Question

MATHEMATICAL CONNECTIONS The surface area \(S\) of a right circular cone with a slant height of 1 unit is given by \(S=\pi r+\pi r^2\), where \(r\) is the radius of the cone. a. Use completing the square to show that $$ r=\frac{1}{\sqrt{\pi}} \sqrt{S+\frac{\pi}{4}}-\frac{1}{2} \text {. } $$ b. Graph the equation in part (a) using a graphing calculator. Then find the radius of a right circular cone with a slant height of 1 unit and a surface area of \(\frac{3 \pi}{4}\) square units.

Short answer

Part a) \(r = \sqrt {S + \pi /4} / \sqrt {\pi} - 1/2\), Part b) radius of a right circular cone with a slant height of 1 unit and a surface area of \(3 \pi /4\), is approximately \(0.203\) units.

Step-by-step solution

Completing the Square

The given formula for the surface area \(S\) of a right circular cone is \(S=r\pi +r^2\pi\). That can be rearranged to \(r^2\pi + r\pi - S = 0\). Now, to complete the square with regard to \(r\), the equation needs to be in the form \((r+d)^2\). First, calculate the square of half the coefficient of \(r\) (which is \(1/2\)). This gives \((1/2)^2 = 1/4 = \pi /4\). Add and subtract this inside the equation: \(r^2\pi + r\pi + \pi /4- \pi /4 - S = 0\). This allows to rewrite the equation in the form \(( r + 1/2 )^2\pi - \pi /4 - S = 0\). Solve for \(r\): \(r = \sqrt {S + \pi /4} / \sqrt {\pi} - 1/2 \))

Graph the Function

This step involves graphing the equation found in the previous step, \(r = \sqrt {S + \pi /4} / \sqrt {\pi} - 1/2\), which represents a square-root function that is shifted on the coordinate plane. For this step, a graphing calculator or a graphing software can be used. This graph will show the relationship between the surface area \(S\) and the respective radius \(r\).

Find a Specific Radius

To find the radius of a right circular cone with a slant height of 1 unit and a surface area of \(3 \pi /4\), the value for \(S\) (which is given as \(3 \pi /4\)) needs to be plugged into the function derived in step 1. This will interpret a specific value for \(r\). Doing the calculations gives \(r = \sqrt {3 \pi /4 + \pi /4} / \sqrt {\pi} - 1/2 \approx 0.203\). Hence, the radius of the right circular cone is approximately \(0.203\) units.

Key concepts

Completing the Square

Understanding completing the square is a foundational skill in algebra that simplifies many processes, including solving quadratic equations and graphing parabolic curves. This method reshapes a quadratic equation into a perfect square trinomial, plus or minus a constant. Here’s why it is so powerful:

Consider a quadratic equation in the standard form \( ax^2 + bx + c = 0 \). By manipulating this equation using the completing the square technique, you can find the vertex of a parabola and readily solve the equation.

A Step-by-Step Look

First, you need to ensure the coefficient of the square term is 1 (by dividing if necessary). Then, you take half of the linear term's coefficient (\(b/2\)), square it, and add it to both sides. This process creates a perfect square trinomial on one side, often leading to an equation of the form \( (x+d)^2 = e \), which is easier to solve.

In our exercise, to find the radius (\(r\)) of the cone, we use completing the square to transform the surface area equation \(S = \(\pi r + \pi r^2\)\) into a form where \(r\) can be clearly isolated. By adding and subtracting \(\pi / 4\), we convert the equation into a perfect square.

Graphing Equations

Now, let’s understand the crux of graphing equations, an essential aspect of visualizing mathematical relationships and functions. By graphing the equation of a geometric shape or function, we can interpret and analyze the relationship between variables.

A graphing calculator or software can test our understanding by allowing us to enter equations and view the corresponding graph. In our given problem, the equation \( r = \sqrt{S + \pi/4} / \sqrt{\pi} - 1/2 \) is a transformed algebraic function relating the surface area \(S\) of a cone to its radius \(r\).

Visual Insights

Graphing this equation depicts how the radius changes with different surface area values. Typically, the graph of a square-root function will rise slowly; the increase in surface area results in a subtle increase in radius. This visualization helps students intuitively understand the proportional relationship between \(S\) and \(r\) for a right circular cone.

Algebraic Functions

Lastly, let's delve into algebraic functions. These are expressions using mathematical operations that relate an input to an output. For every value plugged into the function, there is a corresponding single output, following a rule expressed by the function equation.

In the case of our textbook problem, we deal with a square-root function, which is a specific type of algebraic function. Algebraic functions can include polynomials, rationals, radicals, and more. Each type of function has its own unique properties and graphs accordingly.

Functions at Work

For the surface area of our cone, \(S\), when seen as a function of the radius, \(r\), helps us understand how dimensions of the cone affect its surface area. Likewise, when we graph the function and find specific points, like the radius for a given surface area, we are essentially using the algebraic formula inversely to deduce a measurement from an area, showcasing the versatility of algebra in geometric contexts.