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Math 140 Lecture 15 Angles, arcs, and radians p is approximately 3.14. In answers, use the exact p not the decimal 3.14. A circle of radius r has circumference 2pr. Unit circle circumference = 2pr = 2p1 = 2p. Greek letters: q is “theta”, w is “omega”. DEFINITION. An angle intercepts an arc of length s on a circle of radius r. The angle’s radian measure is q = s /r r q s For unit circles, r = 1 and radian measure = arc length: q = s. Radians are dimensionless and “radians” can be omitted, 10 = 10 radians. `Radians and degrees on the unit circle . p/2 = 90 o p/4= 45 o 1 o o o 0 = 0, 360 = 2p 180 = p -p/4= -45 o Clockwise angles ?are negative. CONVERSION FORMULAS. 180o = p radians Hence the following conversion ratios are 1. radians 180 o 180o 1 and radians 1 radians `18 18 1 18 180 o o o o radians 10 10 radians To convert to radians, conversion factor must have radians on top. To cancel degrees, conversion factor needs degrees on the bottom. To convert between radians and degrees, use the ratio radians 180 o , 180 o radians which cancels the current units. ` Convert 50 o to radians. Which formula cancels degrees and gives radians? (A) 50o radians 180 o o 180 (B) 50o radians ` Convert 5 radians to degrees. Which formula cancels radians and gives degrees? radians (A) 5 radians 180 o 180 o (B) 5 radians radians `Convert 100o to radians. ... 5/9 rad. `Convert 40o to radians. . . . `Convert p/6 radians to degrees. `Convert p/5 radians to degrees. . . If q is in radians, then q = s/r `Find the length of a 36o arc on circle of radius 10 inches. Convert 36o to radians. qr = s s = qr `Find the length of a 30o arc on a circle of radius 12 inches. ... 2 . . Find the length of the arc. . . `Find the degree measure of an angle intercepting a 5 inch arc on a circle of radius 12 inches. `Find the degree measure of an angle which intercepts a 7 inch arc on a circle of radius 18 inches. Find the radian measure q = s/r . ... 75 0 / . . Then convert to degrees. . . Speed DEFINITION. If an object travels a distance d in time t, its linear speed is d/t. E.g., miles/hour. If an object rotates through an angle q in time t, its rotational speed is w = q/t. E.g., radians/second. THEOREM. If a point rotates around a circle of radius r with rotational speed w, then its linear speed is wr. PROOF. Suppose the point rotates through angle q in time t. t w = q/t The distance d it travels = the length of the arc it traces = qr . tits linear speed = d/t = qr/t = (q/t)r = wr . `A point revolves around a circle of radius 10 feet at 4 revolutions per minute. `A point revolves around circle of radius 3 feet at 10 revolutions per minute. ... 60 . Find its rotational speed in radians/min. . Find its linear speed. . What are the units for the linear speed? . `The tip of a clock’s minute hand is 6 inches from the center. How fast is the tip moving? `A point on the equator is 4000 miles from the center of the earth. Distance moved per day? How many inches does it move in one hour? Speed in miles/day? How fast is the tip moving in inches/hour? Speed in miles/hour? ... 1000mi/hr. How fast is the tip moving in inches/minute? Trigonometric functions The unit circle has radius one and center (0, 0). There are four quadrants I, II, III, IV as pictured. An angle is in standard position if its vertex is the origin (0,0) and its initial side is the positive x-axis. The other side of the angle is the terminal side. The terminal side intersects the unit circle at a point Pq. Pq II q I (0,0) III IV DEFINITION. Given a standard-position angle of radian measure q, let (x, y) be the coordinates of Pq. tanq=y/x=slope Pq= (x, y) 1 y = sin q q x=cos q The six trigonometric functions of q are: sin q = y sine csc q = 1/ sin q cos q = x tan q = sin q / cos q cosine tangent sec q = 1/cos q cosecant secant cot q = cos q / sin q cotangent `Draw an angle of p/2 radians in standard position. `sin /2 `cos /2 `tan /2 `A point (x, y) on the unit circle is in the second quadrant and y = 34 . 7 3 Find sin , cos , tan . ... 34 4 7 1 q 3/4 x sin cos tan b Know the sin, cos and tan of: 0, p/6, p/4, p/3, p/2. (0,1) p/2=90o 1 (1/ 2, l3/ 2) p/3=60o o (1/l2, 1/l2) p/4=45 o p/6=30 (l3/2,1/2) 0o (1,0) sin(0) = 0 cos(0) = 1 tan(0) = 0 sin(p/6) = 1/2 cos(p/6) = sin(p/4) = 1/ 2 cos(p/4) = 1/ 2 tan(p/4) = 1 sin(p/3) = l3̄/2 cos(p/3) = 1/2 tan(p/3) = l3̄ sin(p/2) = 1 cos(p/2) = 0 tan(p/2) = undef. bb 3 /2 tan(p/6) = 1/l3̄ p/2 y sin q ___ _ slope of line = = x cos q 1 y = sin q q x =cos q = tan q