Problem 67 - Calculus - AP Edition

Question

A metal rod \(15 \mathrm{cm}\) long and \(5 \mathrm{cm}\) in diameter is to be covered (except for the ends) with insulation that is \(0.1 \mathrm{cm}\) thick. Use differentials to estimate the volume of insulation. [Hint: Let \(\Delta V\) be the change in volume of the rod.]

Step-by-Step Solution

Answer

The estimated volume of the insulation is \\(7.65\pi \, \text{cm}^3 \\).

Step 1: Identify the Required Differential

We need to estimate the volume of the insulation using differentials. This requires calculating the change in volume, \(\Delta V\), when a thin layer of insulation is added to the rod.

Step 2: Define the Original Volume of the Rod

The original volume \(V\) of the rod is that of a cylinder, given by the formula: \[ V = \pi r^2h \] where \(r = 2.5 \, \text{cm}\) (radius) and \(h = 15 \, \text{cm}\) (height). Substitute these values into the equation: \[ V = \pi (2.5)^2(15) = 93.75\pi \, \text{cm}^3 \]

Step 3: Calculate the Radius After Adding Insulation

The insulation adds \(0.1 \, \text{cm}\) to the radius of the rod. Therefore, the new radius, \(r + dr\), is given by: \[ r + dr = 2.5 + 0.1 = 2.6 \, \text{cm} \]

Step 4: Compute the Volume with Insulation

The volume of the rod with insulation, \(V + dV\), is: \[ V + dV = \pi (2.6)^2(15) = 101.4\pi \, \text{cm}^3 \]

Step 5: Calculate the Differential and the Change in Volume

The change in volume, \(\Delta V\), is the difference between the volume with insulation and the original volume: \[ \Delta V = 101.4\pi - 93.75\pi = (101.4 - 93.75)\pi = 7.65\pi \, \text{cm}^3 \] So, the volume of the insulation is estimated to be \(7.65\pi \, \text{cm}^3 \).

Key Concepts

Cylinder Volume

To understand how the volume of a cylinder is calculated, we must familiarize ourselves with the formula. A cylinder's volume is determined by the equation: \[ V = \pi r^2h \]Here, \(r\) stands for the radius of the cylinder, and \(h\) stands for its height. This formula reflects that the volume is essentially the area of the circular base, \(\pi r^2\), multiplied by the height of the cylinder.

• This formula is crucial when dealing with problems involving cylindrical structures.
• Understanding this formula allows us to calculate different aspects like changes in volume.
• In our exercise, the rod is shaped like a cylinder where the radius and height are known, allowing the straightforward calculation of its volume.

By plugging in the given values for radius and height, we easily found the original volume of the rod to be \(93.75\pi \, \text{cm}^3\). With this foundation, we can explore changes in volume due to insulation.

Change in Volume

Change in volume helps us estimate how much space a new layer takes up when added to an existing object. Using the concept of differentials, we can calculate the approximate change in volume, \(\Delta V\), when a small thickness is added.

• This technique involves finding the difference between the volume of the object before and after the addition.
• In the exercise, we calculated the original volume of the metal rod and the volume with insulation.

The change in volume, \(\Delta V\), is obtained by subtracting the original volume \(V\) from the new volume \(V + dV\):\[ \Delta V = (101.4\pi) - (93.75\pi) \]Thus, the rod's insulation contributes an additional volume of \(7.65\pi \, ext{cm}^3\). This clear understanding of how to find \(\Delta V\) using differentials is very useful in practical applications.

Radius and Diameter

In a cylindrical object, understanding the radius and diameter is fundamental. The radius \(r\) is the distance from the center of the circle to any point on its edge, and the diameter \(d\) is twice the radius, spanning across the circle.

• Knowing these measurements is essential for volume calculations.
• The diameter is simply \(d = 2r\), allowing you to convert easily between the two.
• The diameter of the rod provided was 5 cm, resulting in a radius of 2.5 cm, crucial for our volume calculations.

When a new layer is added, like insulation, it increases the radius. In this exercise, the insulation increased the radius from 2.5 cm to 2.6 cm. This makes the recalculation of volume necessary to estimate the changes accurately. Having these basic distance measurements helps ensure precise mathematical work and success in such exercises.