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for the quantity pouring in or out when the depth of water varies by $$1$$ foot; hence find, in gallons, the quantity withdrawn hourly when the depth is reduced from $$14$$ to $$10$$ feet in $$24$$ hours.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