Water is tripping out from a conical funnel at the uniform rate of 2 cm cube per second for tiny hole at the vertex at the bottle when the slant height of the water is 4 cm find the rate of decrease if the slant height of the water  given that the vertical angle of the funnel is 120 degree

To solve this problem, we need to determine how the slant height of the water in the conical funnel changes with time as water drips out. Given the uniform rate of water flow, the goal is to find the rate of change of the slant height at the moment when the slant height is 4 cm. Let's work through this step by step.

### Step 1: Relevant Geometry of the Cone
We are given:
- The vertical angle of the cone is \(120^\circ\).
- The rate of change of volume of water is \(\frac{dV}{dt} = -2 \, \text{cm}^3/\text{s}\) (since the volume is decreasing).
- We need to find the rate of change of the slant height \(l\) when \(l = 4 \, \text{cm}\).

### Step 2: Relationship between Volume and Slant Height

For a cone, the volume is given by:

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

where:
- \(r\) is the radius of the circular cross-section,
- \(h\) is the height of the water inside the cone.

To relate this to the slant height \(l\), we use the geometry of the cone. The cone forms a right triangle where:
- The slant height \(l\) is the hypotenuse,
- The radius \(r\) and height \(h\) form the legs.

Given the vertical angle \(120^\circ\), the half-angle is \(60^\circ\). Therefore, by trigonometry:

\[
\tan(60^\circ) = \frac{r}{h} \quad \Rightarrow \quad r = \sqrt{3}h
\]

Additionally, from the Pythagorean theorem for the right triangle in the cone, we have:

\[
l^2 = r^2 + h^2
\]

Substituting \(r = \sqrt{3}h\):

\[
l^2 = (\sqrt{3}h)^2 + h^2 = 3h^2 + h^2 = 4h^2
\]

Thus, the relationship between \(l\) and \(h\) is:

\[
l = 2h \quad \Rightarrow \quad h = \frac{l}{2}
\]

### Step 3: Express Volume in Terms of \(l\)

Now, substituting \(r = \sqrt{3}h\) and \(h = \frac{l}{2}\) into the volume formula:

\[
V = \frac{1}{3} \pi (\sqrt{3}h)^2 h = \frac{1}{3} \pi (3h^2) h = \pi h^3
\]

Since \(h = \frac{l}{2}\), we get:

\[
V = \pi \left(\frac{l}{2}\right)^3 = \frac{\pi l^3}{8}
\]

### Step 4: Differentiate Volume with Respect to Time

We now differentiate \(V = \frac{\pi l^3}{8}\) with respect to time \(t\):

\[
\frac{dV}{dt} = \frac{\pi}{8} \cdot 3l^2 \frac{dl}{dt}
\]

Substitute \(\frac{dV}{dt} = -2 \, \text{cm}^3/\text{s}\):

\[
-2 = \frac{\pi}{8} \cdot 3l^2 \frac{dl}{dt}
\]

### Step 5: Solve for \(\frac{dl}{dt}\)

When \(l = 4 \, \text{cm}\), substitute this value into the equation:

\[
-2 = \frac{\pi}{8} \cdot 3(4)^2 \frac{dl}{dt}
\]

\[
-2 = \frac{\pi}{8} \cdot 3(16) \frac{dl}{dt}
\]

\[
-2 = \frac{\pi}{8} \cdot 48 \frac{dl}{dt}
\]

\[
-2 = 6\pi \frac{dl}{dt}
\]

Solve for \(\frac{dl}{dt}\):

\[
\frac{dl}{dt} = \frac{-2}{6\pi} = \frac{-1}{3\pi} \, \text{cm/s}
\]

### Final Answer:

The rate of decrease of the slant height when the slant height is 4 cm is:

\[
\frac{dl}{dt} = \frac{-1}{3\pi} \, \text{cm/s}
\]

This means the slant height decreases at a rate of approximately \(-0.106 \, \text{cm/s}\).

---

Would you like more details on any part of this solution? Here are five related questions you might find interesting:

1. How would the rate of change of slant height differ if the rate of water flow was faster?
2. What if the vertical angle of the cone was 90 degrees instead of 120 degrees?
3. How can you derive the formula for the volume of a cone?
4. What are some real-world applications of problems involving rates of change in conical objects?
5. How can the surface area of the cone be used in problems like this?

**Tip:** Always double-check units when working through calculus problems that involve rates of change, as dimensional consistency is key to correct solutions.

Water is tripping out from a conical funnel at the uniform rate of 2 cm cube per second for tiny hole at the vertex at the bottle when the slant height of the water is 4 cm find the rate of decrease if the slant height of the water given that the vertical angle of the funnel is 120 degree

Explore a related rates problem where water drips out of a conical funnel at a uniform rate of 2 cm³ per second. Learn how to calculate the rate of decrease in slant height when the slant height is 4 cm. This problem involves conic geometry, differentiation, and the Pythagorean Theorem.

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