Question

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 6 cm, and the height is 13 cm.
Find the rate at which the water level is rising when the water level is 3 cm.
cm/sec

Answer 1

Rates of change are the change of a quantity over another.

The rate of change in height of the water level is 36.51 cm per second.

Let the length of the top square be s, and the height be h.

The volume of the pyramid is:

$\mathbf{V = \frac{1}{3}s^2h}$

At time t, we have:

$\mathbf{V = \frac{1}{3}s(t)^2h(t)}$

The relationship between the side length and height is:

$\mathbf{s : h=6 : 13}$

Express as fractions

$\mathbf{\frac{s }{ h}=\frac{6 }{ 13}}$

Make s the subject

$\mathbf{s=\frac{6 }{ 13}h}$

So, we have:

$\mathbf{V = \frac{1}{3}s^2(t)h(t)}$

$\mathbf{V = \frac{1}{3} \times (\frac{6 }{ 13}h(t))^2 \times h(t)}$

$\mathbf{V = \frac{1}{3} \times \frac{36}{ 169}h^2(t) \times h(t)}$

$\mathbf{V = \frac{1}{3} \times \frac{36}{ 169}h^3(t)}$

Differentiate

$\mathbf{V'(t) =3 \times \frac{1}{3} \times \frac{36}{ 169}h^2(t) \times h'(t)}$

$\mathbf{V'(t) =\frac{36}{ 169}h^2(t) \times h'(t)}$

Make h'(t) the subject

$\mathbf{h'(t) = \frac{169 \times V'(t)}{36 \times h^2(t)}}$

The water level rises constantly at 70 cm^3/s, and the water level is 3 cm.

So, we have:

$\mathbf{V'(t) = 70}$

$\mathbf{h(t) = 3}$

$\mathbf{h'(t) = \frac{169 \times V'(t)}{36 \times h^2(t)}}$ becomes

$\mathbf{h'(t) = \frac{169 \times 70}{36 \times 3^2}}$

$\mathbf{h'(t) = \frac{169 \times 70}{36 \times 9}}$

$\mathbf{h'(t) = 36.51}$

Hence, the rate of change in height of the water level is 36.51 cm per second.

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