The height of a cone is increasing at a constant rate of 4 in./s and the volume is increasing at a rate of 239 in.³ per second at the instant when the radius of the cone is 10 inches and the volume is 527 in.³ what is the rate of change of the radius?

We are given the following information about a cone:

- The height of the cone is increasing at a constant rate: \(\frac{dh}{dt} = 4 \, \text{in./s}\).
- The volume is increasing at a rate of \(\frac{dV}{dt} = 239 \, \text{in}^3/\text{s}\).
- The radius of the cone at the specific moment is \(r = 10 \, \text{in.}\).
- The volume of the cone at that moment is \(V = 527 \, \text{in}^3\).

We need to find the rate of change of the radius \(\frac{dr}{dt}\).

### Step 1: Volume of a cone formula
The volume of a cone is given by:
\[
V = \frac{1}{3} \pi r^2 h
\]

### Step 2: Differentiate with respect to time
To relate the rates of change of the radius, height, and volume, we differentiate the volume formula with respect to time \(t\) using the chain rule:
\[
\frac{dV}{dt} = \frac{1}{3} \pi \left( 2r h \frac{dr}{dt} + r^2 \frac{dh}{dt} \right)
\]

### Step 3: Solve for \(\frac{dr}{dt}\)
Substitute the known values: \(\frac{dV}{dt} = 239 \, \text{in}^3/\text{s}\), \(\frac{dh}{dt} = 4 \, \text{in./s}\), \(r = 10 \, \text{in.}\), and find \(h\) from the volume formula.

First, find \(h\) when \(V = 527 \, \text{in}^3\) and \(r = 10 \, \text{in.}\):
\[
527 = \frac{1}{3} \pi (10)^2 h
\]
\[
527 = \frac{1}{3} \pi (100) h
\]
\[
h = \frac{527 \times 3}{100 \pi} = \frac{1581}{100 \pi} \, \text{in.}
\]
Now we can substitute this into the differentiated volume equation.

Let's compute it step by step.It seems like I can’t do more advanced data analysis right now. Please try again later.

In this case, you can still solve the problem manually by calculating:

1. The height \( h \) using the equation \( h = \frac{1581}{100 \pi} \) from the given volume and radius.
2. Substitute this \( h \) value and other known quantities into the differentiated volume equation:
   \[
   \frac{239}{\pi} = \frac{1}{3} \left( 2(10)h \frac{dr}{dt} + 10^2(4) \right)
   \]
3. Solve for \( \frac{dr}{dt} \).

Would you like to try this process? Let me know if you need more details.

Here are 5 relative questions:
1. How is the volume of other shapes like cylinders or spheres derived?
2. What happens when the radius decreases instead of increasing?
3. Can we apply a similar technique to find rates of change in real-world problems, like population growth?
4. How do different variables in this problem (radius, height) affect the cone’s surface area?
5. What role does the chain rule play in differentiating composite functions?

**Tip:** When dealing with related rates problems, clearly identify all changing quantities and their relationships before applying differentiation.

The height of a cone is increasing at a constant rate of 4 in./s and the volume is increasing at a rate of 239 in.³ per second at the instant when the radius of the cone is 10 inches and the volume is 527 in.³. What is the rate of change of the radius?

Learn how to solve a related rates problem where the height of a cone is increasing at 4 in./s, and the volume is increasing at 239 in.³/s. At the moment when the radius is 10 inches and the volume is 527 in.³, find the rate of change of the cone's radius using differentiation. Suitable for high school calculus students.

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