TRACT 38.

PROBLEMS.

365

ascertain the Time of Filling it to 6 Feet high , as before inthe last Problem.

Let acdb represent the sluice; and when the tide hasrisen to any height gh, below cd the top of the sluice, with-out the ditches, let ff he the mean height of the water within.And put i = 3 — ab = ac ;

Then \/g : ^eg : : 2 g : 2 \fg(x — z) the velo- A ccity of the water through aefb ; andVg ■ Veg :: fg : f\/g(x—z) the mean vel. through eghf ;theref. 2bzVg(x — z) is the quantity per sec, through aefb;and ^-b[x — z)*/g(x — z) is the same through eghf ;conseq. \bVg X (2„r + z)f(x — z) is the whole throughaghb per second. This quantity divided by the surface A,

gives x (2x -|- z)</ {x — z) — v the velocity per second

with which ef, or the surface of the water in the ditches,rises. Therefore

But as gh rises uniformly 1 foot in 30' or 1800", there-fore 1 : AG : : 1800" : 1800a' =: t the time of the tide rising

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g = 16a;

a = horizontal section of the ditches;x — AG;

Z = AE.

c

E

I>

H

F

B

%

v

, . .. : -- 3a

■ -vi r ( 1 2 i + z) a /(x-k)’

(2x + z)\/(x — z) . x is the fluxional equa. expressing

Z

the relation between x and z ; where m —

3200

or 13A|4 when a = 200000 square feet.

1200 Vs — 231

Now to find the fluent of this equation, assume z =

5 g ix x 4*

+ BX* + CX % + VX* &c. So shall

A 5 H- 4b 7.

a3 + 4ab + 8c

\/(x—z) = X^

5 8 il.

2x + z = 2x + Ax T -i bx t + c.r 1 ' 8cc,

(2 x+z) f(x—z)x

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