EXAMPLE. 157 



Then we have 



AP = x sec , Pp = sec dx, 



T>TV b sec 

 and y = PP = j x\ 



a TTV i b sec 2 

 .*. the area of PPpp = --- ^ z <?. 



v 



The velocity of discharge through this area 



therefore the quantity discharged in an element of time 



sec 2 

 -, 



_ "] _. 

 xdx \dt 



[b sec 2 / s"! , , 

 - ^ V2# ^t J rf#, 



the a;-limits being and z\ and this must equal Kdz, 

 from (2). 



Hence we have, from (3), taking the negative sign, be- 

 cause z decreases as t increases, 



= f 



J 



- Ed > 



15/7T 



15/7T tan 2 



tan 8 



between the limits /i and z. 



