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 cm
Length of the square base = 7 cm
If 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.
