Water is pouring into an inverted cone at the rate of 8 cubic feet per minute. If the height of the cone is 12 feet and the radius of its base is 6 feet, how fast is the water level rising when the water is 4 feet deep? If there is a small hole at the bottom of the cone so that when water is pouring into the cone at the rate of 8 cubic ft per minute, water is also leaving from the bottom hole at the rate of 2 cubic ft per minute. How fast is the water rising when the water reaches 4 ft deep?
Question
Water is pouring into an inverted cone at the rate of 8 cubic feet per minute. If the height of the cone is 12 feet and the radius of its base is 6 feet, how fast is the water level rising when the water is 4 feet deep? If there is a small hole at the bottom of the cone so that when water is pouring into the cone at the rate of 8 cubic ft per minute, water is also leaving from the bottom hole at the rate of 2 cubic ft per minute. How fast is the water rising when the water reaches 4 ft deep?
Solution
The problem involves the application of calculus, specifically related rates.
Step 1: Understand the problem The problem is asking for the rate at which the water level is rising in a cone when the water is 4 feet deep. The cone is filled at a rate of 8 cubic feet per minute, and water is leaving at a rate of 2 cubic feet per minute through a hole at the bottom.
Step 2: Set up the equation The volume V of a cone is given by the formula V = (1/3)πr²h, where r is the radius and h is the height. Since the cone is similar at all times, r/h = 6/12 = 1/2, so r = h/2. Substituting r into the volume formula gives V = (1/3)π(h/2)²h = (1/12)πh³.
Step 3: Differentiate with respect to time Differentiating both sides of the equation with respect to time t gives dV/dt = (1/4)πh²dh/dt.
Step 4: Solve for dh/dt We know that dV/dt = 8 - 2 = 6 cubic feet per minute (the rate at which water is being added minus the rate at which it's leaving). We want to find dh/dt when h = 4. Substituting these values into the equation gives 6 = (1/4)π(4)²dh/dt. Solving for dh/dt gives dh/dt = 6/(π(4)²/4) = 6/π feet per minute.
So, the water level is rising at a rate of 6/π feet per minute when the water is 4 feet deep.
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