Pre-Calc Exam Notes 93

Pre-Calc Exam Notes 93 - must be a rectangle. In...

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Arc Length Section 4.2 93 Example 4.7 The centers of two belt pulleys, with radii of 5 cm and 8 cm, respectively, are 15 cm apart. Find the total length L of the belt around the pulleys. Solution: In Figure 4.2.4 we see that, by symmetry, L = 2 ( h DE + EF + h FG ). D G 3 5 C F 5 A B E 15 Figure 4.2.4 Belt pulleys with radii 5 cm and 8 cm First, at the center B of the pulley with radius 8, draw a circle of radius 3, which is the difference in the radii of the two pulleys. Let C be the point where this circle intersects BF . Then we know that the tangent line AC to this smaller circle is perpendicular to the line segment BF . Thus, ACB is a right angle, and so the length of AC is AC = r AB 2 BC 2 = r 15 2 3 2 = r 216 = 6 r 6 by the Pythagorean Theorem. Now since AE EF and EF CF and CF AC , the quadrilateral AEFC
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Unformatted text preview: must be a rectangle. In particular, EF = AC , so EF = 6 r 6. By formula (4.4) we know that h DE = EA · ∠ DAE and h FG = BF · ∠ GBF , where the angles are measured in radians. So thinking of angles in radians (using π rad = 180 ◦ ), we see from Figure 4.2.4 that ∠ DAE = π − ∠ EAC − ∠ BAC = π − π 2 − ∠ BAC = π 2 − ∠ BAC , where sin ∠ BAC = BC AB = 3 15 = 0.2 ⇒ ∠ BAC = 0.201 rad. Thus, ∠ DAE = π 2 − 0.201 = 1.37 rad. So since AE and BF are parallel, we have ∠ ABC = ∠ DAE = 1.37 rad. Thus, ∠ GBF = π − ∠ ABC = π − 1.37 = 1.77 rad. Hence, L = 2 ( h DE + EF + h FG ) = 2 (5 (1.37) + 6 r 6 + 8 (1.77)) = 71.41 cm ....
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This note was uploaded on 01/21/2012 for the course MAC 1130 taught by Professor Dr.cheun during the Fall '11 term at FSU.

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