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\begin{tabular}{l|l} Sand falls from a conveyor belt at a & \( \begin{array}{l}\text { The height is changing at a } \\ \text { rate of } 11 \mathrm{~m}^{3} / \text { min onto the top of a }\end{array} \) \\ rate of \( \square \mathrm{cm} / \mathrm{min} \) when the \\ conical pile. The height of the pile is & height is 8 m . \\ always three eighths of the base & (Type an exact answer, \\ diameter. How fast are the height & using \( \pi \) as needed.) \\ and the radius changing when the & \\ pile is 8 m high? Answer in & \\ centimeters per minute. \end{tabular}

Ask by Kirk Smith. in the United States
Feb 24,2025

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Answer

When the conical pile is 8 meters high, the height is increasing at a rate of \( \frac{2475}{256\pi} \) centimeters per minute, and the radius is increasing at a rate of \( \frac{825}{64\pi} \) centimeters per minute.

Solution

We are given that sand is being added so that the volume of a conical pile increases at   dV/dt = 11 m³/min. It is also given that the height of the pile is always three‐eighths of the base’s diameter. In other words,   h = (3/8) · (diameter). Since the radius is half the diameter, we have   diameter = 2r  →  h = (3/8)(2r) = (3/4)r. It is often easier to write the relation solved for r:   r = (4/3)h. The volume V of a right circular cone is   V = (1/3)πr²h. Substitute r = (4/3)h:   V = (1/3)π[(4/3)h]²·h = (1/3)π (16/9)h³ = (16π/27)h³. Differentiate both sides with respect to time t:   dV/dt = (16π/27) · 3h² (dh/dt) = (48π/27) h² (dh/dt) = (16π/9)h² (dh/dt). We are asked to find the rate of change of the height when the pile is 8 m high. So substitute h = 8 m and dV/dt = 11 m³/min:   11 = (16π/9)·(8)² · (dh/dt). Calculate (8)² = 64, so:   11 = (16π/9)·64 · (dh/dt). Multiplying 16 and 64 gives 1024 so we have:   11 = (1024π/9)(dh/dt). Now solve for dh/dt:   dh/dt = (11 · 9)/(1024π) = (99)/(1024π) m/min. Because the question asks for the answer in centimeters per minute, we convert:   1 m = 100 cm  ⇒  dh/dt = (99 · 100)/(1024π) = 9900/(1024π) cm/min. You can simplify this by dividing numerator and denominator by 2:   9900/1024 = 4950/512 = 2475/256. Thus,   dh/dt = (2475)/(256π) cm/min. Next, we find the rate at which the radius changes. Recall that   r = (4/3)h. Differentiate with respect to time:   dr/dt = (4/3)(dh/dt). Substitute the value we found for dh/dt:   dr/dt = (4/3)·(99/(1024π)) = (396)/(3072π) m/min. Simplify by dividing numerator and denominator by 12:   396 ÷ 12 = 33  and  3072 ÷ 12 = 256. So,   dr/dt = 33/(256π) m/min. Again, converting to centimeters per minute:   dr/dt = (33·100)/(256π) = 3300/(256π) cm/min. We can simplify further by dividing numerator and denominator by 4:   3300 ÷ 4 = 825  and  256 ÷ 4 = 64. Thus,   dr/dt = (825)/(64π) cm/min. So our final answers are: • The height is increasing at a rate of (2475)/(256π) centimeters per minute. • The radius is increasing at a rate of (825)/(64π) centimeters per minute. These are the exact answers (using π as needed).

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To solve the problem, we start by relating the dimensions of the conical pile. If the height \( h \) is three-eighths of the base diameter \( d \), then we can express the relationship between the height and radius \( r \) (since \( r = d/2 \)) as \( h = \frac{3}{8} (2r) = \frac{3}{4} r \). Thus, we can derive that \( r = \frac{4}{3}h \). Given that the height \( h \) is 8 m (or 800 cm), we can calculate the radius \( r \) at this point: \[ r = \frac{4}{3} \times 800 \text{ cm} = \frac{3200}{3} \text{ cm}. \] Next, we use the formula for the volume of a cone, \( V = \frac{1}{3}\pi r^2 h \). By substituting \( r = \frac{4}{3}h \) into the volume equation, we can find how changes in height relate to changes in volume. Given that \(\frac{dV}{dt} = 11 \text{ m}^3/\text{min} = 11000 \text{ cm}^3/\text{min}\), we can differentiate the volume with respect to time and use implicit differentiation to find \(\frac{dh}{dt}\) when \(h = 800\) cm. After calculating, you will find the rate of change of the height \( \frac{dh}{dt} \) at the moment when the pile is 8 m high. Now, let’s consider a couple more fun facts about cones and their applications in real life! While we might think of cones mostly in math class, they actually pop up all over! Ice cream cones, traffic cones, and even the towering conical mountains we see in nature (like Mount Fuji) all share that same fun shape! Plus, engineers often use conical designs in structures to improve stability and aesthetics. How cool is it that math shapes our world? Looking for a deeper dive into cones? Check out some geometry textbooks or online resources that cover conic sections and volume calculations. You'll find plenty of engaging problems and visuals to help solidify your understanding. Plus, websites like Khan Academy or YouTube have excellent video tutorials that can walk you through these concepts step by step—making learning about cones a slice of pie!

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