McGraw Hill Glencoe Algebra 2, 2012
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McGraw Hill Glencoe Algebra 2, 2012 View details
2. Angles and Angle Measure
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Exercise 47 Page 804

Practice makes perfect
a

A carousel makes 5 rotations per minute. The inner and outer circles formed by riders have a radius of 13.1 and 17.2 feet, respectively. We are asked to find the angle θ expressed in radians through which the carousel rotates in one second.

First, recall that one full revolution is an angle of rotation equal to 2π radians. Since the carousel makes 5 rotations per minute, this gives us the angle through which the carousel rotates in 1 minute. 5( 2π)=10π We also know that the angle of rotation θ is always proportional to the time of rotation. By using these two bits of information we can write a proportion. 1 min/10π=1 s/θ Note that we need to express 1 minute as 60 seconds so that the units are consistent. Then we can solve for the angle θ. 60 s/10π=1 s/θ ⇒ θ = π/6 The angle equals π6.

b

Now, we need to find the difference in the arc lengths between the riders sitting in the outside row and the riders sitting in the inside row during one second.

First, let's recall the formula for the length of an arc s. In the formula r is the radius and θ is the measure of the central angle, expressed in radians. s=rθ Let's now focus on the outer circle. First, in Part B we found that the carousel rotates π6 in 1 second. Also, the radius of the circle equals 17.2 feet. We can substitute these values into the formula for the arc length and simplify.

s_1 = rθ
s_1=( 17.2)( π/6)
s_1=17.2Ï€/6
s_1=9.005898 ...
s_1 ≈ 9.0

The arc is approximately 9.0 feet long. Let's now move on to the second arc. Here the angle of rotation is the same, but the radius equals 13.1 feet.

s_2 = rθ
s_2=( 13.1)( π/6)
s_2=13.1Ï€/6
s_2 = 6.859143 ...
s_2 ≈ 6.9

The second arc is approxiamtely 6.9 feet long. Let's now evaluate their difference. s_1-s_2 &≈ 9.0-6.9 &=2.1 The difference in arc lengths equals approximately 2.1 feet.