NEWTON.

1850. (A). Enunciate and prove Newton's second Lemma. (B). Hence shew that two quantities may vanish in

an infinite ratio to each other.

Taking the figure as constructed in the Lemma, we have it proved by the Lemma that the sum of the parallelograms ab, bc, cd,* &c. vanishes in the limit; and à fortiori the parallelogram ab vanishes. But in the limit the number of these parallelograms is infinite, and therefore the above-mentioned sum is infinitely greater than the parallelogram ab. Hence two quantities may vanish in an infinite ratio to each other.

1849. (4). Enunciate and prove Newton's fourth Lemma.

(B). In Lemma X., if the velocity vary as the square of the time, shew that the space will vary as the cube of the time.

In Lemma x. the time is divided into equal intervals represented by the lines AB, BC, CD, ... (fig. 64); the velocities being represented by lines perpendicular to these, Bb, Cc, Dd.

It is hence shewn that if Ak be the curve passing through all the points b, c, d, &c., when brought indefinitely near to one another,

space in time AD: space in time AK:: area ADd: area AKk.

Let the time AK be divided into the same number of equal parts as AD is divided into.

Then, since the velocity varies as the square of the time in which it is generated, the parallelogram Ab is proportional to AB3, Bc to BC.AC2 or 22.AB3, CD to 3'AB3; and

*See the figure in Goodwin's Newton, Lemma 11.

so on. Similarly, if AB', B'C', &c. are the intervals into which AK is divided, Ab' will be proportional to AB", B'c' to 22.AB", C'd' to 32.AB'3, and so on. Thus the ratio

of each one of the former series of parallelograms to the corresponding one of the latter series is AB3: AB", or (since the number of intervals in AD, AK is the same), AD3 : AK3.

But by Lemma IV. the curvilinear areas ADd, AKk will be in the same ratio.

Therefore, from above,

space in time AD: space in time AK :: AD3 : AK3,

or the spaces are proportional to the cubes of the times.

1851. (A). Enunciate and prove Newton's fourth Lemma.

(B). Apply this Lemma to prove that the area included between a hyperbola and the tangents at the vertices of the conjugate hyperbola is equal to the area included between the conjugate hyperbola and the tangents at the vertices of the hyperbola.

Let the semiaxes CA, CB, (fig. 65) of the hyperbola be divided each into the same number (n) of small equal parts. Let mm' be the (r+1)th part of CB, counting from C; so that

Hence the area of the small parallelogram in P is

Now the conjugate hyperbola is a concentric hyperbola having AM for its transverse axis. Hence, if we divide BCAK into n parallelograms of equal breadth parallel to BC, we shall find the area of the (r+1)th parallelogram from C

Thus we have divided the figs. BCAK, BCAL into the same number of parallelograms, each of which in one figure has ultimately to the corresponding one in the other a ratio of equality. Hence, by the Lemma, the figures themselves have to one another a ratio of equality; and therefore the figures KK', LL', which are four times the former figures, are equal to one another.

1849. (A). Define similar curves.

(B). Shew that all parabolas are similar to each other.

(C). Describe an instrument which is adapted for drawing curves similar to given curves.

Two curves are said to be similar, when there can be drawn in them two distances from two points similarly situated, such that if any two other distances be drawn equally inclined to the former, the four are proportional. (Evans's Newton, Cor. to Lem. v.)

Let P (fig. 66) be any point in a parabola. Draw the tangent PT and the perpendicular PY upon it from the focus. By the property of the parabola,

Let A', S', Y', be points in any other parabola corresponding to A, S, Y; P' a point in this new parabola, such that

Now since the LA'S'P' LASP, therefore the LA'S'Y'-LASY. Hence the triangles A'S'Y', ASY are similar, and therefore

which, in accordance with the above definition, proves the parabolas to be similar.

Again, let it be required to draw a curve similar to a given curve APB

Let BC, bc (fig. 67) be two equal parallel rulers joined together by two equal parallel rulers Bb, Cc on hinges at B, b, C, c. APp a rod moveable about a hinge at A.

If now we suppose BC, bc, and APP to have each a longitudinal slit, then, by making a pin passing through both rods APP, BC at P to move along the curve BPA, a pencil passing through the intersection of APp, be will trace out the curve bp.A similar to BPA. For, since BP is parallel to bp, therefore

1851.

AP: Ap: AB: Ab.

(A). The spaces described from rest by a body under the action of any finite force are in the beginning of the motion as the squares of the times in which they are described.

(B). If the force vary as the time from rest, prove that the velocity will vary as the square of the time.

Let the same system be pursued in (B) as is adopted in the Lemma; i.e. let AB, BC, CD, &c. represent equal intervals into which the whole time AK is divided.

Let the lines Bb, Cc, Dd, (fig. 68) at right angles to AK represent the magnitudes of the force at the end of the times AB, AC, AD, &c.... Since, by hypothesis, the force varies as the time from rest, the points b, c, d, &c. will all lie in a straight line through A.

If we complete the rectangles Ab, Bc, Cd, &c., these will represent the velocities generated in the intervals AB, BC, CD,... on the supposition that the force remains the same during any interval as it is at the end of such interval. Hence, by reasoning similar to that in the Lemma, the limit of the sum of all the rectangles, i.e. the triangle AKk, will represent the whole velocity actually generated in the time AK.

Thus we get

velocity in time AK: velocity in time AD:: area AKk: area ADd :: AK: Ak2;

i.e. the velocity a square of the time.

1850.

(A). If a body move in free space under the action of a central force, the velocity at any point of the orbit varies inversely as the perpendicular let fall from the centre upon the tangent.

(B). If lines proportional to the Earth's velocity, and always parallel to the direction of its motion, be drawn from a fixed point, shew that the extremities of these lines will trace out a circle.

Let the lines be drawn from the focus, and let p' be the length of any one of them. Then, since p' is by hypothesis proportional to the Earth's velocity, and by (4), the Earth's velocity is inversely proportional to the perpendicular (p) from the focus on the direction of the Earth's motion, i.e. the tangent to the orbit, p' must be proportional to

pp' is constant.

1

therefore
>
p

We have then two lines always at right angles to one another, such that their lengths are symmetrically related; i.e.