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Water is drained out of tank, shaped as an inverted right circular cone that has a

radius of 16cm and a height of 8cm, at the rate of 4 cm³/min. At what rate is the
depth of the water changing at the instant when the water in the tank is 6 cm
deep? Show all work and include units in your answer. Give an exact answer
showing all work and include units in your answer. (10 points)

User Sobolevn
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1 Answer

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Final answer:

To find the rate at which the water depth is changing in an inverted conical tank, we use related rates calculus considering the volume of the cone and the proportional relationship of tank dimensions. The depth of water is decreasing at a rate of -1 / (18π) cm/min when the depth is 6 cm.

Step-by-step explanation:

The student is asking about the rate at which the depth of water is changing in an inverted conical tank when water is being drained at a consistent rate. Since we have a geometric shape where the dimensions are related, we can use related rates calculus to find the rate at which the water depth is changing.

Let's define the variables involved in the problem: V for volume, r for radius, h for water depth, and t for time. The volume of a cone is given by the formula V = (1/3)πr^2h. Because the cone is right circular and inverted, there's a proportional relationship between the water depth and radius which can be expressed as r/h = R/H, where R = 16 cm and H = 8 cm are the radius and height of the entire tank. As such, r = 2h.

Now we differentiate V with respect to t.

dV/dt = (1/3)π(2h)^2dh/dt = (4/3)πh^2dh/dt

We know that dV/dt is -4 cm³/min (negative because the volume is decreasing). When h = 6 cm, we substitute the values to find dh/dt:

-4 = (4/3)π(6)^2dh/dt

Then we solve for dh/dt:

dh/dt = -4 / ((4/3)π(6)^2)

dh/dt = -1 / (18π) cm/min

Therefore, the water depth is decreasing at a rate of -1 / (18π) cm/min at the instant when the depth is 6 cm.

User Oberfreak
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