Continuous ie that lim s p s p 0 this looks obvious

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continuous, i.e. that lim { S ( P 0 ) - S ( P ) } = 0. This looks obvious enough in the larger figure, but less so in such a case as is shown in the smaller figure. Indeed it is not possible to proceed further, with any degree of rigour, without a careful analysis of precisely what is meant by the length of a curve. It is however easy to see what the formula must be. Let us suppose that the curve has a tangent whose direction varies continuously, so that
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[VI : 146] DERIVATIVES AND INTEGRALS 305 φ 0 ( x ) is continuous. Then the assumption that the curve has a length leads to the equation { S ( x + h ) - S ( x ) } /h = { PP 0 } /h = ( PP 0 /h ) × ( { PP 0 } /PP 0 ) , where { PP 0 } is the arc whose chord is PP 0 . Now PP 0 + PR 2 + RP 0 2 = h r 1 + k 2 h 2 , and k = φ ( x + h ) - φ ( x ) = 0 ( ξ ) , where ξ lies between x and x + h . Hence lim( PP 0 /h ) = lim p 1 + [ φ 0 ( ξ )] 2 = p 1 + [ φ 0 ( x )] 2 . If also we assume that lim { PP 0 } /PP 0 = 1 , we obtain the result S 0 ( x ) = lim { S ( x + h ) - S ( x ) } /h = p 1 + [ φ 0 ( x )] 2 and so S ( x ) = Z p 1 + [ φ 0 ( x )] 2 dx. Examples LIV. 1. Calculate the area of the segment cut off from the parabola y = x 2 / 4 a by the ordinate x = ξ , and the length of the arc which bounds it. 2. Answer the same questions for the curve ay 2 = x 3 , showing that the length of the arc is 8 a 27 ( 1 + 9 ξ 4 a 3 / 2 - 1 ) . 3. Calculate the areas and lengths of the circles x 2 + y 2 = a 2 , x 2 + y 2 = 2 ax by means of the formulae of §§ 145 146 .
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[VI : 146] DERIVATIVES AND INTEGRALS 306 4. Show that the area of the ellipse ( x 2 /a 2 ) + ( y 2 /b 2 ) = 1 is πab . 5. Find the area bounded by the curve y = sin x and the segment of the axis of x from x = 0 to x = 2 π . [Here Φ( x ) = - cos x , and the difference between the values of - cos x for x = 0 and x = 2 π is zero. The explanation of this is of course that between x = π and x = 2 π the curve lies below the axis of x , and so the corresponding part of the area is counted negative in applying the method. The area from x = 0 to x = π is - cos π + cos 0 = 2; and the whole area required, when every part is counted positive, is twice this, i.e. is 4.] 6. Suppose that the coordinates of any point on a curve are expressed as functions of a parameter t by equations of the type x = φ ( t ), y = ψ ( t ), φ and ψ being functions of t with continuous derivatives. Prove that if x steadily increases as t varies from t 0 to t 1 , then the area of the region bounded by the corresponding portion of the curve, the axis of x , and the two ordinates corresponding to t 0 and t 1 , is, apart from sign, A ( t 1 ) - A ( t 0 ), where A ( t ) = Z ψ ( t ) φ 0 ( t ) dt = Z y dx dt dt. 7. Suppose that C is a closed curve formed of a single loop and not met by any parallel to either axis in more than two points. And suppose that the coordinates of any point P on the curve can be expressed as in Ex. 6 in terms of t , and that, as t varies from t 0 to t 1 , P moves in the same direction round the curve and returns after a single circuit to its original position. Show that the area of the loop is equal to the difference of the initial and final values of any one of the integrals - Z y dx dt dt, Z x dy dt dt, 1 2 Z x dy dt - y dx dt dt, this difference being of course taken positively.
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