m.

Ex. 2. To find which pair of conjugate diameters of an ellipse make the least angle with each other.

Let m, n, be these diameters, P the angle which they form with each other. Then by ( Art. 59, Conic Sec.) we have mn sin P=ab, and

ab

flux sin P ma+n=a* +6. Therefore sin P=

and.

n (a + bn -ab (a* +6_-2 n)

=0; consequently — ab (a* +6°—2 -* )=0. ni (a* +6%1%) Now since ab cannot be equal to zero, we must have the factor a' +62-2 n = 0, whence n =

(a'+6)

b

2 Hence the equal conjugate diameters of an ellipse are those which by their intersection form the least angle as required. The sine of that angle is 2 ab .

2 tan U Let := =tan U, we shall have sin P-2 tan U

1+tan 'L seca U 2 sin U cos U=sin 2 U; therefore the angle P is equal to that formed by the two lines drawn from the two extremities of the conjugate axis to either extremity of the transverse axis. Ex. 3. Of all triangles constructed on the

M 1 same base AB, and having the same perimeter, which has the greatest surface.

Let the semi-perimeter=9, the base AB=a, the side AM = x, MB will = 29 a-X. Therefore calling y the surface, we shall have

А

a

시

y=v{9(9–a) (9–)(+5—9)}

1

+

=0

and therefore 2 log y=log 9 +log (9-a) +log (9-x)+log (a +1-9);
consequently
2 y

and ly y
y el
at 1-9

at -9 9-
Hence since y cannot=0, we must have the second factor

2 equal to zero, and therefore a +1-939–4, or 2 q-a-t=MB=t, and consequently the triangle required is isosceles.

From this it follows that among all isoperimetrical triangles, or triangles having the same perimeter, the one which has the greatest surface is equilateral. For if AMB is the triangle required, it is evident that it must have a greater surface than any other isoperimetrical triangle AMB, constructed on the same base AB; therefore AM -- NB. In the same manner it may be proved that AM=AB.

49. Hitherto we have only considered the maximum or minimum of the function of a single variable r. To find in what cases any function Y of two variables x and y becomes a maximum or a minimum, we may employ the following method.

1

a

Let us suppose that y has already the value adapted to render the function Y a maximum or minimum ; we shall then only have to find the proper value of x, that is to say we must take the fuxion of the function Y supposing c only to vary, and equate the co-efficient of i to zero Pursuing a similar mode of reasoning, we shall find that to obtain y we must take the fluxion of the function Y making only y to vary, and then equate the co-efficient of y to zero. Hence it follows that if Y is represented generally by Pi+Qý, we must have P=0, and Q=0, two equations which will give the values of x and y proper to render the function Y a maximum or a minimum.

It is easy to see that this same reasoning applies, whatever be the number of variable quantities of which Y may represent a function. Hence in general to determine those values of the variable quantities which will render the function Y a maximum or minimum, we must take the entire flucion of Y, and equate to zero the co-efficient of the fiuxion of each variable, which will give as many equations as there are unknown quantities.

Ex. I. Let it be required to divide a given number a in three parts, whose product may be a maximum.

Calling x and y two of these parts, the third will be expressed by *-y, and we shall have for the product xy (ar-y). The ant

amx fluxion of this expression is = (a_21y) yz + (a-2y—) zý. Equating separately to zero the co-efficient of ;, and that of y, we shall have a—2x - Y=o=a-24.4., Hence y=x=ja. Consequently the given number must be divided into three equal parts.

Ex. 2. Let it now be required to determine among all isoperimetrical triangles, that which has the greatest surface. We have before resolved this problem, but indirectly Let x, y, be two of its sides, 24 the perimeter, then 2q----y

––

will De the other side, and the surface will be v {9(9-x) (9-4) (= +9-9)}. This must be a maximum; if we call it Y, we shall have 2 log Ylog =log (9-) + log (9-y) + log (x+y-9). Therefore Y= Y: 1

1
(
-) +

.), equating to zero 4+/-3 1-- )

(+4–9 7y the co-efficient of 3 and also that of y, we have x+y-q=-=-x; hence ==y=29=29–5–y. Consequently the triangle is equilateral, as we before found.

Examples for Practice. 1. Of all the squares which may be inscribed in a given square, which is the least?

2. Of all fractions, which is the one that exceeds its mth power by the greatest possible quantity ?

a

ਨੂੰ ਦੇ+ਨੂੰ

3

а

3. Required that number x of which the xth root is a maximum

4. I am desirous of constructing a cylindrical measure of a given capacity, and of which the internal surface may be a minimum. What must be the ratio between the height of the measure and the diameter of its base ?

5. Among all the cylinders which may be inscribed in the same sphere, which is that whose convex surface is a marimum ?

6. Which among these cylinders has the greatest solid content ?

7. What must be the dimensions of the greatest cylinder that can be inscribed in a given cone?

8. Of all triangles standing on the same base, and inscribed in the same circle, which is the greatest ?

9. Which, on the contrary, will be the least of all the triangles that may be circumscribed about the same circle ?

OF VANISHING FRACTIONS.

50. We sometimes meet with algebraic expressions in the form of fractions, which on certain suppositions become equal to . Such

xa? for example, is the quantity

when r=a. Now though in appearance indeterminate, these results are nevertheless susceptible of determinate values ; and the following is a method for ascertaining them.

P Let be a fraction whose numerator and denominator are func

Q tions of x, both of which become zero when x=a. To find the value of this fraction, substitute r+r in place of r in P and Q,

; then making x=a in this latter fraction,

Q+Q it will become .; and this will be the value of the proposed frac

Q tion, on the supposition of x=a+, or of r=a, provided however

a

and we shall have P+P

that the terms of the fraction

do not again destroy one another

upon making x=a.

r?a* Ex. l. Required the value of_ when t=a? Here P=r

P 2x di -a', and Q=r-a; therefore_

2r=2a.

Q Ex, 2. Let there be the geometrical progression 1, ??, ?,...." of

Taking the fluxions as directed, we shall find=(n+1).x"_1=

=x

n, as is evident. Ex. 3. Let there be proposed the quantity V (2a os-x4)—a) a'r

4- å x3 which becomes when r=a. Taking the fluxions and proceeding

a3_2 x3

Va's as before, we shall have✓(2a r-x+) 3.0

16 ga, the value

a

3 4

of the proposed quantity.

51. But if it happens that on substituting a instead of in.

this

a

n+1

1-X

;

fraction also becomes.o; we must treat it in the same manner as the first, and so on, till we arrive at a value of which one term at least is finite. Ex. If we take the fluxion of the identical equation x++ory

after dividing both sides by we shall have st 2 *? +372...+2=

n7"
*+n2+2_- (n+1) +++1

which becomes (1 – x)

nt1 when r=1.

P 1-2" (x+1)*+n (n+2)x
Therefore

but this Q

-2 (1-1) new expression gives also.. on substituting 1 for x; we must therefore take the fluxions of its numerator and denominator sepa

-1.29-(n+1)' + n(n+1)(n+2).x" which rately, and we shall have

2 on making a=1, gives n (n+1) for the sum of the arithmetical

x. progression 1, 2, 3,...n. Ex. 2. In the quadratrix y= =-* tan comeand this expression bes

sin .. when x=a. Therefore y=

a flux cot C

2

>

a

CX

comes

cit

Examples for Practice.

ar? + acé--2 acx 1. Required the value of the fraction

when r=4 b.xl_2 bcx + bc

a* _6* 2. What is the value of the fraction

when :=0.

Note. By these principles we may in each particular case find the indeterminate values of o xoo,

and of co-.

For o xoo becomes since 0 = We also reduce to this same process co- by

b supposing that the first co arises from and the second from

1 For example, if x=1, we have

(because log

a

avelog o , log 3

1-1 l=o). Taking the fluxion of

log *

as directed, it becomes-l.