Mathematics: 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.

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An inverted pyramid is being filled with water at

wereallmadhere111

08.01.2022

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 = $\frac{1}{3}(\text {Area of the base})\times (h)$

V = $\frac{1}{3}(x^{2} )\times (h)$

It is given that rate of water is filling with 70 cubic centimeters per second.

$\frac{dV}{dt}=70$

From two similar triangles in the figure attached,

$\frac{x}{h}=\frac{7}{10}$

$x=\frac{7h}{10}$

By replacing the value of h,

V = $\frac{1}{3}(\frac{7h}{10})^{2}h$

V = $\frac{1}{3}(\frac{49h^{2} }{100})h$

V = $\frac{1}{3}(\frac{49h^{3}}{100})$

Now we integrate the equation with respect to time 't'

$\frac{dV}{dt}=\frac{d}{dt}(\frac{1}{3}\times \frac{49h^{3} }{100})$

70 = $\frac{49h^{2} }{100}\times \frac{dh}{dt}$

$\frac{dh}{dt}=\frac{100\times 70}{49h^{2}}$

For h = 7 cm

$\frac{dh}{dt}=\frac{70\times 100}{49\times 49}$

$\frac{dh}{dt}=2.91$

Therefore, rate of change in height of the water level is 2.91 cm per second.

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