Question

question 7 extra credit
an inverted pyramid is being filled with water at a constant rate of 35 cubic
centimeters per second. the pyramid, at the top, has the shape of a square with
sides of length 3 cm, and the height is 5 cm.
since the base is square, the volume is given by ( v=\frac{1}{3}b^{2}h ).
hint: substitute for ( b ) in the volume formula using the ratio of ( b ) to ( h ).
find the rate at which the water level is rising when the water level is 2 cm.
cm/sec
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Explanation:

Step1: Find the relationship between \(b\) and \(h\)

By similar - triangles, \(\frac{b}{h}=\frac{3}{5}\), so \(b = \frac{3}{5}h\)

Step2: Substitute \(b\) into the volume formula

Substitute \(b=\frac{3}{5}h\) into \(V=\frac{1}{3}b^{2}h\). Then \(V=\frac{1}{3}(\frac{3}{5}h)^{2}h=\frac{1}{3}\times\frac{9}{25}h^{2}\times h=\frac{3}{25}h^{3}\)

Step3: Differentiate \(V\) with respect to \(t\)

Differentiate both sides of \(V = \frac{3}{25}h^{3}\) with respect to \(t\) using the chain rule. \(\frac{dV}{dt}=\frac{3}{25}\times3h^{2}\frac{dh}{dt}=\frac{9}{25}h^{2}\frac{dh}{dt}\)

Step4: Solve for \(\frac{dh}{dt}\)

We know that \(\frac{dV}{dt}=35\) (the rate of change of volume). When \(h = 2\), we substitute into \(\frac{dV}{dt}=\frac{9}{25}h^{2}\frac{dh}{dt}\)

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Answer:

\(\frac{875}{36}\text{ cm/sec}\)