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arc AM pa (29%)* = AN. chord AM (10) If p, r, be the perpendicular on the tangent and the radius vector at any point of a curve, then ?
u noint of a curve, then ” will be the perpendicular on the tangent at the corresponding point of the curve which is the locus of the extremity of p.
Let x, y, be the co-ordinates of the first curve, a, b, of the second; then p being the perpendicular on the tangent, its equation is aw + By = pi = a* + B?,
(1) since a, b, are the co-ordinates of the extremity of p. But the line being a tangent, this equation will hold when we put & + dw and y + dx for x and y; we then have adx + ßdy = 0.
(2) Now if V = 0 be the equation connecting a and B, that is to say, the equation to the locus of the extremity of p, and if P be the perpendicular on the tangent of that curve,
[ dᏤ , dᏤ
Now differentiating (1) considering x, y, a, ß, as variables, and paying attention to (2), we have
(x - 2a) da + (y – 2B) dß = 0.
1 (3) – (4) = 0 gives, on equating to zero the coefficients of each differential,
(11) To find the least polygon of a given number of sides which will circumscribe a given oval figure.
Let AB, BC, CD, (fig. 26) be consecutive sides of the polygon. Produce AB, DC to meet in E, which take as origin, the axes being EA, ED. Then the position of BC must be such as to make BEC a maximum.
Now calling as before the intercepts of the tangent x,, yo,
x and y being the co-ordinates of the point of contact P.
The area BEC = 1 x,y, sin E, therefore
is to be a maximum, (neglecting the negative sign). Differentiate with respect to X,
I dy dy dre dx
« dx) dx dy The last factor alone gives a solution. From it we have
That is, EM = 1 EB = MB, and hence also CP= PB, or CB is bisected at the point of contact. As the same condition holds for every side of the polygon, it follows that, when the polygon circumscribing an oval is a minimum, each side is bisected at the point of contact. Hence we see that of all the parallelograms which circumscribe an ellipse, those are least which have their sides parallel to conjugate diameters.
(12) The degree of a curve being n, there cannot be more than n(n − 1) tangents drawn to it from one point. Let U = C
(1) be the equation to the curve, then the equation to the tangent is
,du du du du
dx dy dx dy and the condition that this tangent shall pass through a given point a, b, is
- du du du du
The equations (1) and (2) being combined together will give the values of x and y at the points of contact; and as both equations are of n dimensions in & and y, (since u is
du du of n dimensions and and of n - 1, and therefore
dx dy du du
of n dimensions), it would appear that the redx dy sulting equation is of the degree n’, and therefore that there are as many tangents passing through the point. But the degree of the equation can always be reduced; for we may combine (2) with any multiple of (1), and the result of the elimination between the new equation and either of the others will still give us the co-ordinates of the point of contact. Multiply (1) by n and subtract it from (2), then we have du du
Now by a property of homogeneous functions, if v be homogeneous of n dimensions in x and y,
I do do
- = nu.
"dx 'dy This then will be true of the terms of n dimensions in u, and they will therefore disappear from the second side of the equation (3), which will thus be reduced to (n - 1) dimensions,
du du since and are only of that degree. Hence the combination of (1) with (3) will rise only to the degree n (n − 1), which therefore represents the greatest number of tangents which can be drawn from a given point to a curve of n dimensions. Waring had fixed the limit at n’, as it at first sight appears to be; the preceding process of reduction is due to Bobillier, Annales de Gergonne, Vol. xix. p. 106. It is to be observed that though n (n − 1) is the greatest number of tangents which can be drawn, it seldom reaches that limit, since the final equation generally involves impossible roots which refer to tangents drawn to the branches of the curve which do not lie in the plane xy. Since n (n − 1) is essentially even, it may happen that for certain positions of the point all the roots are impossible; a result which is geometrically apparent, inasmuch as from the interior of an oval curve, such as the ellipse, no tangents can be drawn to the part of the curve which lies in the plane of xy.
Asymptotes. As an asymptote is a line which, intersecting the axes at a finite distance from the origin, is a tangent to the curve at an infinite distance, it appears that if X, or yo remain finite when x or y are infinite, their values will determine the position of the asymptote.
A more convenient method however is that first given by Stirling, in his Lineæ Tertii ordinis Newtonianæ, p. 48.
If y = f (x) be the equation to the curve, and if we can expand f(x) in descending powers of V, so that
then when x = 00, the terms involving negative powers of vanish, and the equation to the curve coincides with that to another curve the equation to which is
y = Am 20m + Am1203-* + &c. + 0,8 + do. This then is the general equation to a curvilinear asymptote, the nature of which will depend on the degree of the highest power of which is involved in it. The most important case is that in which the equation is reduced to
y = ax + do, that is, in which the asymptote is a straight line.
This method fails when the asymptote is parallel to the axis of y, as in that case the coefficient of a would be infinite: but asymptotes of this kind are visible by a simple inspection of the equation to the curve when it is put under the form y = f (x). For the value of y being infinite for the abscissa corresponding to the asymptote, we have only to find what values of w will make f (x) = 0, or to make the denominator of f(w) vanish, since no finite value of x in the numerator can make f(x) = 60. These values of w being found, the ordinates drawn through them are asymptotes to the curve. (13) Let the equation to the curve be
yü = ax + x3.
dy 2ax + 3.02 Then
do sy* '
2ax + 323 3 (y3 – x) – 2 aw? and y = y 3y2
3 y But from the equation to the curve, 3 (43 – 019) = 3ax?, therefore
To find the value of when w and y are infinite, we have from the original equation
=> +1 = 1 when and y are infinite.