Applications of Functions

In this lesson, we solve practical geometry problems by expressing quantities such as area, perimeter, volume, chord length, and distance as functions of a variable.

Problem 1: Area of a Right Triangle

A right triangle has one side \( x \) and hypotenuse 10 meters. Find the area as a function of \( x \).

Solution

The area of a right triangle with legs \( x \) and \( y \) is:

\[ A = \frac{1}{2}xy \]

Using the Pythagorean theorem:

\[ 10^2 = x^2 + y^2 \] \[ y = \sqrt{100 - x^2} \]

Substitute into the area formula:

\[ A(x) = \frac{1}{2}x\sqrt{100 - x^2} \]

Problem 2: Perimeter of a Rectangle

A rectangle has area \(100 \text{ cm}^2\) and width \(x\). Find the perimeter as a function of \(x\).

Solution

\[ 100 = xy \] \[ y = \frac{100}{x} \] \[ P = 2(x + y) \] \[ P(x) = 2\left(x + \frac{100}{x}\right) \]

Problem 3: Area of a Square

Find the area of a square as a function of its perimeter \(x\).

Solution

\[ x = 4L \] \[ L = \frac{x}{4} \] \[ A = L^2 \] \[ A(x) = \left(\frac{x}{4}\right)^2 = \frac{x^2}{16} \]

Problem 4: Volume of a Cylinder

A right circular cylinder has radius \(r\) and height \(2r\). Find the volume as a function of \(r\).

Solution

\[ V = \pi r^2 h \] \[ V(r) = \pi r^2 (2r) \] \[ V(r) = 2\pi r^3 \]

Problem 5: Length of a Chord as a Function of Arc Length

A circle has radius \( r = 10 \) cm. Express the chord length \(L\) as a function of arc length \(s\).

Chord and arc relationship diagram

Solution

\[ \sin\left(\frac{a}{2}\right) = \frac{L/2}{r} \] \[ L = 20 \sin\left(\frac{a}{2}\right) \] \[ s = ra = 10a \] \[ a = \frac{s}{10} \] \[ L(s) = 20 \sin\left(\frac{s}{20}\right) \]

Problem 6: Distance as a Function of \(x\)

Distance composed of two right triangle hypotenuses

Solution

\[ d_1 = \sqrt{x^2 + 9} \] \[ d_2 = \sqrt{25 + (7 - x)^2} \] \[ d(x) = \sqrt{x^2 + 9} + \sqrt{25 + (7 - x)^2} \]

Exercises

  1. Express the area \(A\) of a disk in terms of its circumference \(C\).
  2. The width of a rectangle is \(w\). Express the area \(A\) in terms of perimeter \(P\) and width \(w\).

Answers

\[ A = \frac{C^2}{4\pi} \] \[ A = \frac{1}{2}w(P - 2w) \]

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