Kinetic Derivation of Tangent EquationAuthor(s): A. Latham BakerSource: The American Mathematical Monthly, Vol. 8, No. 5 (May, 1901), pp. 111-115Published by: Mathematical Association of AmericaStable URL: http://www.jstor.org/stable/2968617 .

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KINETIC DERIVATION OF TANGENT EQUATION.

By A. LATHAM BAKER, University of Rochester, Rochester, N. Y.

The following investigation is, I am afraid, more interesting than valu- able, though it gives a very simple way of getting the equation of a tangent. But there is evidently an underlying principle which I have not succeeded in getting at, which would, if found, undoubtedly be valuable.

Since a curve can be colnsidered as generated by the motion of a point, we can consider this point as the tracing point of a mechanism so adjusted as to trace the proper curve. This tracing point can be imagined to be the common point of various sets of x and y arms, each set composed of a certain numnber of x points, or of y points.

Now if, for example, the mechanism be set for the curve represented by the equation ax2-1+by2 -c, and we stop the action of an x and a y point, leaving an x and a y point current, the new curve will be a straight line in continuation of the old motion. That is, the straight line will be a tangent line.

The algebraic equivalent of this would be found by writing

axx1 + byyl ~c

which would consequently be the equation of the tangent line, found by stopping one pair of co6rdinates (indicated by the subscripts) and letting the remaining pair remain current.

In the same way, from the algebraic mechanism

axrn + by" =c

we get the equation for the tangent line

axi mxtbylml m- y-c.

Extending this thought to more complicated forms, we have the following divisions of algebraic mechanisms, using the word simple to indicate sets o terms containing onily x's or only y's, and compound for terms containing both x's and y's.

A. TERMS SIMPLE AND COMPOSED OF THE SAME NUMBER OF TRACING POINTS,

that is of the same degree, as in the examples already given, ax2 +by2 c. In these we adopt a current pair of co'ordinates and make all the others fixed, viz.,

ax,x+by1y-c. Example. x+3y=a. To have a fixed point (point of tangency) we must

have two mechanismns compounded, viz.,

x+3y-I x+3y 2a,

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112

whence, leaving one x and one y current, we get x+3y+x, +3y1, 2a. B. TERMS SIMPLE, BUT WITH AN UNEQUAL NUMBER OF TRACING POINTS, with

the subdivisions as to variable terms. B1. TWO TERMS ONLY, exampley =x2. B2. MORE THAN TWO TERMS BUT EACH TERM BALANCED BY ITS MATE IN THE

OTHER LETTER.

B3. MORE THAN TWO TERMS BUT THE TERMS NOT BALANCED.

B1. As in A, we make one pair of tracing points (co6rdinates) current, but here the potency of the tracing point is inversely proportional to the number required to constitute a term, and the number of terms from which a current y point must be taken to balance a current x point must be directly proportional to the number of points in the term (degree of the term). Thus in y x3 +b, one y must be balanced by three x's. Hence, taking three mechanisms together, we get

y+2yW3x3 +3b.

Now having current one y and three x's, we get y+2yjz3xx12+3b. Similarly in y2-=X3, two y's must be balanced by three x's and we must

take 2y2 +y2=3x3. Leaving current two y's and three x's, we get 2yy1 +y12 =3xx12. Again in y2 4px, one x balances two y's, hence 2y2--4px+ 4px. Leaving

current one x and two y's we get 2yy1 z4px+4pxj. B2. Here we put each set- of homogeneous terms equal to a summand of

the absolute term, operate on these equations by A-B1, and add the results. Example. y+y2=X+X2 +C.

y-x=p y+y1 (x+x1l)=2p y2 _x2=q 2Yli -2xx1 ~2q

whence y+y1 +2yy1j=x+x, +2xxl +2c. B,. Here we take the suni of the y's (or x's) and break it up into the slim

of term of degree similar to the original, each summand being put equal to a term of the x's (or y's), and apply A -B2. Or take the sets of homogeneous terms and place them equal to surnmands of the constant term; operate on these sum- mands by A-B2 and add the results.

Example. y+y2=_X y=p y2 q

Whence y+YlP1p+Pl

2yyl=q+(1l. Whence y+y1j+22yyj1 x+x1.

Example. y+y2 X3. y 3. y+2y, 3pp,2 y2= q3 2yy1 +y12=3qq12

Whence y+2y1 +2yyl +y12 3xx12.

Example. yx-x3-x2+x. p=X3, p+p 1 3xx12-pi

q _ X2, q+ql =-2xxj r=, r+r1-x+X1

Whence y + y1 3xxl2 -2xx1 +x+x1 -x13.

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Example. y2-X3-X2. P X3 2pp tp 12-3xx q2 _X2 2qqj==-2xx1

q12 =-x12

Whence 2yyl +y12 =3xx12-2xxl -x13. Here q12 is added so that we can suil up to y's.

Example. y+y2= =X3. Y'P, y+y V-2p y2-q, 2yy1l=2q

-X3 r, x13-3xx12 =2r Whence y+y1 +2yy1+x13--3xX 12 =O.

Example. y+y2-X3. Y P, y+2y -3p y2 q, yi2 2yy j3q,

-X3-r, -3xx 12=3r Whence y+2?, +y12 + 2yyl -3xx12 O0.

Example. y x- X2. p-=X3 p+p1 3XX12-X13 q--X2 q+q,1 -2xx1

Whence y+y =3xx12 -X1-2xx1.

Example. Y+Y2 r-x2. y_-_p2 Y+Yl =-2pp y 2_-q 2 2yyl -2qqj

Whence Y+Yj + 2yy, =-2xx1.

Example. y+y*2-x3-X2+X. p+p2 X'3, p+2p1+2ppl+p12 =3xx12 q+q2=-X2, q+ql +2qql -2xxj

r+r2 =-x, r+ 2rr, + r-=x+x1 Whence y+2yyl +y1 =3xx,2-2xx1 x+x1 -pt -p12 -3xx12-2xx1 +x-x13

+X 2

Example. y+/2 -X3 -X2 +X. Y-x_P, y+yP-x-xl 2p y2 +X2 zq, 2(Qyy +xxl )=2q

-X3-r, -3XX12-x13=2r

Whence y+y,-x-x1 +2yy1 +2xx1j3xx12 .

The value for 2r is found as follows. Suppose r variable say z, then z+2z--3X3, z+2z1=--3xx,2, but once z is really constant this becomes 2z1 -=2r

--3XX 12-X la.

C. TERMS COMPOUND, BUT OF EQUAL DEGREE.

C. THE X'S AND y'S BALANCED.

C2. THE X'S AND Y'S NOT BALANCED.

C,. As in A make one pair current, taking each element from a different term.

Example. xy 1. xy+xy=2, so that a current y may be balanced by a current x and leave a fixed point. Whence xy1 +x,y-2.

Example. X2yg-c. x2y2+X2y2 =2c. Whence xxyl 2+yyyx12=2c.

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Example. X3y- =C. x3y3 +x3y3--2c. Whence x13y12y+xx,2yi1-=3c. C2. As in B3, break up the absolute ternm into summands, each summand

being equal to one of the class of terms, and proceed as in B1, making one pair current, selecting its constituents from a number of terms inversely proportional to the weight of the elements selected, operating upon the terms as classified in A-C1.

Example. x2y3 c. Here two x's inust be balanced by three y's. Hence 2xx 1y13 +3yy2 X12 X 5c.

Example. x2y+xy2=a. x2y-a1, 2xx1y1 +X12y-3a,. Xy f42 Xal 2xyl 1 y+xy12 3a2.

Whence x(2x1y1 +y'12)+y(2x1yl +? 12) ~3a. D. TERMS COMPOUND AND NOT OF THE SAME DEGREE.

D1. X'S AND Y'S NOT BALANCED IN EACH TERM.

D2. X'S AND Y'S BALANCED IN EACH TERM.

D1. Decompose into surnmands as in B3, el and add the results. Example. x2,f+x2y3=a. x2y-=P, x12y+2x, y1x=-3p

X2y3-q 3x12y 2+2x,y13ax5q Whence x,2y+2xlylx+3x,2y12y +y2xjy1,3x 3a+2q=3a+2x,2yl3.

Example. x2y+xly-c. x2yz _p, x12y+2xxl Yi 3p x3yq, xi13 +3xx12y1-4q

Whenice xi3y+3xx12yl +X12 y+2xxjyj Y c+X13y1.

Example. X3y +Xy=c. x3y=p, x13y+3xx12y, 4p xy q, x1y+y,x=2q

Whence x13y+3xx12y1 + xly+ylX r2c+2p 2c+ 2x,3yl. D2. In D1 owing to the uniequal weight of x and y in the terms we are

compelled by C2 to take a numnber of terms proportional to the degree of the term before we can operate. In D2 however, owing to the cancellations we must specifically designate this requirement or it will be overlooked. Hence, treat each term by itself and combine the results, each multiplied by the degree of its term.

Examnple. xy+x2y2 -,a. x1y+xy1 +2(x,2yly+xxlyl2 )=2p+4q-2a + 2x 2y12.

Example. X4y44+xy-- a. x4y4 4(x,4y13y +y4x 3x) 8p xy--(q, x1 y+Xy I2q

Whence 4x14 yi3 y+4lXy14x13 +x1y+xyl -8p + 2q-2a+6x,4y144

Example. xy+x3y3 =a. xy-p, x1y+xy I2p -x3y3-q, 3(xx12y13 +X13y12y) 3.2q

Whence x,y+ylx+3xx,2y13 +3x,3y12y=2p+6q=2a+4x:y1 +4x13y1 3.

Example. X2y2 +x3y3=a. 2(xl Xy12 +Xl2yy1 ) + 3(XX,2y13 +X 13yy12 ) 4p+ 6q=4a+2x13 y13

E. COMPOSITES OF A, B, C, D.

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Make each homogeneous set of terms equal to a summand of the absolute term. Operate on these by A-D, and add the results.

Example. xy+x+y=a. xy=p, x Y+xyI-- 2p x+y =q, x+x1 +Y+Y1 =2q

Whence xly+xy,+x+y+xl+y,---2x,y,+2x,+2y,, and x1y+xyl+x+y -x1y1 -a.

Example. xy+xz 1. xy=p, xy? +xly=-2p x-q, x +x =2q

Whence x1y+xy1 +x+x,=2(p-+q)=2.

Example. x2+y2+xyzz1. X2 +y2 p- 2(xxi +yy) 2p xy q, xy1 +xjy=2q

Whence xy1 +x, y+2xx1 +2yy, -=2.

Example. x2 +xy--a. x2 I 2xxl -2p xy=q, xy1 +x,y=--2q

Whence x1y+xy, +2xxt =2a.

PROOF THAT FOR MAXIMUM CURRENT THE EXTERNAL AND INTERNAL RESISTANCES SHOULD BE EQUAL.

By JAMES S. STEVENS, Professor of Physics, University of Maine, Orono, Me.

If we have a cells to connect we may take m series with n cells in each series. Then mnn=a.

By formula for Ohm's law,

C- nE nr + R m

where r and R are respectively the internal resistance of each cell and the total external resistance.

Dividing numerator and denominator by n we have

C~ E r R m n

For maximum current it is necessary to make -r - - a minimum. m n

The expression takes the following form:

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