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Thus, again, the results are imaginary. The problem, however, is impossible. For, if any given area is to be inclosed in rectangular form, the perimeter will be the least when the figure is a square. But the square root of 50 exceeds 7; hence the field will have a perimeter of more than 28 rods, and can not be inclosed by a fence 24 rods long. We conclude, therefore,
That imaginary roots indicate impossible conditions in the problem.
PROBLEM OF THE LIGHTS.
313. To illustrate more fully the rules of algebraic interpretation, we present for discussion the following general
PROBLEM.—Find upon the line which joins two lights, A and B, the point which is equally illuminated by them; admitting that the intensity of a light at any given distance, is equal to its intensity at the distance 1, divided by the square of the given distance.
Let a represent the intensity of the light A at the distance 1, and b the intensity of the light B at the distance 1.
Let c denote the distance AB, between the two lights. Assume A as the origin of distances, and regard all distances measured from A toward the right as positive.
Finally, let C denote the point of equal illumination, and let x represent the distance of this point from A. Then – must be the distance of the same point from B. That is,
BC=(-3. But by the conditions of the problem, the intensity of the light
a A at the distance x is and the intensity of the light B at the
6 distance c-a is
But these intensities are equal, because
C represents the point of equal illumination; hence we have the equation,
va By reducing this equation, we obtain two values of x, as follows:
Vamb Since the two values of x are real, and also unequal, we conclude,
That there are two points of equal illumination on the line AB, or on this line produced.
This is evidently the conclusion to which we ought to arrive by an algebraic solution of the problem, in order to satisfy its conditions in a general manner. For, whatever may be the relative intensities of the two lights, there must always be one point of equal illumination between them. And if the lights are of unequal intensities, there must be another point of equal illumination, in the prolongation of the line, on the side of the lesser light.
We will now discuss the values of x, under several hypotheses.
1st. Suppose a > b. In this case, both values of x are positive; therefore, both points of equal illumination are situated to the right of A. The first value of a is less than
is less than
vatvo unity, being a proper fraction. This value of x is also greater thau one half of c; for we have Va=ya;
(1) and since a >b, Va+yb < 2va.
(2) Dividing (1) by (2), we have,
va 1 Va+
2 and consequently,
Vatro> ; Hence, the first point of equal illumination is at C, between A and B, but nearer B than A. The second value of x is greater than c; for,
a-16 than unity, being an improper fraction. Hence, the second point is at C', in the prolongation of the line beyond B.
These conclusions are evidently correct. For, the supposition that a is greater than b, implies that B is the feebler light; both points should therefore be nearer B than A.
2d. Suppose a <b.
The first value of x is positive. It is, moreover, less than one half of c. For we have Vu=va;
(1) and since a <b, Va+vb> 27a. Dividing (1) by (2), we obtain,
Vatroz and therefore,
+ Hence, the first point of equal illumination falls between A and B, and nearer A than B, as it should, because A is the lesser light.
The second value of x is negative, since the denominator,ya-yb, is negative.
Now in the statement of this problem, we considered distances reckoned from A toward the right as positive; hence, according to the rule for interpreting negative results, previously established, (182), we must consider the negative result in this case, as a distance to be reckoned from A toward the left. Hence, the second point will be situated to the left of A, at C". And this is as it should be, because A, under the present supposition, is the lesser light.
3d. Suppose a =
= b. In this case, the first value of x is positive, and equal to Hence the first point of equal illumination is midway between A and B.
cya The second value of x is
This result indicates that
0 there is no other point of equal illumination in the line AB, or in AB produced, at a finite distance from A.
These conclusions are obviously correct. For, under the present supposition, the two lights are equally intense. Hence any point, to be equally illuminated by them, must be equally distant from them; and the only point which fulfills this condition is the point midway between them.
If, however, we consider a and b as two varying quantities, at first unequal, but continually approaching equality, then the second value of x will become greater and greater by degrees, until it reaches infinity. Under these conditions, the second point of equal illumination will continually recede from A, moving toward the right or toward the left, according as a is greater, or less than b, until it is finally removed to an infinite distance. In this view of the case, it is sometimes said that there are two points of equal illumination, under the hypothesis, a = b; one point being at an infinite distance from A. 4th. Suppose a = b and c = 0.
0 The first value of x reduces to 0; hence the first point
21a is situated at A.
0 The second value of x is the symbol of indetermination;
Ö (188, 4). This result shows that there are an infinite number of other points equally illuminated by the two lights.
These interpretations are evidently correct. For, as the lights, under the present hypothesis, are equally intense, and both situated at A, every point in space must be equally illuminated by them
5. Suppose c= 0, and a b or a < b.
Both values of w now reduce to 0; and the common rule for interpreting zero might lead us to suppose that the two points of
equal illumination coincide with the point A. But this conclusion is not strictly correct; for it is obvious that when two lights, of unequal intensities, occupy the same place, there is no point in space equally illuminated by them; not even the point in which they are both situated.
Let us return to the original equation (m), which truly represents the conditions of the problem. If we put c=0, the result is
an equation which can not be satisfied by any value of x whatever, while a > b or a < b. For by suwstituting any value for x we shall always obtain two unequal fractions. If x=0, the two members are two unequal infinities.
We conclude, therefore, that under the supposition, c= 0, while a and b are unequal, the problem fails altogether, and is impossible.
Thus we learn that zero may be the answer to a possible, or an impossible problem. And whenever we obtain this symbol as the result of a solution, we must not interpret it on the assumption that the thing required in the problem is possible; but we must first determine whether the conditions are rational or absurd, by considering the nature of the problem, or by substituting zero in the original equation.
PROBLEMS PRODUCING QUADRATIO EQUATIONS.
314. It will be found that some of the following problems may be solved by a single unknown quantity, while others require two. Still others may be conveniently solved by means of either one or two letters. It is left to the judgment and skill of the learner to discover the mode of solution, in each example, which is most simple.
1. It is required to divide the number 14 into two such parts, that 9 times the quotient of the greater divided by the less, may
be equal to 16 times the quotient of the less divided by the greater.
Ans. 8 and 6.