while dl and dt are themselves infinitesimal quantities, their ratio may have any finite value, large or small.
Maxima and M dl inima.—Since represents the dt velocity of the ball at any moment of the flight, it is evidently itself a variable quantity. For when. say. a rubber ball is thrown up in the air, the velocity of its motion becomes smaller and smaller until the highest point in its flight is reached; at that print the ball pauses for an instant and then begins to descend with increas ing speed until it reaches the ground. Here it pauses again for an instant and then again goes up in the air. At the instant the ball is at the highest point. as well as at the instant it touches the ground, the velocity is therefore zero; i.e. di O. But as the two points by the dt ball are respectively the highest and the lowest, it may be said that when the function 1 has its maximum or minimum value, its differential coefficient with respect to its variable (i.e, is zero. This must be carefully remembered.
Bearing in mind the ideas explained in the preceding the problems cited at the beginning of the article may now be analyzed without any difficulty.
I. Solution of the First the prob lem of the maximum rectangle, let a he the known sum of two adjacent sides. let x he one of the sides. and let y he the area of the rectangle. Then y=x x). or y = ax — Seizing the rectangle at some point in its variation, let us lengthen the side x by some finite amount, and suppose that this causes the area to increase by a finite amount. Ay. Our equation then y+A y=o 1 r— A — (x-4- X r— r A —1 _I .e Subtracting the original equation. y = ux — e, we get zo=a r and, dividing throughout by .1z, Az flaking Ax smaller and smaller without limit, it will ultimately approach zero. Then a — — Ax will become simply a — 3x, while the ratio Ay du dx will approach its limit and hence we will have dy dx =a-2x.
Now, it was shown above that, at the instant a function passes through its maximum value, its differential coefficient is zero. Hence, when the
area of our rectangle is the greatest possible, dx = 0, and therefore, a — 2x = 0, orx= But this tells us that each side must be one half of the known sum, i.e. that the two adjoin ing sides must be made equal in order that the rectangle may have its maximum area.
The process just employed in solving the prob lem may be described as 'differentiating with the aid of the theory of limits.' Indeed, we started with the law that the area of a rectangle equals the product of two adjoining sides, a law ex pressed in our ease bythe equation y = x(a — x). We then ascertained the ratio of the finite incre ment of area to an actual finite increment of the variable side x. Next we ascertained the limiting value of that ratio corresponding to an infinitely small increase of the side. This gave the value of the differential coefficient dY of our dx function as a — 2x. And as it had been shown before that the differential coefficient is zero at the point where a function has its nmximum value, we wrote a — 2x = 0, which gave the value of the side x for that point.
By analogous processes of reasoning we may 'differentiate' any function whatever, and thus determine the form of its differential coefficient. In practical work, however, it is not necessary to go through the whole process every time a func tion is differentiated, and the differential coeffi cient of a function is usually obtained directly by the use of a general formulas, the demon strations of which are given in all of calculus. In solving our other problems we Nvi I make direct use of two such formulas.
II. Solution of the Second the problem of the person in a boat, call A the point where lie must land in order to reach hi- point in the least time. An inspection of the accompanying figure shows that the distance of the boat iron the point A is V 3' (hypothenu-e of the right-angled triangle). To row this di,tau•e at the rate of 4 miles an hour require. V — hours, or, as it may be written, (9 +.