### Applications Using Radicals

Mathematicians and physicists have studied the motion of pendulums in great detail because this motion explains many other behaviors that occur in nature. This type of motion is called **simple harmonic motion** and it is important because it describes anything that repeats periodically. Galileo was the first person to study the motion of a pendulum, around the year 1600. He found that the time it takes a pendulum to complete a swing doesn’t depend on its mass or on its angle of swing (as long as the angle of the swing is small). Rather, it depends only on the length of the pendulum.

The time it takes a pendulum to complete one whole back-and-forth swing is called the **period** of the pendulum. Galileo found that the period of a pendulum is proportional to the square root of its length: \begin{align*}T = a \sqrt{L}\end{align*}. The proportionality constant, \begin{align*}a\end{align*}, depends on the acceleration of gravity: \begin{align*}a = \frac{2 \pi}{\sqrt{g}}\end{align*}. At sea level on Earth, acceleration of gravity is \begin{align*}g = 9.81 \ m/s^2\end{align*} (meters per second squared). Using this value of gravity, we find \begin{align*}a = 2.0\end{align*} with units of \begin{align*}\frac{s}{\sqrt{m}}\end{align*} (seconds divided by the square root of meters).

Up until the mid \begin{align*}20^{th}\end{align*} century, all clocks used pendulums as their central time keeping component.

#### Real-World Application: Pendulums

Graph the period of a pendulum of a clock swinging in a house on Earth at sea level as we change the length of the pendulum. What does the length of the pendulum need to be for its period to be one second?

The function for the period of a pendulum at sea level is \begin{align*}T = 2 \sqrt{L}\end{align*}.

We start by making a table of values for this function:

\begin{align*}L\end{align*} | \begin{align*}T = 2 \sqrt{L}\end{align*} |
---|---|

0 | \begin{align*}T = 2 \sqrt{0} = 0\end{align*} |

1 | \begin{align*}T = 2 \sqrt{1} = 2\end{align*} |

2 | \begin{align*}y = 2 \sqrt{2} = 2.8\end{align*} |

3 | \begin{align*}y = 2 \sqrt{3} = 3.5\end{align*} |

4 | \begin{align*}y = 2 \sqrt{4} = 4\end{align*} |

5 | \begin{align*}y = 2 \sqrt{5} = 4.5\end{align*} |

Now let's graph the function. It makes sense to let the horizontal axis represent the length of the pendulum and the vertical axis represent the period of the pendulum.

We can see from the graph that a length of approximately \begin{align*}\frac{1}{4}\end{align*} meters gives a period of 1 second. We can confirm this answer by using our function for the period and plugging in \begin{align*}T = 1 \ second\end{align*}:

\begin{align*}T & = 2 \sqrt{L} \Rightarrow 1 = 2 \sqrt{L}\end{align*}

\begin{align*}&\text{Square both sides of the equation:} && 1 = 4L\\ &\text{Solve for} \ L: && L = \frac{1}{4} \ meters\end{align*}

#### Real-World Application: TV Screens

“Square” TV screens have an aspect ratio of 4:3; in other words, the width of the screen is \begin{align*}\frac{4}{3}\end{align*} the height. TV “sizes” are traditionally represented as the length of the diagonal of the television screen. Graph the length of the diagonal of a screen as a function of the area of the screen. What is the diagonal of a screen with an area of \begin{align*}180 \ in^2\end{align*}?

Let \begin{align*}d =\end{align*} length of the diagonal, \begin{align*}x =\end{align*} width

Then 4 \begin{align*}\times\end{align*} height = 3 \begin{align*}\times\end{align*} width

Or, height = \begin{align*}\frac{3}{4}x\end{align*}.

The area of the screen is: \begin{align*}A =\end{align*} length \begin{align*}\times\end{align*} width or \begin{align*} A = \frac{3}{4} x^2\end{align*}

Find how the diagonal length relates to the width by using the Pythagorean theorem:

\begin{align*}x^2 + \left(\frac{3}{4} x \right)^2 & = d^2\\ x^2 + \frac{9}{16}x^2 & = d^2\\ \frac{25}{16}x^2 & = d^2 \Rightarrow x^2 = \frac{16}{25}d^2 \Rightarrow x = \frac{4}{5}d\end{align*}

Therefore, the diagonal length relates to the area as follows: \begin{align*} A = \frac{3}{4} \left( \frac{4}{5}d \right)^2 = \frac{3}{4} \cdot \frac{16}{25}d^2 = \frac{12}{25}d^2\end{align*}.

We can also flip that around to find the diagonal length as a function of the area: \begin{align*}d^2 = \frac{25}{12} A\end{align*} or \begin{align*}d = \frac{5}{2 \sqrt{3}} \sqrt{A}\end{align*}.

Now we can make a graph where the horizontal axis represents the area of the television screen and the vertical axis is the length of the diagonal. First let’s make a table of values:

\begin{align*}A\end{align*} | \begin{align*}d = \frac{5}{2 \sqrt{3}} \sqrt{A}\end{align*} |
---|---|

0 | 0 |

25 | 7.2 |

50 | 10.2 |

75 | 12.5 |

100 | 14.4 |

125 | 16.1 |

150 | 17.6 |

175 | 19 |

200 | 20.4 |

From the graph we can estimate that when the area of a TV screen is 180 \begin{align*}in^2\end{align*} the length of the diagonal is approximately 19.5 inches. We can confirm this by plugging in \begin{align*}A = 180\end{align*} into the formula that relates the diagonal to the area: \begin{align*}d = \frac{5}{2\sqrt{3}} \sqrt{A} = \frac{5}{2 \sqrt{3}} \sqrt{180} = 19.4 \ inches\end{align*}.

Radicals often arise in problems involving areas and volumes of geometrical figures.

#### Real-World Application: Pool Dimensions

A pool is twice as long as it is wide and is surrounded by a walkway of uniform width of 1 foot. The combined area of the pool and the walkway is 400 square feet. Find the dimensions of the pool and the area of the pool.

Make a sketch:

Let \begin{align*}x =\end{align*} the width of the pool. Then:

Area \begin{align*}=\end{align*} length \begin{align*}\times\end{align*} width

Combined length of pool and walkway \begin{align*}= 2x + 2\end{align*}

Combined width of pool and walkway \begin{align*}= x + 2\end{align*}

\begin{align*}\text{Area} = (2x + 2)(x + 2)\end{align*}

Since the combined area of pool and walkway is \begin{align*}400 \ ft^2\end{align*} we can write the equation

\begin{align*}(2x + 2)(x + 2) = 400\end{align*}

\begin{align*}\text{Multiply in order to eliminate the parentheses:} && 2x^2 + 4x + 2x + 4 & = 400\\ \text{Collect like terms:} && 2x^2 + 6x + 4 &= 400\\ \text{Move all terms to one side of the equation:} && 2x^2 + 6x - 396 & = 0\\ \text{Divide all terms by 2:} && x^2 + 3x - 198 & = 0\\ \text{Use the quadratic formula:} && x & =\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\\ && x & = \frac{-3 \pm \sqrt{3^2 - 4(1)(-198)}}{2(1)}\\ && x & = \frac{-3 \pm \sqrt{801}}{2} = \frac{-3 \pm 28.3}{2}\\ && x & = 12.65 \ feet\end{align*}

(The other answer is negative, so we can throw it out because only a positive number makes sense for the width of a swimming pool.)

So the dimensions of the pool are: \begin{align*}length = 12.65\end{align*} and \begin{align*}width = 25.3\end{align*} (since the width is 2 times the length)

That means that the area of just the pool is \begin{align*}A = 12.65 \cdot 25.3 \to 320 ft^2\end{align*}

Check by plugging the result in the area formula:

Area \begin{align*}= (2(12.65) + 2)(12.65 + 2) = 27.3 \cdot 14.65 = 400 \ ft^2.\end{align*}

The answer checks out.

### Example

#### Example 1

The volume of a soda can is \begin{align*}355 \ cm^3\end{align*}. The height of the can is four times the radius of the base. Find the radius of the base of the cylinder.

Make a sketch:

Let \begin{align*}x =\end{align*} the radius of the cylinder base. Then the height of the cylinder is \begin{align*}4x\end{align*}.

The volume of a cylinder is given by \begin{align*}V = \pi R^2 \cdot h\end{align*}; in this case, \begin{align*}R\end{align*} is \begin{align*}x\end{align*} and \begin{align*}h\end{align*} is \begin{align*}4x\end{align*}, and we know the volume is 355.

Solve the equation:

\begin{align*}355 & = \pi x^2 \cdot(4x)\\ 355 & = 4 \pi x^3\\ x^3 & = \frac{355}{4 \pi}\\ x & = \sqrt[3]{\frac{355}{4 \pi}} = 3.046 \ cm\end{align*}

Check by substituting the result back into the formula:

\begin{align*}V = \pi R^2 \cdot h = \pi (3.046)^2 \cdot (4 \cdot 3.046) = 355 \ cm^3\end{align*}

So the volume is \begin{align*}355 \ cm^3\end{align*}. **The answer checks out.**

### Review

- If a certain model of a laptop has a diagonal of 15.4 inches and a length of 14.35 inches, find the width.
- If a certain model of a laptop has a width of 12.78 inches and an area of 114.25 inches squared, find the diagonal.
- The acceleration of gravity can also given in feet per second squared. It is \begin{align*}g = 32 \ ft/s^2\end{align*}at sea level.
- Graph the period of a pendulum with respect to its length in feet.
- For what length in feet will the period of a pendulum be 2 seconds?

- The acceleration of gravity on the Moon is \begin{align*}1.6 \ m/s^2\end{align*}.
- Graph the period of a pendulum on the Moon with respect to its length in meters.
- For what length, in meters, will the period of a pendulum be 10 seconds?

- The acceleration of gravity on Mars is \begin{align*}3.69 \ m/s^2\end{align*}.
- Graph the period of a pendulum on the Mars with respect to its length in meters.
- For what length, in meters, will the period of a pendulum be 3 seconds?

- The acceleration of gravity on the Earth depends on the latitude and altitude of a place. The value of \begin{align*}g\end{align*} is slightly smaller for places closer to the Equator than places closer to the poles and the value of \begin{align*}g\end{align*} is slightly smaller for places at higher altitudes that it is for places at lower altitudes. In Helsinki the value of \begin{align*}g = 9.819 \ m/s^2\end{align*}, in Los Angeles the value of \begin{align*}g = 9.796 \ m/s^2\end{align*} and in Mexico City the value of \begin{align*}g = 9.779 \ m/s^2\end{align*}.
- Graph the period of a pendulum with respect to its length for all three cities on the same graph.
- Use the formula to find for what length, in meters, will the period of a pendulum be 8 seconds in each of these cities?

- The aspect ratio of a wide-screen TV is 2.39:1.
- Graph the length of the diagonal of a screen as a function of the area of the screen.
- What is the diagonal of a screen with area \begin{align*}150 \ in^2\end{align*}?

For 8-10, rationalize the denominator.

- The volume of a soup can is \begin{align*}452 \ cm^3\end{align*}. The height of the can is three times the radius of the base. Find the radius of the base of the cylinder.
- The volume of a spherical balloon is \begin{align*}950 \ cm^3\end{align*}. Find the radius of the balloon. (Volume of a sphere \begin{align*}= \frac{4}{3} \pi R^3\end{align*}).
- A rectangular picture is 9 inches wide and 12 inches long. The picture has a frame of uniform width. If the combined area of picture and frame is \begin{align*}180 \ in^2\end{align*}, what is the width of the frame?

### Review (Answers)

To view the Review answers, open this PDF file and look for section 11.5.