Sidebilder
PDF
ePub

To find the point of contrary flexure, we have

pa do P

(p being the I upon the tangent). w po d0+ dp Hence substituting

2 MP
pe

4 Mo + p
dp
8 M3 – 2 Mp*

= 0 at the point of inflexion, (see de

(4 M + p) Simpson, new edit.),

::p* – 4M2 -- 0

IN 2M

M M
Hence 0 =

which equations will give the p 2 M

2

or

[ocr errors]

position of the point of contrary flexure.

17. To find that curve whose normal has the same relation to the distance between the origin of abscissæ, and the intersection of the normal with the axis, that the ordinate of a parabola has to its corresponding abscissa, we have

y'? = px' in the parabola, and :: (Normal)' = p. (abscissa + subnormal) in the curve required. y

p d.ro

d.x Hence

pdx 2ydy

2

= d.c

[merged small][merged small][merged small][ocr errors][merged small][merged small][merged small][merged small][merged small][ocr errors]

Now if r be the radius of a 0, and the origin of abscissæ be distant from its circumference by the interval

y,' = - (2r + a) a + 2 . (1 + a) x, - x?. The curve is therefore a circle.

a, we have

[ocr errors]

18. To trace the curve whose equation is 6.5* + y)*

y? By the solution of a quadratic, we have

In 87+1 2.ro - 1

[merged small][merged small][ocr errors][merged small]

Let x = 0. Then

y = 0, or the curve passes through the origin of abscissæ A, (see Fig. 1.)

Let x = $1= AB or Ab, Then

y = 0, or the curve passes through B and b, and y is imaginary for every greater value of x.

Also since the pairs of values of y, ( + y) are the same for the positive as for the equal negative values of x, the curve will consist of two equal and similar ovals AMBA, AmbA, intersecting, (the curve is therefore quadrable) in A.

Again

dy
= 72.

8.x2 +1
dx

8x9 +1 X( 8x2 + 1 - 2x2 - 1)! = tan 0 (e being the inclination of the tangent at any point to the line of abscissæ.)

Hence when the curve is parallel to the axis, tan. 0 = tan. 0 = 0, or

2.x 8x2 +1=0

2.x

or x =

Again, since

dy
dr

when x = 0, the point at A is multiple,

and it will be found, by the usual methods, that this point is double; that there is at A for each branch a point of single inflexion, and that the curve there cuts the axis at an angle of 45°.

To find at what angle the curve cuts the axis when x = a maximum, or when x = + 1, we have

tan. A =
dy

= 00 = tan. 90°.
d.x

:: the angle required = 90°

Making AM = p, and Z MAB = 0, y = PM = p. sin. « } and substituting in the given equation

I = AP =p.cos. Q we get the polar equation p = cos. 20.

(a) If BM = 2, then we get

de

and the arcs of the curve will be found

dz =

[ocr errors]

St

xdx = those of the elastic curve, whose equation is y =

Nithe abscissa x corresponding to p.

This curve is called the Lemniscata of James Bernoulli.

It is the locus of the intersection of the perpendicular from the centre upon the tangent of an equilateral hyperbola, whose semiaxes are unity. For in the equilateral hyperbola

a 62
pa

bcos.? Q -- a' sin.”
1

when a b= 1
cos. 20

a' b? and p* (the perpendicular) =

po - a + b* :: p = cos. 20, (See equat. a). See Prob. I, Vol. II.

[ocr errors]

19. To find the equation of the curve whose subtangent = b + y, we have

ydx
dy

=b+ y

[ocr errors]

= bl.y + y + c :: I. (y en ) = x - C and y' e' = etc = et = c' e*, the equation required.

ec

[blocks in formation]

20. The subcontrary section of an oblique cone is a circle. Let a m 6 (Fig. 14,) be the section made by a plane passing through the cone parallel to the circular base AB, AVB being the triangular section of the cone made by a plane passing through the vertex and centre of the base. Also let A'm B' be the subcontrary section made by a plane I plane AVB, and inclined to BV by the angle A'B'V = A, and intersecting amb in mp.

Join Vm, Vp and produce them to M and P respectively;

join PM

Then since the planes amb, A'm B', are I plane AVB, their intersection mp is I ab. Also because PM is parallel to pm, (being the intersections of a plane with two parallel planes,) it is I plane AVB, and :: I AB. Again, from similar triangles,

AP : ap :: VP: vp :: MP : mp and BP : bp :: VP : vp :: MP : mp :: AP.BP: ap .bp :: MP? : mp? But AP. BP = MP (by prop. of O)

:: ap . bp = mp3 Also since 2 pB'b = LA= La, and the vertical Lare equal, the A A'ap, B'bp are similar,

:. A'p : ap :: pb : B'p

:: A'p. B'p = ap. bp = mp,
and :. the subcontrary section A'm B’ is a circle.

21. To find the curve by whose revolution round its line of abscissæ, a solid is generated = of its circumscribing ey

3

5

linder.

Let r and y be its co-ordinates, x beginning at the vertex.
Then, by the question
Say'da ту* xx base of solid x alt.)

5

3

( :: y'da = ydy x x +

[blocks in formation]

3dy d.r

y and l. y3 = l. r + l. c= l. :: y3 = cx is the equation to the curve, which is, therefore, the cubic parabola.

са

22. To construct the spiral whose arc is the measure of the ratio between the ordinates which intercept it.

If CAP = 0, AP = pq and CP = 2,

AC being = 1 (Fig. 15)
Then z = 1.1

[ocr errors][merged small][merged small][merged small]

But dz = dpa + pa do?, as we readily learn from the figure.

dp .. dp* t på d0=

[merged small][merged small][ocr errors][merged small][ocr errors][merged small][merged small][merged small][ocr errors][merged small]

udu

du

+ N u

u su Nu 1 + sec -2 + c N1-p2 + sec.

1

tc P

P Let 0 = 0, then p= 1, and c=0 and :. A = sec. -1

1 N1 - p2, the equation of the curve ;

P

[ocr errors]

which being analogous to that of the involute of a circle whose radius is unit, we will attempt the construction by the aid of that curve.

« ForrigeFortsett »