Proposition 85

Table of Contents

PROPOSITION 85 Problem

According to the preceding hypothesis of gravity and resistance if the curve MA (Fig. 84) is given, upon which the descent is completed, to find the curve for the ascents with this property, that the time of each ascent is equal to the time of the preceding descent.

Solution

On putting as before g for the force acting and the exponent of the medium is equal to k, for the curve MA let the abscissa AP = x, the arc AM =s and for the curve sought AN, the abscissa AQ = t, and the arc AN = r. A height equal to b is put corresponding to a certain speed of descent acquired at A, with which speed the body completes the following ascent on the curve AN . [p. 425] With these in place the height corresponding to the speed at M is equal to

…

and the height corresponding to the speed at N in the ascent is equal to :

…

Therefore the time of descent along the arc MA is equal to …

and the time of the ascent along the arc AN is equal to :

…

which two times, upon integrating, there is put

…

must be equal. According to this being obtained, put … and

…

where X, T, S and R are such functions that vanish on putting x, t, s and r = 0.

Hence this is brought about in order that the two integrals become equal to each other, if after integration there is put X = b and T = b. But S and R as both X and T are clearly quantities not depending on b and they must have the same relation between each other, whatever value b may have. Therefore what is sought is satisfied, if R is such a function of T, as S of X. Or on taking R = S there also corresponds T = X. Now .. hence on making R = S then

…

Moreover since with this in place it is necessary that X = T or e dx = e dt , then it follows that t = e k dx. While now then …

Hence from which the construction of the curve is known, since by taking the arc .. to this there corresponds the abscissa …

The equation for the curve AN is more conveniently found from the given ..

equation between s and x. For since

if the values are put in place of s and x, then the equation is produced between t and r for the curve AN sought. Q.E.I.

Corollary 1

Since it is the case that r = 2kl (2 − e 2k ) , then it follows that Now since it must be the case that dr > dt, lest the curve AN becomes imaginary, the curve AN is real up to that point, and in correspondence

Corollary 2

Now e is always greater than unity; from which it follows, that everywhere ds > dx, and from that

Whereby if the given curve is real, then the sought curve is always real also.

Corollary 3

Since it is the case that s = −2kl (2 − e 2k ) , then thus an easier relation can be taken in the computation between ds and dx.

Corollary 4

From the solution of the problem it is likewise apparent, how the inverse problem or this shall be solved. For if the ascending curve AN is given, or the equation between t and r, from that the equation between x and s is formed with the help of the equation

Corollary 5

Towards investigating the form of the curve around the point A , s and r are placed very small and then e 2 k = 1 , hence dr becoming equal to ds and dt = dx. From which it is evident that the lowest parts of the curves MA and NA are similar and equal to each other.

Example 1

Let the given line of the descents be the right line MA (Fig. 85) at some inclination, in order that s = αx or ds = αdx . Now since we have this equation is obtained between t and r for the curve sought : and the integral of this is …

Which equation converted into a series gives : and thus at the lowest point A it is the case that dr = αdt . This curve has a horizontal r tangent somewhere, which place is found on putting dt = 0; now then 2 = e 2 k or r = 2kl 2 , to which there corresponds αt = 2k − 2kl 2 .

If r becomes greater than 2kl 2 , the value of dt is made negative and thus the curve again descends, while – dt becomes equal to dr; but this happens, if the equation is [p. 429] But since α cannot be less than 1, then if α = 1 , then the vertical tangent from

A stands apart, and if α > 1 , then there is no vertical tangent beyond the horizontal tangent. But otherwise [?] beyond the tangent there is obtained a vertical, where it is the case that …

Hence in this case, in which the given line is vertical or α = 1 , then r = 0 or the tangent at A is vertical.

Corollary 6

If s denotes the whole length of the arc of descent, r expresses the whole arc described on the curve AN (Fig. 84) from the following ascent. Whereby if the arc of the descent s is given, the arc of the ascent can be found : For since when we put T = X, the letters s and r denote the whole arcs of the descent and of the ascent.

Example 2

Let the given curve MA be the tautochrone of the descents, that we have previously found for the same hypothesis of the resistance; the curve AN has this property, that all the ascents are also completed in equal times, clearly the same by which the descents are completed on MA. Whereby the curve AN is a tautochrone of the ascents with the curve MA found before now continued. Moreover so that this can be shown from that calculation, we take the equation for the tautochrone curve of the descents, which is (719) [p. 430]

Since now with these substituted, … or …

Which equation is formed from the former, if t is put in place of x and – r for s. Whereby this curve AN has been continued, and with MA the tautochrone of the ascents.

Corollary 7

Hence with the arc of the descent s given on the tautochrone of the descents, then the arc of the following ascent is the tautochrone of the ascents : And if the descent begins from the cusp of the tautochrone, the place of this is given by : ..

(729), and the arc of the ascent is : as we have found above (732).

Scholium

Thus, since the tautochrone according to this hypothesis of the resistance satisfies the question and the curve is continued, hence we can seize the handle and more continuous curves can be investigated, of which the two branches of the curves in turn MA and AN are able to be sustained – which we present in the following proposition.