Problem 97 - College Algebra and Calculus: An Applied Approach

Question

Solve for the indicated variable. Lateral Surface Area of a Cylinder Solve for \(h\) in \(A=2 \pi r h\).

Step-by-Step Solution

Answer

The solution to the problem is \(h = \frac{A}{2 \pi r}\).

Step 1: Identify the Given Formula

The given formula is \(A = 2 \pi r h\) which is used to calculate the Lateral Surface Area of a cylinder.

Step 2: Solve for \(h\)

To isolate \(h\), divide both sides of the equation by \(2 \pi r\). This will give: \(h= \frac{A}{2 \pi r}\).

Key Concepts

Lateral Surface Area

Understanding the lateral surface area is essential when dealing with three-dimensional shapes like cylinders. In the context of cylinders, the lateral surface area refers to the area of the sides of the shape—excluding the top and bottom circles. Imagine wrapping a label around a can, that label covers the lateral surface area of the cylindrical can.

The formula to calculate this area is expressed as \( A = 2 \times \pi \times r \times h \), where \( A \) represents the lateral surface area, \( r \) is the radius of the base circle, and \( h \) is the height of the cylinder. This relationship is critical to know as it ties together the elements of a cylinder's geometry with its surface size.

Cylinder

A cylinder is a basic geometric shape that has two parallel circular bases connected by a curved surface at a fixed distance from each other. That fixed distance is the height \( h \), and the two parallel circles have a radius \( r \). It's like a tube with two congruent circles on each end.

When dealing with problems involving cylinders, comprehending the parts of the cylinder is vital. The two bases contribute to the total surface area, but they are not considered in the lateral surface area calculations. It’s important to differentiate between the total surface area and the lateral surface area, as the latter only concerns the side surface, which excludes the area of the circles.

Algebraic Manipulation

Algebraic manipulation involves rearranging equations to solve for a particular variable. In our example, we are looking to solve for \( h \) from the equation \( A = 2 \times \pi \times r \times h \). To do this, we manipulate the equation in a way that isolates \( h \).

The process is straightforward. First, identify that \( h \) is being multiplied by \( 2 \), \( \pi \) (pi), and \( r \). To isolate \( h \), you would divide both sides of the equation by \( 2 \pi r \) since dividing by a number effectively 'undoes' the multiplication of that number. The result would be \( h = \frac{A}{2 \pi r} \), giving us the height \( h \) in terms of \( A \) and \( r \), allowing us to calculate it when given specific values of lateral surface area and radius.