1 gative after the state when its fluxion is = 0, it follows that at this state the expression is not a minimum, but a maximum. Again, taking the expression 2.3 - ax?, its fluxion 3.x?* 2arx=(2x - 2)rx=0;this divided by rigives 3.x - 2a=0, and x = 'a, its true value when the fluxion of x3 ara is equal to nothing. But now to know whether the given expression be a maximum or a minimum at that time, take r a little less than ļa in the value of the fluxion (3x – 2a) xé, xi and this will evidently be negative; and again, taking r a little more than ža, the value of 3x 2a, or of the fluxion, is as evidently positive. Therefore the fluxion of 23 - ar? being negative before that fluxion is = 0, and positive after it, it follows that in this state the quantity x3 a.x? admits of a minimum, but not of a maximum. 56. SOME EXAMPLES FOR PRACTICE. Exam. 1. To divide a line, or any other given quantity a, into two parts, so that their rectangle or product may be the greatest possible. Exam. 2. To divide the given quantity a into two parts such, that the product of the m power of one, by the na power of the other, may be a maximum. Exam. 3. To divide the given quantity a into three parts such, that the continual product of them all may be a maximum EXAM, 4. To divide the given quantity a into three parts such, that the continual product of the 1st, the square of the 2d, and the cube of the 3d, may be a maximum. Exam. 5. To determine a fraction such, that the difference between its m power and n power shall be the greatest possible. Exam. 6. To divide the number 80 into two such parts, X 7 and y, that 2.x2 + xy + 3y may be a minimum. Exam. 7. To find the greatest rectangle that can be ina scribed in a given right-angled triangle. Exam. 8. To find the greatest rectangle that can be inscribed in the quadrant of a given circle. EXAM. 9. To find the least right-angled triangle that can circumscribe the quadrant of a given circle. Exam. 10. To find the greatest rectangle inscribed in, and the least isosceles triangle circumscribed about, a given semiellipse. Exam. 11, a Exam. 11. To determine the same for a given parabola. EXAM. 12. To determine the same for a given hyperbola. Exam. 13. To inscribe the greatest cylinder in a given cone; or to cut the greatest cylinder out of a given cone. EXAM. 14. To determine the dimensions of a rectangular cistern, capable of containing a given quantity a of water, so as to be lined with lead at the least possible expense. Exam. 15. Required the dimensions of a cylindrical tankard, to hold one quart of ale measure, that can be made of the least possible quantity of silver, of a given thickness. EXAM. 16. To cut the greatest parabola from a given a cone. Exam. 17. To cut the greatest ellipse from a given cone. Exam.18. To find the value of r when x is a mininum. a The METHOD OF TANGENTS; OR, TO DRAW TAN GENTS TO CURVES. 57. The Method of Tangents, is a method of determining the quantity of the tangent and subtangent of any algebraic curve; the equation of the curve being given. Or, vice versa, the nature of the curve, from the tangent given. 58. Let dae be another ordinate, indefinitely near to de, meeting the curve, or tangent produced in e; and let Ea be parallel to the axis ad. Then is the elementary triangle Ees similar to the triangle TDE; and therefore a gative after the state when its fluxion is = 0, it follows that at this state the expression is not a minimum, but a maximum. Again, taking the expression 2.3 – ax?, its fluxion 3x25 – 2ar:= (3x – 2a)xx=0; this divided by xx gives 3x – 2a=0, and .r = ja, its true value when the fluxion of x3 axl is equal to nothing. But now to know whether the given expression be a maximum or a minimum at that time, take x a little less than sa in the value of the fluxion (3x – 2a) xi, ) and this will evidently be negative; and again, taking r a little more than ļa, the value of 3.x 2a, or of the fluxion, is as evidently positive. Therefore the fluxion of 213 – axı being negative before that fluxion is = 0, and positive after it, it follows that in this state the quantity x3 a.x? admits of a minimum, but not of a maximum. a a 56. SOME EXAMPLES FOR PRACTICE. Exam. 1. To divide a line, or any other given quantity a, into two parts, so that their rectangle or product may be the greatest possible. Exam. 2. To divide the given quantity a into two parts such, that the product of the m power of one, by the ni power of the other, may be a maximum. Exam. 3. To divide the given quantity a into three parts such, that the continual product of them all may be a maximum. EXAM, 4. To divide the given quantity a into three parts such, that the continual product of the 1st, the square of the 2d, and the cube of the 3d, may be a maximum. ExAM. 5. To determine a fraction such, that the difference between its m power and n power shall be the greatest possible. Exam. 6. To divide the number 80 into two such parts, that 2 r2 -+- xy + 3y? may be a minimum. Exam. 7. To find the greatest rectangle that can be ina scribed in a given right-angled triangle. Exam. 8. To find the greatest rectangle that can be in. scribed in the quadrant of a given circle. Exam. 9. To find the least right-angled triangle that can circumscribe the quadrant of a given circle. Exam. 10. To find the greatest rectangle inscribed in, and the least isosceles triangle circumscribed about, a given semiellipse. EXAM, 11, and دز a a Exam. 11. To determine the same for a given parabola. EXAM. 12. To determine the same for a given hyperbola. Exam. 13. To inscribe the greatest cylinder in a given cone; or to cut the greatest cylinder out of a given cone. Exam. 14. To determine the dimensions of a rectangular cistern, capable of containing a given quantity a of water, so as to be lined with lead at the least possible expense. EXAM. 15. Required the dimensions of a cylindrical tankard, to hold one quart of ale measure, that can be made of the least possible quantity of silver, of a given thickness. ExAM. 16. To cut the greatest parabola from a given cone. Exam. 17. To cut the greatest ellipse from a given cone. EXAM. 18. To find the value of r when r* is a mininum. a THE METHOD OF TANGENTS; OR, TO DRAW TAN GENTS TO CURVES, a 57. The Method of Tangents, is a method of determining the quantity of the tangent and subtangent of any algebraic curve; the equation of the curve being given. Or, vice versa, ; the nature of the curve, from the tangent given. If ae be any curve, and E be any point in it, to which it is required to draw a tangent TE. Draw the ordinate Ed: then if we can deter y mine the subtangent TD, limited be AXD tween the ordinate and tangent, in the axis produced, by joining the points T, E, the line Te will be the tangent sought. 58. Let dae be another ordinate, indefinitely near to DE, meeting the curve, or tangent produced in é; and let Ea be parallel to the axis ad. Then is the elementary triangle Ees similar to the triangle Tre; and therefore Hence the value of the curve, fron the fluent of each of these, gives the four following forms, in series, viz. putting d = 2r the diameter, the curve is + &c) y, 3.x? 3.5.x3 % = (1 + + 2.3d + 2.4.5d2 + &c) v dr, 2.4.6.7013 go2 394 (1 t 3.500 + 46 78 &c) t, 3/35 - gos) + &c)r. 2.353 2.4.555 ر- تی = ( + = Now, it is evident that the simplest of these series, is the third in order, or that which is expressed in terms of the tangent. That form will therefore be the fittest to calculate an example by in numbers. And for this purpose it will be convenient to assume some arc whose tangent, or at least the square of it, is known to be some small simple number. Now, the arc of 45 degrees, it is known, has its tangent equal to the radius; and therefore, taking the radius. r=1, and consequently the tangent of 45', or 1, = 1 also, in this case the arc of 45° to the radius 1, or the arc of the quadrant to the diameter 1, will be equal to the infinite series 1 +}-+ &c. But as this series converges very slowly, it will be proper to take some smaller arc, that the series may converge faster; such as the arc of 30 degrees, the tangent of which is = V's, or its square t = { : which being substituted in the series, the length of the arc of 30° comes out 1 (1. + &c) V. Hence, to com3.3 5.92 7.33 9.34 pute these terms in decimal numbers, after the first, the sheceeding terms will be found by dividing, always by %, and these quotients again by the absolute numbers 3, 5, 7, 9, &c; and lastly, adding every other term together, into two sums, the one the sum of the positive terms, and the other the sum of the negative ones; then lastly, the one sum taken from the other, leaves the length of the arc of 30 degrees; which being the 12th part of the whole circumference when the radius is l, or the 6th part when the diameter is 1, consequently 6 times that are will be the length of the whole circumference to the diameter 1. Therefore, multiplying the first term V f by 6, the product is ✓ 12 = 3.4641016; and hence the operation will be conveniently made as follows: + Terms 1 + |