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dt =

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2c

Now

g 2 de

CV (R'— 5)
and t =
x{PR - - R: sin.-13

+C

R
Let t = 0, when p = R and we have
c=2 x R .

2
1
{W (R-8) + R.
?

sin.-1 p
2c

2

R
Let = 0. Then the whole time to the centre is
R’m
Rim

R.

1
2.
Х

W2u
R

T=

2c

Again, the Periodic Time in a circle whose radius is

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540. Let the distance of the vertex of the parabola from its focus be a, then the velocity in the curve is

с

V (453)

1

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р

and the velocity parallel to the axis
: vel. in curve :: dx : ds

:: dr : (da+ dy)
:: 1:(1+

dyo

dra But y: = 4ax. Therefore dy

Qar

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Again, if ß be the velocity with which the system moves, and y' the ordinate of the curve-in-space at the end of the time t, the whole increment of y' during dt is

dy' = Bdt dy according as the velocity I to the axis arising from the motion in the curve, is in the same or opposite direction to that of the system.

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+ 2?)}
and by p. 172, of the same book

2.ctat s dx (1 + a + 1 . x+x+) =

a

a

4

2

+") XV (1+a+*x+r?) 4-(a+a) SV(1+a+2x+.29

)

dx

+

Х

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Hence, putting a +

a

+

And making y' = 0 when x = 0, we get

= A, and V (1+a+2+ + x= x v=* ***Ax-(077) 1. (20 + A + 2x)} +C

==={4-433.in

* {{2s++ -) +0+*+34) (a-a)"

:: y' =

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1

1

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2x +a+a+21/(1+a+à «+x)

a +

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the equation to the required curve in space.

541. The velocity in a circle, when the force is F, and radius = 1, is

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where he is the absolute force, and or that same function of the radius (raccording to which the central force attracts at different distances.

Hence, at the same distance, and for the same law of force denoted by q, we have

voc N cc (absol. force}).

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543. Let a, b be the given semi-axes, and T the given periodic time. Then since the force tends to the focus, if , and o be the polar co-ordinates, it is easily shewn (see Newton, Prop. 1.) that the areas described by ç are proportional to the times of description; and because the area of the ellipse is wab, we have T:mab :: dt :

pdo

2

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2παο

3
And v =

Х
T

(1-e)
276

2πα
Х
TV (1-e?)

T

3 which gives the velocity at any point of the ellipse.

2are

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P

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F = c2 dp

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545. If u and v'denote the velocity in the curve, and that in a circle at the same distance, we have generally (438)

de

dpp Hence when v = v

de dp

P P
And = lp + c
But when ? =R, let p = P; then

R
C = IR - IP =1


And we have

R b=lxpx

P

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546. If (a – ) be the space described in a time towards a centre where the force is F, we have

rdv =

Fdp :: v= - ? Fdę

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