# Corresponds to a nonnegative value of r hence θ

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Chapter 7 / Exercise 10
Microeconomics: A Contemporary Introduction
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corresponds to a nonnegative value of r . Hence, θ varies from 0 to 2 π . By (1) we obtain L = 2 π 0 2 ( 1 cos θ) dθ Now, 1 cos θ = 2 sin 2 (θ/ 2 ) , and on the interval 0 θ π , sin (θ/ 2 ) is nonnegative, so that 2 ( 1 cos θ) = 4 sin 2 (θ/ 2 ) = 2 sin (θ/ 2 ) there. The graph is symmetric, so it suffices to compute the integral for 0 θ π , and we have L = 2 π 0 2 ( 1 cos θ) dθ = 2 π 0 2 sin (θ/ 2 ) dθ = 8 sin θ 2 π 0 = 8 r = cos 2 θ In Exercises 31 and 32, express the length of the curve as an integral but do not evaluate it. 31. r = ( 2 cos θ) 1 , 0 θ 2 π solution We have f (θ) = ( 2 cos θ) 1 , f (θ) = − ( 2 cos θ) 2 sin θ , hence, f 2 (θ) + f (θ) 2 = ( 2 cos θ) 2 + ( 2 cos θ) 4 sin 2 θ = ( 2 cos θ) 4 ( 2 cos θ) 2 + sin 2 θ = ( 2 cos θ) 2 4 4 cos θ + cos 2 θ + sin 2 θ = ( 2 cos θ) 2 5 4 cos θ Using the integral for the arc length we get L = 2 π 0 5 4 cos θ( 2 cos θ) 2 dθ. r = sin 3 t , 0 θ 2 π In Exercises 33–36, use a computer algebra system to calculate the total length to two decimal places. 33. The three-petal rose r = cos 3 θ in Figure 20 solution We have f (θ) = cos 3 θ , f (θ) = − 3 sin 3 θ , so that f (θ) 2 + f (θ) 2 = cos 2 3 θ + 9 sin 2 3 θ = cos 2 3 θ + sin 2 3 θ + 8 sin 2 3 θ = 1 + 8 sin 2 3 θ Note that the curve is traversed completely for 0 θ π . Using the arc length formula and evaluating with Maple gives L = π 0 f (θ) 2 + f (θ) 2 = π 0 1 + 8 sin 2 3 θ dθ 6 . 682446608 The curve r = 2 + sin 2 θ in Figure 23 35. The curve r = θ sin θ in Figure 24 for 0 θ 4 π y x 5 5 5 10 FIGURE 24 r = θ sin θ for 0 θ 4 π .
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Chapter 7 / Exercise 10
Microeconomics: A Contemporary Introduction
McEachern Expert Verified
May 12, 2011 774 C H A P T E R 11 PARAMETRIC EQUATIONS, POLAR COORDINATES, AND CONIC SECTIONS solution We have f (θ) = θ sin θ , f (θ) = sin θ + θ cos θ , so that f (θ) 2 + f (θ) 2 = θ 2 sin 2 θ + ( sin θ + θ cos θ) 2 = θ 2 sin 2 θ + sin 2 θ + 2 θ sin θ cos θ + θ 2 cos 2 θ = θ 2 + sin 2 θ + θ sin 2 θ using the identities sin 2 θ + cos 2 θ = 1 and 2 sin θ cos θ = sin 2 θ . Thus by the arc length formula and evaluating with Maple, we have L = 4 π 0 f (θ) 2 + f (θ) 2 = 4 π 0 θ 2 + sin 2 θ + θ sin 2 θ dθ 79 . 56423976 r = θ , 0 θ 4 π Further Insights and Challenges 37. Suppose that the polar coordinates of a moving particle at time t are (r(t), θ(t)) . Prove that the particle’s speed is equal to (dr/dt) 2 + r 2 (dθ/dt) 2 . solution The speed of the particle in rectangular coordinates is: ds dt = x (t) 2 + y (t) 2 (1) We need to express the speed in polar coordinates. The x and y coordinates of the moving particles as functions of t are x(t) = r(t) cos θ(t), y(t) = r(t) sin θ(t) We differentiate x(t) and y(t) , using the Product Rule for differentiation. We obtain (omitting the independent variable t ) x = r cos θ r ( sin θ) θ y = r sin θ r ( cos θ) θ Hence, x 2 + y 2 = ( r cos θ sin θ ) 2 + ( r sin θ + cos θ ) 2 = r 2 cos 2 θ 2 r rθ cos θ sin θ + r 2 θ 2 sin 2 θ + r 2 sin 2 θ + 2 r rθ sin 2 θ cos θ + r 2 θ 2 cos 2 θ = r 2 cos 2 θ + sin 2 θ + r 2 θ 2 sin 2 θ + cos 2 θ = r 2 + r 2 θ 2 (2) Substituting (2) into (1) we get ds dt = r 2 + r 2 θ 2 = dr dt 2 + r 2 dt 2 Compute the speed at time
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