PROPOSITION IV.

To find the Derivative of the Segmental Area of any continuous curve referred to rectangular coördinates.

Let Y be segmental area in question, and K its increment; we have (403), (146),

Y

The derivative of the segmental area of any continu- (404) ous curve is equal to the ordinate of that curve regarded as a function of the abscissa. [Y' = y = fx.]

PROPOSITION V.

To find the area of the Ellipse.

Let Y be an elliptical segment embraced by the semiminor axis and the abscissa x; we have (404), (203),

Y

the area sought, no constant being added, since Y2-0 = 0.

If we make b = a, there results

x5

a2

2

a4 Fig. 71.

+

ασ

x5

2

5a4

+(-1) (-2) 2.3

(405)

5a4

(1 − 1) (1 − 2) 2.3

Making x = a and multiplying by 4, we find

(Ellipse), b = 4( 1 − ↓ • ‡ +↓ +(-1)

2

(11)
2

(410)

Ellipse Circumscribing Circle: ba: 26: 2a

:: Minor Axis Major Axis.

(Circle), = b2;

... (411), (Ellipse)a,b : (Circle), :: a : b.

PROPOSITION VI.

Fig. 73.
(412)

The Parabola is two-thirds the circumscribing rect- (4132)

angle.

Scholium. It is obvious that the quadrature of the circle and ellipse can only be obtained approximately, while that of the parabola is exact.

PROPOSITION VII.

To find the proximate area of any continuous curve.

Suppose the equation of the curve, referred to rectangular coordinates, to be

but, if the area Y become = 0 when x = 0, which condition is always admissible, since the origin may be taken at pleasure, we have

29

Y = А ̧x+‡Ãμx2 +‡Ã1⁄2ï3 +‡A3x1 + ... . But, since three points, Po, P1, P2, determine with considerable accuracy a curve of moderate extent, we will take the foot of the first ordinate, yo, as the origin, and the abscissas, x1 = h, x2 = 2h, so that the corresponding ordinates, Yo, Y, Y29 shall be equally distant from each other; .. making x = 0, h, 2h, we have (414)

and

y1 = A + A1 h + A2 • h2,
Y2=A+A1 2h+A2 4h2;

Y2 — Y1 = A1 • h + A2 • 3h2 ;
Y2-2y1+Yo= A2 • 2h2 ;

but when x = 2h, we have (415)

Y = A。 • 2h+§A, • 4h2 +‡A2 • 8h3 = 2h(A。 + A ̧h + 1§A2 • h2) ;

...

or

Y=2h[y1+ • †(Y2 −2y1+Yo)],

Y = th(tyo+2y1 + y2).

If we continue the same method

of admeasurement and notation, we

n-2

(416)

.. Y2+Y4+Y6+...+Y2n = 3h(zyo +2y1 +Y2+QY3+Y4

+2y5+...+Y2n-2+2y2n−1 + y2n).

(417)

This beautiful and useful theorem is due to Simpson. The student should enunciate it in common language.

EXERCISES.

1o. Required the diameter of a circle having ten linear chains in circumference to every square chain in area.

2o. A square plate of silver, 3 inches on the side, is worth $4. What is the value of the greatest circle that can be cut from it? 3°. Had the plate in 2° been an equilateral triangle, what would have been its value?

4°. The two sides including the right angle of a right angled triangle, are three and four rods; what is the area of the circumscribing circle?

5°. Determine a circle circumscribing an isosceles triangle, the two equal sides of which, including an angle of 36°, are 15'15 chs. each?

6°. The equal sides of an isosceles triangle embrace an angle of 473°, and the area of the inscribed circle is one acre. Determine

the triangle.

7°. A circular plate of brass, 20 inches in diameter, is worth $3.75. What is the value of the three greatest and equal circles that can be cut from it?

8°. Required the area of a circular segment embraced by an arc and its chord, the length of which is 5'87 chs., and the breadth 1'35? 9°. The dimensions of an elliptical fish-pond are 10 and 15 rods. What is its area?

10°. The ordinate of a parabolic segment is 3 chains, and the corresponding abscissa 7 chains. Required the area.

11°. Required the area of an elliptical segment embraced between the semiminor axis and an ordinate = 5'657 chains, and having a breadth of one chain.

12°. Wishing to ascertain the cross section of a river 100 yards from water's edge to water's edge, I take soundings every 10 yards, and find them to be in yards:

yo

y1 = 0, y1 = 12, y2 = 20'1, y1 = 25'3, y1 = 28'4, y = 29‘9, y = 29'3, y1 = 26'1, y ̧ = 20‘9, y, = 12'8, Y10 = 0.

BOOK THIRD.

SURVEYING.

SECTION FIRST.

Description and Use of Instruments.

To determine the boundaries of lands, to delineate them in maps, and to compute their areas, constitute the Art of Surveying.

The instruments employed in determining the boundaries of lands are, the Chain, for measuring the lengths of lines, the Surveyor's Cross to determine right angles, and the Azimuth Compass, or the Theodolite, for fixing the inclinations of lines to each other and to the meridian.

The Chain.

Gunter's chain, as has already been observed, is 4 rods or 66 feet in length, and centesimally divided by a hundred links, each, consequently, equal to 7.92 inches.

It is a maxim in land surveying that every instrument, whether for measuring lines or angles, must be used in a horizontal position; for it is the base, or the projection of the field upon the same horizontal plane, that is desired.

The projections of the bounding lines are usually obtained by carrying the chain in a horizontal position, as

represented in fig. 77, where 1, 2, 3,

successive positions of the chain.

..., are the

Fig. 77.

When the inclination of the ground is too

great to admit of a whole chain, a half or a quarter may be taken, and in all cases the proper position of the elevated extremity should be determined by a plumb line or by the dropping of a stone.