematical tools that help sum up infinitesimal parts to find whole amounts. This is especially useful in problems where we deal with continuous data.

• An integral can provide the total sum of incremental changes.
• In our exercise, the slices of water inside the cone are summed.
• The solution sets up an integral to calculate the total work required to pump water.

When we encounter a problem involving physical applications, like removing water from a tank, integrals allow us to add up the work needed for every incremental slice of that water. By setting the bounds of integration, we're specifying from where to where we want to accumulate these quantities, allowing us to calculate the total with precision.

Volume of a Cone

Calculating the volume of a cone is essential when dealing with problems related to capacity and storage. The formula to find the volume of a cone is given by: \[ V = \frac{1}{3} \pi r^2 h \]where \(r\) is the radius of the base, and \(h\) is the height. This particular shape reminds us that the capacity of a cone is only a third the capacity of a cylinder with the same base and height.

• This formula helps to determine how much material or substance can fit inside.
• In the provided exercise, this calculation informs us of the total water volume in the cone.

Understanding the dimensions of the cone in the exercise allows us to derive the limits and variables needed for further calculus steps, such as setting up an integral to assess the work required for varying tasks.

Work and Energy

Work and energy are fundamental concepts in physics, linking the exertion of force to the displacement of objects. Work is calculated as the product of force and distance in the direction of the force. In calculus, this often translates into integrating force with respect to distance.

• The work needed in our exercise depends on lifting slices of water to a certain height.
• The gravitational force involved is due to the weight of water.
• We integrate to include contributions from all parts of the water in the tank.

Applying the concept of work and energy to this problem helps in determining the necessary effort required to pump water out from a certain depth to above the tank. By multiplying the volume of each slice by the force of gravity and the vertical distance it must be moved, and integrating these elements, we find the entire work required.

Physics Applications

There are numerous physics applications where calculus, and specifically integrals, play a crucial role. Understanding how objects and fluids behave under various forces can help solve practical, real-world problems.

• Calculus helps us determine quantities that vary continuously, like the water in a tank.
• It provides a way to sum infinitesimally small slices through integration.
• Work problems, like lifting water to a higher level, are classic applications of calculus in physics.

In this exercise, integrating concepts from physics into calculus becomes essential. The set-up for calculating work to pump water is based on real-world applications, combining geometry (analysing the cone) with physical principles (force and movement) to fully grasp the problem. This illustrates the effectiveness of calculus for solving complex tasks in physics and engineering.