An inverted pyramid is being filled with water at a constant rate of 70 cubic centimeters per second. The pyramid, at the top, has the shape of a square with sides of length 7 cm, and the height is 10 cm. Find the rate at which the water level is rising when the water level is 4 cm.

Answer:Rate of change in height of the water level is 2.91 cm per second.Step-by-step explanation:Height of the inverted pyramid = 10 cmLength of the square base = 7 cmIf water is filled up to the level of h cm then the volume of water up to height h will be V = [tex]\frac{1}{3}(\text {Area of the base})\times (h)[/tex]V = [tex]\frac{1}{3}(x^{2} )\times (h)[/tex]It is given that rate of water is filling with 70 cubic centimeters per second.[tex]\frac{dV}{dt}=70[/tex]From two similar triangles in the figure attached,[tex]\frac{x}{h}=\frac{7}{10}[/tex][tex]x=\frac{7h}{10}[/tex]By replacing the value of h,V = [tex]\frac{1}{3}(\frac{7h}{10})^{2}h[/tex]V = [tex]\frac{1}{3}(\frac{49h^{2} }{100})h[/tex]V = [tex]\frac{1}{3}(\frac{49h^{3}}{100})[/tex]Now we integrate the equation with respect to time 't'[tex]\frac{dV}{dt}=\frac{d}{dt}(\frac{1}{3}\times \frac{49h^{3} }{100})[/tex]70 = [tex]\frac{49h^{2} }{100}\times \frac{dh}{dt}[/tex][tex]\frac{dh}{dt}=\frac{100\times 70}{49h^{2}}[/tex]For h = 7 cm[tex]\frac{dh}{dt}=\frac{70\times 100}{49\times 49}[/tex][tex]\frac{dh}{dt}=2.91[/tex]Therefore, rate of change in height of the water level is 2.91 cm per second.