Trigonometric Applications and Models MATH 160, Precalculus J. Robert Buchanan Department of Mathematics Fall 2011
Objectives In this section we will learn to: solve real-world problems involving right triangles, solve real-world problems involving directional bearings, solve real-world problems involving harmonic motion.
Applications Involving Right Triangles Example Find the unknown angles and sides of the following right triangle. c 35 A b C
Solution By the Pythagorean Theorem a 2 + b 2 = c 2 (25) 2 + b 2 = (35) 2 b = (35) 2 (25) 2 = 10 6. Using right triangle trigonometry sin A = 25 35 = A = arcsin 5 7 0.7956 rad 45.58. Finally, since π = A+B+C = 0.7956+B+ π 2 = B 0.7752 rad 44.42.
Application: Distance An observer in a lighthouse 350 feet above sea level observes two ships directly offshore. The angles of depression to the ships are 4 and 6.5 respectively. How far apart are the ships? Hint: draw a picture and include some right triangles.
Solution For the ship whose angle of depression is 4 : cot 4 = x 4 350 = x 4 5005.23 feet. For the ship whose angle of depression is 6.5 : cot 6.5 = x 6.5 350 = x 6.5 3071.91 feet. Thus the ships are 5005.23 3071.91 = 1933.32 feet apart.
Trigonometry and Bearings Definition In surveying and navigation a bearing measures the acute angle that a path or line makes with a fixed north-south line. W E S 35 E 35 S
Examples: Bearings 80 45 W E W E S N 80 W S N 45 E
Application: Navigation A boat leaves a dock in Myrtle Beach, SC and heads toward Freeport in the Bahamas at a bearing of S 1.4 E, The boat averages a speed of 20 knots (nautical miles (nm) per hour) over the 428 nm trip. 1 How long will it take the boat to make the trip? 2 How far east and south is the boat after 12 hours?
Application: Navigation A boat leaves a dock in Myrtle Beach, SC and heads toward Freeport in the Bahamas at a bearing of S 1.4 E, The boat averages a speed of 20 knots (nautical miles (nm) per hour) over the 428 nm trip. 1 How long will it take the boat to make the trip? t = d v = 428 20 = 21.4 hours 2 How far east and south is the boat after 12 hours?
Application: Navigation A boat leaves a dock in Myrtle Beach, SC and heads toward Freeport in the Bahamas at a bearing of S 1.4 E, The boat averages a speed of 20 knots (nautical miles (nm) per hour) over the 428 nm trip. 1 How long will it take the boat to make the trip? t = d v = 428 20 = 21.4 hours 2 How far east and south is the boat after 12 hours? After 12 hours, the boat has traveled 240 nm. It has traveled x = 240 sin 1.4 5.86 nm east, and y = 240 cos 1.4 239.93 nm south.
Air Navigation In air navigation, bearings are measured in degrees clockwise from north. 60 W E W E 225 S S
Harmonic Motion Remarks: Objects that vibrate, oscillate, rotate, or move in a wave motion are said to exhibit harmonic motion. If an object s motion can be modeled by a sine or cosine function, the object is said to exhibit simple harmonic motion.
Harmonic Motion Remarks: Objects that vibrate, oscillate, rotate, or move in a wave motion are said to exhibit harmonic motion. If an object s motion can be modeled by a sine or cosine function, the object is said to exhibit simple harmonic motion. Definition A point that moves on a coordinate line is said to be in simple harmonic motion if its distance d from the origin at time t is given by either d = a sin ωt or d = a cos ωt where a and ω are real numbers with ω > 0. The motion has amplitude a, period 2π ω, and frequency ω 2π.
Example: Wave Motion A buoy oscillates in simple harmonic motion as waves go past. It is noted that the buoy moves a total of 3.5 feet from its low point to its high point, and that it returns to its high point every 10 seconds. Write an equation that describes the motion of the buoy if its high point is at t = 0.
Example: Wave Motion A buoy oscillates in simple harmonic motion as waves go past. It is noted that the buoy moves a total of 3.5 feet from its low point to its high point, and that it returns to its high point every 10 seconds. Write an equation that describes the motion of the buoy if its high point is at t = 0. The amplitude of the simple harmonic motion is 3.5/2 = 1.75 feet. The period of the motion is 10 seconds, thus 10 = 2π ω = ω = 2π 10 = π 5. Since the high point is reached at t = 0 the model for the simple harmonic motion is ( π ) y = 1.75 cos 5 t.
Illustration 1.5 1.0 0.5 5 10 15 20 0.5 1.0 1.5 ( π ) y = 1.75 cos 5 t.
Example Given the equation for simple harmonic motion, d = 1 2 cos(20πt) find the maximum displacement frequency value of d when t = 3 least positive value of t for which d = 0
Example Given the equation for simple harmonic motion, d = 1 2 cos(20πt) find the maximum displacement d max = 1 2 frequency value of d when t = 3 least positive value of t for which d = 0
Example Given the equation for simple harmonic motion, find the d = 1 2 cos(20πt) maximum displacement d max = 1 2 20π frequency 2π = 10 value of d when t = 3 d = 1 2 cos(60π) = 1 2 least positive value of t for which d = 0 π 1 = 20πt = t = 2 40
Homework Read Section 4.8. Exercises: 1, 5, 9, 13,..., 57, 61