Question

the diagram shows a water container in the form an inverted pyramid , which is such that when the height of the water level is h cm the surface of the water is a square of side 1/2 h cm
1) express the volume of water in the container in terms of h .
[note the volume of a pyramid having a base area A and vertical height h is 1/3Ah.
water is steadily dripping into the container at a constant rate of 20 cm cube per minute.
2)find the rate in cm per minute,at which the water level is rising when the height of the water level is 10 cm.

Answer 1

Answer:(1)$\frac1{12}h^3$ (2) 0.8

$(1)Formula\;of\;volume:\;V=\;\frac13\times A\times h\\Formula\;of\;area\;of\;square:\\A=side\times side=\frac12h\;\times\;\frac12h\;=\;\frac14h^2\\V=\frac13\times A\times h=\frac13\times\frac14h^2\times h=\frac1{12}h^3\\(2)\frac{\operatorname dV}{\operatorname dt}=20\\Find\;\frac{\operatorname dh}{\operatorname dt},\;when\;h\;=10.\\\frac{\operatorname dV}{\operatorname dt}=\frac{\operatorname dV}{\operatorname dh}\cdot\frac{\operatorname dh}{\operatorname dt}\\\frac{\operatorname dV}{\operatorname dh}=\frac{\operatorname d\frac1{12}h^3}{\operatorname dh}=\frac14h^2\\\frac{\operatorname dV}{\operatorname dt}=\frac14h^2\cdot\frac{\operatorname dh}{\operatorname dt}\\Substitute\;\frac{\operatorname dV}{\operatorname dt}=20,\;h=10.\\20=\frac1410^2\frac{\operatorname dh}{\operatorname dt}\\\frac{\operatorname dh}{\operatorname dt}=\frac45=0.8$