Problem 23 - The Calculus with Analytic Geometry

Question

An open cylindrical tank is to have an outside coating of thickness \(\frac{1}{8}\) in. If the inner radius is \(6 \mathrm{ft}\) and the altitude is \(10 \mathrm{ft}\), find by differentials the approximate amount of coating material to be used.

Step-by-Step Solution

Answer

About 3.93 cubic feet of coating material.

Step 1 - Understand the Given Problem

The problem involves an open cylindrical tank where you are asked to find the approximate amount of coating material due to an added thickness using differentials. The inner radius is 6 ft, the height is 10 ft, and the coating thickness is \(\frac{1}{8}\) in.

Step 2 - Convert Thickness to Feet

Convert the coating thickness from inches to feet: \(\frac{1}{8}\) in = \(\frac{1}{8} \times \frac{1}{12}\) ft = \(\frac{1}{96}\) ft.

Step 3 - Use Differential to Approximate Increase in Volume

The differential formula for the volume \( dV \) of a cylinder is \( dV = 2 \pi r h \,dr + \pi r^2 \,dh \). Since the altitude does not change, \( dh = 0 \). So, \( dV = 2 \pi r h \,dr \).

Step 4 - Plug in the Values

Substitute \( r = 6 \) ft, \( h = 10 \) ft, and \( dr = \frac{1}{96} \) ft into the differential formula: \( dV = 2 \pi (6) (10) (\frac{1}{96}) \) = \( \frac{120 \pi}{96} \) = \( \frac{5 \pi}{4} \) cubic feet.

Step 5 - Calculate the Final Volume

Perform the calculation: \( \frac{5 \pi}{4} \approx 3.93 \) cubic feet.

Key Concepts

differentials in calculus

Differentials in calculus help us approximate small changes in variables. Imagine you have a small disk. If you add a tiny layer of thickness to the disk, differentials help you understand how much volume has been added. This method applies not just to disks, but to cylinders too, like in our problem. Differentials let us approximate these small changes without doing complicated calculations.

When we use differentials, it means we're looking at how a function changes as its variables change a little. This is very useful in practical problems where exact calculation might be difficult. In this problem, we are adding a thin layer to a cylindrical tank and want to know how much material is needed. Using differentials simplifies this.

The differential formula for volume (\( dV \)) of a cylinder considers small changes in radius (\( dr \)) and height (\( dh \)). Here, only the radius changes because the coating adds to the tank's radius, not height. So, our differential equation focuses on this change.

cylinder volume

The volume of a cylinder is straightforward to calculate if you know the radius and height. The formula is:
\( V = \pi r^2 h \).

But what happens if we slightly change the size of the cylinder, like adding a thin coating of material around it? That's what our exercise is about. For small changes like this, we use differentials.

In the context of our problem, the initial inner radius of the tank is 6 feet and the height is 10 feet. Now, we add a coating of thickness \(1/8 \) inch. Before finding the new volume, we must convert the thickness from inches to feet: \(1/8 \) inch = \(1/96 \) foot.

Adding this thickness changes the radius slightly. To find out how much this changes the volume, we use the differential formula for the volume of a cylinder:
\[ dV = 2 \pi r h dr \]
The height (\( h \)) does not change here.

calculus applications

Calculus is more than just abstract math. It has real-world applications, like approximating changes in geometric shapes. Our problem demonstrates how calculus helps in practical scenarios.

Imagine industries dealing with tanks, pipes, or any cylindrical objects. Knowing how a small change in dimension affects something can save material and cost. Here, differentials provide a quick way to calculate the needed amount of coating material without excessive detailed measurements.

By applying differentials, we estimate the extra volume due to a coating. We found that adding a small thickness to the tank could be approximated using the formula:
\[ dV = 2 \pi r h dr \]
After calculating, we discovered the new volume would be about 3.93 cubic feet. This rounded number comes from easier steps, useful for engineers and everyday planning. Hence, calculus offers simplified yet accurate results, beneficial for various fields.