An inverted pyramid is being filled with water at a constant rate of 50 cubic centimeters per second. The pyramid, at the top, has the shape of a square with sides of length 6 cm, and the height is 14 cm. Find the rate at which the water level is rising when the water level is 6 cm.

1. the first recommend step is to make a sketch of the situation. We draw the triangle representing the pyramid and we label the sides. (as the picture attached)

2. What we need to find is the change of the height of the water with respect the time (dh/dt), so we need to find an equation that allow us to find that. We start with the equation of the volume of the pyramid which is [tex]V=\frac{b^2h}{3}[/tex] . To find (dh/dt) it is necessary to take the derivative of this equation, however, the equation involves two variables (b and h). We only can derive if the equation is in terms of only one variable. therefore, we need to find another equation to do this.

3. Applying ratios of triangle, we can find and expression that relates b with h. To apply this ratio, it is necessary to split the triangle drawn in step 1, so we will form a right triangle. To apply the ratio of triangle we follow the next formula: [tex]\frac{height1 }{base1} =\frac{height2}{base2}[/tex]

Where the subscript 1 is referring to the triangle 1 and the 2 to the triangle 2.

Replacing the values and isolating "b" we get...

[tex]\frac{14}{3} =\frac{h}{b} \\\\b= \frac{3h}{14}[/tex]

4. Now we can replace the equation 2 into 1 and we get the volume of the pyramid in terms of only one variable (h)

5. Take the derivative of the last equation

6. isolate dh (dh is what we need to find, dh means how the height changes with respect the time)