JUST THE MATHS UNIT NUMBER 13.1 INTEGRATION APPLICATIONS 1 (Second moments of an arc) by A.J.Hobson 13.1.1 Introduction 13.1. The second moment of an arc about the y-axis 13.1.3 The second moment of an arc about the x-axis 13.1.4 The radius of gyration of an arc 13.1.5 Exercises 13.1.6 Answers to exercises
UNIT 13.1 - INTEGRATION APPLICATIONS 1 SECOND MOMENTS OF AN ARC 13.1.1 INTRODUCTION Suppose that C denotes an arc (with length s) in the xy-plane of cartesian co-ordinates, and suppose that δs is the length of a small element of this arc. Then the second moment of C about a fixed line, l, in the plane of C is given by lim δs C h δs, where h is the perpendicular distance, from l, of the element with length δs. h l δs C 13.1. THE SECOND MOMENT OF AN ARC ABOUT THE Y-AXIS Let us consider an arc of the curve whose equation is y = f(x), joining two points, P and Q, at x = a and x = b, respectively. 1
y δy P δs Q O a δx x b The arc may be divided up into small elements of typical length, δs, by using neighbouring points along the arc, separated by typical distances of δx (parallel to the x-axis) and δy (parallel to the y-axis). The second moment of each element about the y-axis is x times the length of the element; that is, x δs, implying that the total second moment of the arc about the y-axis is given by lim δs C x δs. But, by Pythagoras Theorem, δs ( ) (δx) + (δy) δy = 1 + δx, δx so that the second moment of arc becomes lim δx x=b x=a x ( ) δy 1 + δx = δx b a x ( ) 1 + dx. dx Note: If the curve is given parametrically by x = x(t), y = y(t),
then, using the same principles as in Unit 13.4, we may conclude that the second moment of the arc about the y-axis is given by ) t ( dx ± x + t 1 dt ( ) dt, dt according as dx dt is positive or negative. 13.1.3 THE SECOND MOMENT OF AN ARC ABOUT THE X-AXIS (a) For an arc whose equation is y = f(x), contained between x = a and x = b, the second moment about the x-axis will be b a y ( ) 1 + dx. dx Note: If the curve is given parametrically by x = x(t), y = y(t), then, using the same principles as in Unit 13.4, the second moment of the arc about the x-axis is given by ) t ( dx ± y + t 1 dt ( ) dt, dt according as dx dt is positive or negative. 3
(b) For an arc whose equation is x = g(y), contained between y = c and y = d, we may reverse the roles of x and y in section 13.1. so that the second moment about the x-axis is given by d c y ( ) dx 1 +. d y S δy δs c R O δx x Note: If the curve is given parametrically by x = x(t), y = y(t), then, using the same principles as in Unit 13.4, we may conclude that the second moment of the arc about the x-axis is given by ) t ( dx ± y + t 1 dt ( ) dt, dt according as dt is positive or negative and where t = t 1 when y = c and t = t when y = d. 4
EXAMPLES 1. Determine the second moments about the x-axis and the y-axis of the arc of the circle whose equation is x + y = a, lying in the first quadrant. Solution y a O x Using implicit differentiation, we have x + y dx = and hence, dx = x y. The second moment about the y-axis is therefore given by a But x + y = a and, hence, x 1 + x a y dx = x x y + y dx. second moment = a ax y dx. 5
Making the substitution x = a sin u gives second moment = π a 3 sin u du = a 3 π 1 cos u du = a 3 [ u ] sin u π 4 = πa3 4. By symmetry, the second moment about the x-axis will also be πa3 4.. Determine the second moments about the x-axis and the y-axis of the first quadrant arc of the curve with parametric equations Solution x = acos 3 θ, y = asin 3 θ. y O x Firstly, we have dx dθ = 3acos θ sin θ and dθ = 3asin θ cos θ. Hence, the second moment about the y-axis is given by π x 9a cos 4 θsin θ + 9a sin 4 θcos θ dθ, which, on using cos θ + sin θ 1, becomes π a cos 6 θ.3a cos θ sin θ dθ 6
π = 3a 3 cos 7 θ sin θ dθ [ ] π = 3a cos8 θ 8 = 3a3 8. Similarly, the second moment about the x-axis is given by π ) ( dx y + dθ ( ) dθ = dθ π a sin 6 θ.(3a cos θ sin θ) dθ = 3a 3 π [ sin sin 7 θ cos θ dθ = 3a 3 8 ] π θ 8 = 3a3 8, though, again, this second result could be deduced, by symmetry, from the first. 13.1.4 THE RADIUS OF GYRATION OF AN ARC Having calculated the second moment of an arc about a certain axis it is possible to determine a positive value, k, with the property that the second moment about the axis is given by sk, where s is the total length of the arc. We simply divide the value of the second moment by s in order to obtain the value of k and, hence, the value of k. The value of k is called the radius of gyration of the given arc about the given axis. Note: The radius of gyration effectively tries to concentrate the whole arc at a single point for the purposes of considering second moments; but, unlike a centroid, this point has no specific location. 7
EXAMPLES 1. Determine the radius of gyration, about the y-axis, of the arc of the circle whose equation is x + y = a, lying in the first quadrant. Solution y O a x From Example 1 in Section 13.1.3, we know that the Second Moment of the arc about the y-axis is equal to πa3 4. Also, the length of the arc is πa, which implies that the radius of gyration is πa 3 4 πa = a.. Determine the radius of gyration, about the y-axis, of the first quadrant arc of the curve with parametric equations x = acos 3 θ, y = asin 3 θ. 8
Solution y O x From Example in Section 13.1.3, we know that dx dθ = 3acos θ sin θ and dθ = 3asin θ cos θ and that the second moment of the arc about the y-axis is equal to 3a3 8. Also, the length of the arc is given by a π This simplifies to ) ( dx + dθ ( ) dθ = dθ π 9a cos 4 θsin θ + 9a sin 4 θcos θ dθ. π 3a cos θ sin θ dθ = 3a [ sin θ ] π = 3a. Thus, the radius of gyration is 3a 3 8 3a = a. 9
13.1.5 EXERCISES 1. Determine the second moments about (a) the x-axis and (b) the y-axis of the straight line segment with equation lying between x = and x = 3. y = x + 1,. Determine the second moment about the y-axis of the first-quadrant arc of the curve whose equation is lying between x = and x =. 5y = 4x 5, 3. Determine, correct to two places of decimals, the second moment, about the x-axis, of the arc of the curve whose equation is lying between x =.1 and x =.5. y = e x, 4. Given that x 1 dx = 1 ( x x 1 ln(x + x 1) ) + C, determine, correct to two places of decimals, the second moment, about the x-axis, of the arc of the curve whose equation is lying between x = and x = 1. y = 8x, 5. Verify, using integration, that the radius of gyration, about the y-axis, of the straight line segment defined by the equation from x = to x = 1 is 1 3. y = 3x +, 1
6. Determine the radius of gyration about the x-axis of the arc of the circle given parametrically by from θ = to θ = π 4. 7. For the curve whose equation is show that x = 5 cos θ, y = 5 sin θ, 9y = x(3 x), dx = 1 x x. Hence, determine, correct to three significant figures, the radius of gyration, about the y-axis, of the first quadrant arch of this curve. 13.1.6 ANSWERS TO EXERCISES 1.. 3. 4. 5. 6. 7. (a) 9 5 Second moment = k = 5 9 5.78 1.9 8.59 (b) 1 5. 1 3 15(π ) 1π k 1.68 Length = 1..13 11