Page:Calculus Made Easy.pdf/65

of the bottom is $200$ feet. Find an expression for the quantity pouring in or out when the depth of water varies by $1$ foot; hence find, in gallons, the quantity withdrawn hourly when the depth is reduced from $14$ to $10$ feet in $24$ hours.

The volume of a frustum of pyramid of height $H$, and of bases $A$ and $a$, is $V={\dfrac {H}{3}}(A+a+{\sqrt {Aa}})$. It is easily seen that, the slope being $45^{\circ }$, if the depth be $h$, the length of the side of the square surface of the water is $200+2h$ feet, so that the volume of water is

${\dfrac {h}{3}}[200^{2}+(200+2h)^{2}+200(200+2h)]=40,000h+400h^{2}+{\dfrac {4h^{3}}{3}}$.

${\dfrac {dV}{dh}}=40,000+800h+4h^{2}=$ cubic feet per foot of depth variation. The mean level from $14$ to $10$ feet is $12$ feet, when $h=12$, ${\dfrac {dV}{dh}}=50,176$ cubic feet.

Gallons per hour corresponding to a change of depth of $4$ ft. in $24$ hours ${}={\dfrac {4\times 50,176\times 6.25}{24}}=52,267$ gallons.

(13) The absolute pressure, in atmospheres, $P$, of saturated steam at the temperature $t^{\circ }$ C. is given by Dulong as being $P=\left({\dfrac {40+t}{140}}\right)^{5}$ as long as $t$ is above 80°. Find the rate of variation of the pressure with the temperature at $100^{\circ }$ C.