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It will be observed that I have assumed the (bra and the dy*s to be the same in both functions: and the result must necessarily involve the ratio of these to each other. If however dx = m dy, where m is a constant multiplier, the result will be independent of the infinitesimal variations.

The following is a case in which the preceding process of evaluation is frequently required. Let r(x,y) = c be an implicit function of two variables; whence after differentiation we have

dy W

-f =; (44)

dx ,dv\


and suppose x0, yo to be values of x and y which simultaneously satisfy the given equation, and are also such that = 0,

and (^-) =0: then becomes indeterminate, and is to be

xdy'o dx

evaluated according to the preceding method. Let us take the total differentials of the numerator and the denominator; then (44) becomes


dy ^dx'
dx ,rfFv


= - ^ J (45)

(dxiy) dx + W> dy and let us suppose all these second derived-functions not to

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whence we have -~ = 0, and = 3, and = — 1. ax

Other examples will occur subsequently in which -rjj has to be evaluated in a similar way, when x = y = 00 .

140.] Again, let it be required to find in ascending powers of h and k the value of F (x + h, y + k), v (x, y) and all its partial derived-functions up to those of the nth order inclusive being finite and continuous for all values of x and y between the values x and x 4 h, y and y + k; h and k being finite increments of x and y.

Let us make the same substitutions in terms of t as in Art. 137. Then by Maclaurins Theorem, Art. 58, (13), we have

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Ex. 2. The equatiou to a conic is

r(x,y) A.r8 + nj?y + cy24 Y.x + ay K = 0;

it is required to prove that (^p = 0, = 0, are the equations to two diameters which are conjugate to diameters parallel to the axes of x and y respectively.

d F \ d F \

In the first place, it is manifest that (^j = 0, and [ j^) = 0

are two equations of the first degree in x and y, and therefore represent straight lines. Let the origin of coordinates be moved to a point (/«, k) by replacing x and y by x -f /* and y + k respectively. And let it be observed that the given equation is only of two dimensions in terms of x and y; and that therefore all the partial derived-functions of it after the second vanish. In Y(x, y) let x and y be replaced by x-\-h and y + k; and let us expand in powers of x and y; and we have

T(x + h,y+k) = 0= F(A,A) + (jh)x + [^)y

Now if h and k are such that = 0, (52) has no term involving the first power of x, and therefore is not changed when x is substituted for + x. And i^^j = 2aa + + E, which

is an expression of one dimension in terms of h and k, and therefore represents a straight line, of which h and k are the current coordinates: this line therefore bisects all chords of the conic which are parallel to the axis of x, and is therefore a diameter conjugate to the diameter which is parallel to the axis

of x. Similarly, if (^) = 0, (52) has no term involving the

first power of y; and therefore 0> which is equal to

2ca + Ba + G = 0, and represents a straight line, is the equation to a diameter of the conic which bisects all chords parallel to the axis of y; and is therefore conjugate to the diameter which

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