Sand is falling from a conveyor onto a sand
pile that has the shape of a cone where the
radius is always three times the height. If
sand is being poured at the rate of 20 cu.
ft./min. how fast is the base radius changing
at the moment when there is 200 cu. ft of
sand in the pile?

To solve this problem, we'll use related rates and the geometric relationship between the volume of a cone and its dimensions.

### Step 1: Understand the given relationships.
- The volume \( V \) of a cone is given by:
  \[
  V = \frac{1}{3} \pi r^2 h
  \]
  where \( r \) is the radius of the base and \( h \) is the height.
  
- We're told that the radius \( r \) is always three times the height \( h \):
  \[
  r = 3h
  \]

### Step 2: Substitute the relationship into the volume formula.
Substituting \( r = 3h \) into the volume formula:
\[
V = \frac{1}{3} \pi (3h)^2 h = \frac{1}{3} \pi \times 9h^2 \times h = 3\pi h^3
\]

### Step 3: Differentiate the volume with respect to time.
Given that sand is falling at a rate of 20 cubic feet per minute, we have:
\[
\frac{dV}{dt} = 20 \, \text{cu. ft./min.}
\]

Now, differentiate \( V = 3\pi h^3 \) with respect to time \( t \):
\[
\frac{dV}{dt} = 9\pi h^2 \frac{dh}{dt}
\]

### Step 4: Solve for \(\frac{dh}{dt}\).
We rearrange the equation to solve for \(\frac{dh}{dt}\):
\[
\frac{dh}{dt} = \frac{\frac{dV}{dt}}{9\pi h^2} = \frac{20}{9\pi h^2}
\]

### Step 5: Find the height \( h \) when \( V = 200 \) cu. ft.
Using \( V = 3\pi h^3 \) and setting \( V = 200 \) cu. ft.:
\[
200 = 3\pi h^3
\]
\[
h^3 = \frac{200}{3\pi}
\]
\[
h = \sqrt[3]{\frac{200}{3\pi}}
\]

### Step 6: Calculate \(\frac{dh}{dt}\) at this height.
Now substitute the value of \( h \) into \(\frac{dh}{dt}\):
\[
\frac{dh}{dt} = \frac{20}{9\pi \left(\sqrt[3]{\frac{200}{3\pi}}\right)^2}
\]

### Step 7: Relate \( \frac{dr}{dt} \) to \( \frac{dh}{dt} \).
Since \( r = 3h \), differentiating with respect to time \( t \) gives:
\[
\frac{dr}{dt} = 3 \frac{dh}{dt}
\]

### Step 8: Calculate \(\frac{dr}{dt}\).
\[
\frac{dr}{dt} = 3 \times \frac{20}{9\pi \left(\sqrt[3]{\frac{200}{3\pi}}\right)^2}
\]

This simplifies to:
\[
\frac{dr}{dt} = \frac{20}{3\pi \left(\sqrt[3]{\frac{200}{3\pi}}\right)^2}
\]

### Step 9: Simplify and provide the final answer.
Simplifying further:
\[
\frac{dr}{dt} = \frac{20}{3\pi \times \frac{200^{2/3}}{(3\pi)^{2/3}}} = \frac{20 \times (3\pi)^{2/3}}{600^{2/3} \times 3\pi}
\]

Finally:
\[
\frac{dr}{dt} = \frac{(3\pi)^{2/3}}{10\cdot 200^{2/3}} \, \text{ft/min}
\]

Thus, the radius of the base of the sand pile is changing at this rate when there are 200 cubic feet of sand in the pile.

Would you like more details on any step or have any questions?

---

Here are some related questions to consider:
1. How does the shape of the sand pile affect the rate of radius change?
2. What happens to \(\frac{dr}{dt}\) as the volume of sand increases indefinitely?
3. How would the solution change if the radius was not directly proportional to the height?
4. Can you derive the relationship between radius and height for a different type of cone, like one with a fixed angle?
5. What if the rate of sand being poured changes over time? How would you modify the approach?

**Tip:** Always check if there are proportional relationships between dimensions; they simplify the problem significantly.

Sand is falling from a conveyor onto a sand pile that has the shape of a cone where the radius is always three times the height. If sand is being poured at the rate of 20 cu. ft./min., how fast is the base radius changing at the moment when there is 200 cu. ft of sand in the pile?

Learn how to solve a related rates problem involving a cone-shaped sand pile where the radius is always three times the height. Find out how fast the radius is changing when 200 cubic feet of sand is in the pile, with a pouring rate of 20 cu. ft./min. Perfect for students in advanced calculus.

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