We are dealing with a related rates problem where sand forms a conical pile. The rate at which the volume of sand is increasing is given, and we are asked to find how fast the height and radius of the pile are changing when the height is 7 meters.

Given:

• The rate of change of volume $\frac{dV}{dt} = 10 \, \text{m}^3/\text{min}$.
• The height $h$ of the cone is always $\frac{3}{8}$ of the base diameter. This implies $h = \frac{3}{4}r$, where $r$ is the radius of the base.

Volume of a cone:

The formula for the volume of a cone is: $V = \frac{1}{3} \pi r^2 h$ Substituting $h = \frac{3}{4}r$ into the volume formula: $V = \frac{1}{3} \pi r^2 \left( \frac{3}{4}r \right) = \frac{\pi}{4} r^3$

Step 1: Differentiate with respect to time $t$

To find how fast the radius $r$ is changing, we differentiate the volume equation $V = \frac{\pi}{4} r^3$ with respect to time $t$: $\frac{dV}{dt} = \frac{\pi}{4} \cdot 3r^2 \cdot \frac{dr}{dt}$ $\frac{dV}{dt} = \frac{3\pi}{4} r^2 \frac{dr}{dt}$

Step 2: Solve for $\frac{dr}{dt}$

We are given $\frac{dV}{dt} = 10 \, \text{m}^3/\text{min}$, so we substitute that into the equation: $10 = \frac{3\pi}{4} r^2 \frac{dr}{dt}$ Now we need to find $r$ when $h = 7 \, \text{m}$. From $h = \frac{3}{4}r$, we can solve for $r$: $7 = \frac{3}{4}r \quad \Rightarrow \quad r = \frac{7 \times 4}{3} = \frac{28}{3} \, \text{m}$ Substitute $r = \frac{28}{3}$ into the differentiated volume equation: $10 = \frac{3\pi}{4} \left( \frac{28}{3} \right)^2 \frac{dr}{dt}$ Simplify: $10 = \frac{3\pi}{4} \times \frac{784}{9} \frac{dr}{dt}$ $10 = \frac{3\pi \times 784}{36} \frac{dr}{dt}$ $10 = \frac{2352\pi}{36} \frac{dr}{dt}$ $10 = \frac{196\pi}{3} \frac{dr}{dt}$ Solve for $\frac{dr}{dt}$: $\frac{dr}{dt} = \frac{30}{196\pi} = \frac{15}{98\pi} \, \text{m/min}$ Convert to centimeters per minute: $\frac{dr}{dt} = \frac{15}{98\pi} \times 100 \, \text{cm/min}$ $\frac{dr}{dt} \approx \frac{1500}{98\pi} \approx 4.87 \, \text{cm/min}$

Step 3: Find $\frac{dh}{dt}$

Since $h = \frac{3}{4}r$, we differentiate both sides: $\frac{dh}{dt} = \frac{3}{4} \frac{dr}{dt}$ Substitute $\frac{dr}{dt} \approx 4.87 \, \text{cm/min}$: $\frac{dh}{dt} = \frac{3}{4} \times 4.87 \approx 3.65 \, \text{cm/min}$

Final Answer:

• The radius is changing at approximately $4.87 \, \text{cm/min}$.
• The height is changing at approximately $3.65 \, \text{cm/min}$.

Would you like further clarification or details on any step?

Related Questions:

1. How would the solution change if the height was half the base diameter?
2. What happens if the sand falls at a faster rate, say $20 \, \text{m}^3/\text{min}$?
3. How does the relationship between height and radius affect the rate of change in the volume?
4. Can you calculate the rate of change of the surface area of the cone at the same height?
5. How does the rate of change of the radius compare to the rate of change of the height for different cone dimensions?

Tip:

For related rates problems, always identify the key relationships (equations) between variables before differentiating, as it simplifies the calculations later.