# 9 297 6 a f θ ln θ f θ 1 θ dy dx θ e 1 e sin e

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Chapter 7 / Exercise 19
Nature of Mathematics
Smith
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9 . 297 6. (a) f ( θ ) = ln θ, f 0 ( θ ) = 1 θ dy dx θ = e = 1 e sin e + (ln e ) cos e 1 e cos e - (ln e ) sin e = sin e + e cos e cos e - e sin e 1 . 0193 (b) f ( θ ) = ln θ, f 0 ( θ ) = 1 θ dy dx θ =4 = 1 4 sin 4 + (ln4)cos4 1 4 cos 4 - (ln4) sin 4 = sin 4 + 4(ln 4) cos 4 cos 4 - 4(ln4) sin 4 ≈ - 1 . 2366 7. (a) | r | is a maximum when sin 3 θ = ± 1. This occurs when 3 θ = π/ 2 + or θ = π/ 6 + kπ/ 3 for any integer k . We have f ( θ ) = sin 3 θ so f 0 ( θ ) = 3 cos 3 θ . Thus the slope of the tangent line is: dy dx θ = π 6 + 3 = 0 + sin( π 2 + ) cos( π 6 + 3 ) 0 - sin( π 2 + ) sin( π 6 + 3 ) = sin( π 2 + ) cos( π 6 + 3 ) - sin( π 2 + ) sin( π 6 + 3 ) = - cos( π 6 + 3 ) sin( π 6 + 3 ) Here, the first terms of the numerator and denominator are always 0 since both have a factor of cos( π 2 + ). At θ = π 6 + 3 , r = sin( π 2 + ) so the point in question is either cos π 6 + 3 , sin π 6 + 3 or - cos π 6 + 3 , - sin π 6 + 3 . In either case, the slope of the radius connecting the point to the origin is sin( π 6 + 3 ) cos( π 6 + 3 ) which is the negative recipri- col of the slope of the tangent line found above. Therefore the tangent line is per- pendicular to the radius connecting the point to the origin. (b)
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Chapter 7 / Exercise 19
Nature of Mathematics
Smith
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552 CHAPTER 9. PARAMETRIC EQUATIONS AND POLAR COORDINATES -1 y 2 3 -3 1 0 5 0 2 -4 -5 -2 1 -1 t -2 4 f ( θ ) = 3 cos 3 θ sin θ + sin 3 θ cos θ r = sin 3 θ x = r cos θ ; y = r sin θ x = sin 3 θ cos θ ; y = sin 3 θ sin θ dy = 3 cos 3 θ sin θ + sin 3 θ cos θ ... (1) For horizontal tangents, dy = 0 3 cos 3 θ sin θ + sin 3 θ cos θ = 0 To find the locations of all horizontal tan- gents, we have to find the roots of (1) Clearly, θ = 0 , π 2 , π are the solutions of (1) from the graph f ( θ ) = 3 cos 3 θ sin θ + sin 3 θ cos θ f 0 ( θ ) = 6 cos 3 θ cos θ - 10 sin 3 θ sin θ We can find the other two solutions by Newton Raphson’s Method θ 1 = θ 0 - f ( θ 0 ) f 0 ( θ 0 ) , Where θ 0 is the initial guess. θ 0 = π 4 = 3 . 14 4 = 0 . 785 f (0 . 785) = - 0 . 99998 , f 0 (0 . 785) = - 7 . 9998 θ 1 = 0 . 785 - ( - 0 . 99998) ( - 7 . 9998) = 0 . 659 Repeating the above process for θ 0 = 3 π 4 θ 1 = θ 0 - f ( θ 0 ) f 0 ( θ 0 ) θ 0 = 3 π 4 = 2 . 3562 f (2 . 356) = 1 , f 0 (2 . 356) = - 8 θ 1 = 2 . 356 - 1 ( - 8) θ 1 = 2 . 356 + 0 . 125 = 2 . 481 So the roots of this equation are θ = 0 , 0 . 659 , π 2 , 2 . 481 , π The correspoding points are: for θ = 0 , ( x, y ) = (sin 3 θ cos θ, sin 3 θ sin θ ) for θ = π , ( x, y ) = (0 , 0) for θ = 0 . 659 , ( x, y ) = (0 . 73 , 0 . 56) for θ = π 2 , ( x, y ) = (0 , - 1) for θ = 2 . 481 , ( - 0 . 73 , 0 . 56) Therefore at four points (0 , 0) , (0 , - 1) and (0 . 73 , 0 . 56) , ( - 0 . 73 , 0 . 56) there are horizontal tangents. Concavity of the polar curve: r = sin 3 θ u = 1 r = 1 sin 3 θ = cos ec 3 θ Concavity occurs when u + d 2 u 2 > 0 du = - 3 cos ec 3 θ cot 3 θ d 2 u 2 = 9 cos ec 3 θ (cos ec 2 3 θ + cot 2 3 θ ) Therefore u + d 2 u 2 = cos ec 3 θ + 9 cos ec 3 θ (cos ec 2 3 θ + cot 2 3 θ ) = cos ec 3 θ 1 + 9(cos ec 2 3 θ + cot 2 3 θ ) ... (2) We have to find the nature of the curve at θ = 0 . 659 . Substituting θ in (2) u + d 2 u 2 = (1 . 0885) 1 + 9(cos ec 2 3 θ + cot 2 3 θ ) The terms in the square bracket are pos- itive because they are square terms u + d 2 u 2 > 0 Therefore u + d 2 u 2 > 0 , and so the curve is concave at θ = 0 . 659 Substituting for θ = 0 . 2481 in (2) we get u + d 2 u 2 > 0 Therefore at (0 . 73 , 0 . 56) and ( - 0 . 73 , 0 . 56) the curve is concave At θ =0, π 2 , π concavity is not possible 0.8 -2.0 2 -2 0 -0.8 -1.6 -0.4 0.0 2.0 -1 0.4 1.2 -1.2 1 1.6 8. (a) | r | is a maximum when cos 4 θ = ± 1. This