A reservoir is in the form of a frustum of a cone with upper base of radius 9 ft and lower base of radius 4 ft and altitude of 10 ft. The water in the reservoir is x ft deep. If the level of the water is increasing at 4 ft/min, how fast is the volume of the water in the reservoir increasing when its depth is 2 ft ?

# Related Rates : Frustrum of a Cone given the radius and heigth

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When the depth of water is 2 ft

$r = 2 \left( \dfrac{9 - 4}{10} \right) + 4$

$r = 5 ~ \text{ft}$ ← radius of water surface

$Q = vA$

$\dfrac{dV}{dt} = \dfrac{dh}{dt} \times \pi r^2$

$\dfrac{dV}{dt} = 4 \times \pi (5^2)$

$\dfrac{dV}{dt} = 100\pi ~ \text{ft}^3/\text{min}$ ←

answer