Question

Oil from a conical funnel is dripping at the rate of \(5\ \text{cm}^3/\text{s}\). If the radius and height of the funnel are 10 cm and 20 cm respectively, then the rate at which the oil level drops when it is 5 cm from the top is

• A. \( \frac{8}{45\pi} \) cm/s
• B. \( -\frac{8}{45\pi} \) cm/s
• C. \( -\frac{4}{45\pi} \) cm/s
• D. \( \frac{4}{45\pi} \) cm/s

Hint:

In related rates problems involving cones, first use similarity of triangles to express \(r\) in terms of \(h\), then differentiate volume with respect to time.

Answer 1

The Correct Option is D

Step 1: Write the given data.
Radius of cone:
\[ R=10\ \text{cm} \]
Height of cone:
\[ H=20\ \text{cm} \]
Oil is dripping at the rate:
\[ \frac{dV}{dt}=-5\ \text{cm}^3/\text{s} \]

Step 2: Find height of oil level from vertex.
The oil level is 5 cm from the top. Since total height is 20 cm, height of oil from the vertex is:
\[ h=20-5=15\ \text{cm} \]

Step 3: Use similarity of triangles.
For the cone:
\[ \frac{r}{h}=\frac{R}{H}=\frac{10}{20}=\frac{1}{2} \]
So,
\[ r=\frac{h}{2} \]

Step 4: Write volume of oil.
Volume of cone is:
\[ V=\frac{1}{3}\pi r^2h \]
Substitute \(r=\frac{h}{2}\):
\[ V=\frac{1}{3}\pi\left(\frac{h}{2}\right)^2h \]
\[ V=\frac{\pi h^3}{12} \]

Step 5: Differentiate with respect to time.
\[ \frac{dV}{dt}=\frac{\pi}{12}\cdot 3h^2\frac{dh}{dt} \]
\[ \frac{dV}{dt}=\frac{\pi h^2}{4}\frac{dh}{dt} \]

Step 6: Substitute values.
At \(h=15\),
\[ -5=\frac{\pi(15)^2}{4}\frac{dh}{dt} \]
\[ -5=\frac{225\pi}{4}\frac{dh}{dt} \]

Step 7: Solve for \( \frac{dh}{dt} \).
\[ \frac{dh}{dt}=\frac{-5\cdot 4}{225\pi} \]
\[ \frac{dh}{dt}=-\frac{20}{225\pi} \]
\[ \frac{dh}{dt}=-\frac{4}{45\pi} \]
Since the question asks the rate at which the oil level drops, we take the positive magnitude:
\[ \frac{4}{45\pi}\ \text{cm/s} \]
Final Answer:
\[ \boxed{\frac{4}{45\pi}\ \text{cm/s}} \]