The volume of a cone decreases at a rate of 4 cm^3 s^-1. How fast is the radius when its radius is 5cm and its height is 8cm?

To find how fast the radius of a cone is changing when its radius is 5 cm and its height is 8 cm, we can use related rates and the volume formula for a cone. The volume \(V\) of a cone is given by:

\[ V = \frac{1}{3} \pi r^2 h \]

Given that the volume decreases at a rate of 4 cm\(^3\)/s, we can write:

\[ \frac{dV}{dt} = -4 \text{ cm}^3/\text{s} \]

We need to find the rate at which the radius is changing, \(\frac{dr}{dt}\), when \(r = 5 \text{ cm}\) and \(h = 8 \text{ cm}\).

First, differentiate the volume formula with respect to time \(t\):

\[ \frac{dV}{dt} = \frac{d}{dt} \left( \frac{1}{3} \pi r^2 h \right) \]

Since \(h\) is also a function of time, we apply the product rule:

\[ \frac{dV}{dt} = \frac{1}{3} \pi \left( 2r h \frac{dr}{dt} + r^2 \frac{dh}{dt} \right) \]

We need to express \(\frac{dh}{dt}\) in terms of \(\frac{dr}{dt}\). We can use the fact that for similar cones, the ratio of height to radius remains constant. Thus:

\[ \frac{h}{r} = \text{constant} \]

Differentiate both sides with respect to time:

\[ \frac{d}{dt} \left( \frac{h}{r} \right) = 0 \]

\[ \frac{r \frac{dh}{dt} - h \frac{dr}{dt}}{r^2} = 0 \]

\[ r \frac{dh}{dt} = h \frac{dr}{dt} \]

\[ \frac{dh}{dt} = \frac{h}{r} \frac{dr}{dt} \]

Substitute this into the differentiated volume equation:

\[ \frac{dV}{dt} = \frac{1}{3} \pi \left( 2r h \frac{dr}{dt} + r^2 \left( \frac{h}{r} \frac{dr}{dt} \right) \right) \]

\[ \frac{dV}{dt} = \frac{1}{3} \pi \left( 2r h \frac{dr}{dt} + r h \frac{dr}{dt} \right) \]

\[ \frac{dV}{dt} = \frac{1}{3} \pi \left( 3r h \frac{dr}{dt} \right) \]

\[ \frac{dV}{dt} = \pi r h \frac{dr}{dt} \]

Now, substitute the known values \( \frac{dV}{dt} = -4 \text{ cm}^3/\text{s} \), \( r = 5 \text{ cm} \), and \( h = 8 \text{ cm} \):

\[ -4 = \pi \cdot 5 \cdot 8 \frac{dr}{dt} \]

\[ -4 = 40 \pi \frac{dr}{dt} \]

\[ \frac{dr}{dt} = \frac{-4}{40 \pi} \]

\[ \frac{dr}{dt} = \frac{-1}{10 \pi} \]

So, the radius is decreasing at a rate of \(\frac{1}{10 \pi}\) cm/s.

Would you like more details or have any questions?

Here are 5 questions you can ask next:
1. How do we apply related rates to different geometrical shapes?
2. Can you explain more about the product rule in calculus?
3. How can we use similar triangles in related rates problems?
4. What are some real-world applications of related rates?
5. How do we find the rate of change of height in similar problems?

**Tip:** When solving related rates problems, always start by writing down what you know and what you need to find, then differentiate the relevant equations with respect to time.

Learn how to find the rate of change of the radius of a cone using related rates in calculus. This problem involves calculating derivatives, applying volume formulas of cones, and understanding the concept of related rates. Suitable for advanced high school students.

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