Question

1. Water is flowing out of a conical funnel through its apex at a rate of 12 cubic inches per minute. If the tunnel is initially full, how long will it take for it to be one-third full? What is the height of the water level? Assume the radius to be 6 inches and the altitude of the cone to be 15 inches.

2. Two similar cones have volumes 81 pi over 2 inches cube and the slant height of the bigger cone is 7.5 inches. Find the integer solution to the height of the smaller cone.

3. If the slant height of the cone is 16 inches and the total area is 120 pi square inches, find the height of the cone.

4. What is the height if a right circular cone having a slant height of 6 square root of 10 units and a base radius of 6 units?

5. Find the ratio of the slant height to the radius of a right circular cone in which the volume and lateral area are numerically equal. Assume the altitude of the cone to be 9 units.

Answer 1

Answer on Question #58137 – Math – Geometry

1. Water is flowing out of a conical funnel through its apex at a rate of 12 cubic inches per minute. If the funnel is initially full, how long will it take for it to be one-third full? What is the height of the water level? Assume the radius to be 6 inches and the altitude of the cone to be 15 inches.

Solution:

The volume $V$ of the cone is given by $V = \frac{1}{3} Ah$, where $A = \pi r^2$ is the area of the base, $h$ is the altitude of the cone and $r$ is the base radius. Thus $V = \frac{\pi}{3} r^2 h \cong 565.487$ cubic inches.

If water is flowing out at the rate of $v$ we obtain:

$$V(t) = V - vt$$

where $t$ is the time of flowing out. Thus in $t$ minutes there remains $V_1$ cubic inches of water in the funnel.

$$V_1 = \frac{1}{3}V = V - vt$$

Then $t = \frac{2}{3v} V = \frac{2}{3 \cdot 12}565.487 \cong 31.416$ minutes or $t \cong 31$ minutes 25 seconds

If the funnel is one-third full we have new values of the volume, base radius $r_1$ and altitude $h_1$

$$V \to V_1 = \frac{1}{3}V, \quad r \to r_1, \quad h \to h_1$$

The linear size of the funnel becomes $\alpha$ times smaller: $r = \alpha r_1, h = \alpha h_1$.

So

$$V_1 = \frac{\pi}{3} r_1^2 h_1 = \frac{\pi}{3} \frac{r^2}{\alpha^2} \frac{h}{\alpha} = \frac{V}{\alpha^3} = \frac{V}{3}$$

Hence $\alpha^3 = 3$ and the height of the water level is $h_1 = \frac{h}{\sqrt[3]{3}} \cong \frac{15}{1.442} \cong 10.402$ inches.

Answer:

It will take about 31 minutes and 26 seconds for funnel to be one-third full. And the height of the water level will be 10.402 inches

2. Two similar cones have volumes 81 pi over 2 inches cube and the slant height of the bigger cone is 7.5 inches. Find the integer solution to the height of the smaller cone.

**Solution:**

We have $V_{b} + V_{s} = \frac{81\pi}{2}, V_{s} < V_{b}$

Hence $V_{b} > \frac{81\pi}{4}$

where $V_{b}$ is the volume of the bigger cone and $V_{s}$ is the volume of smaller one:

$$V _ {b} = \frac {1}{3} \pi R ^ {2} H \text{ and } V _ {s} = \frac {1}{3} \pi r ^ {2} n$$

$R$ is the radius of the bigger cone, $H$ is its height

$r$ is the radius of the smaller cone, $n$ is integer number, its height

If $L$ is the slant height of the bigger cone, then $R^2 + H^2 = L^2$ and we can write down

$$V _ {b} = \frac {1}{3} \pi (L ^ {2} - H ^ {2}) H > \frac {8 1 \pi}{4}$$ $$\text{or } (7. 5 ^ {2} - H ^ {2}) H > \frac {2 4 3}{4} \Rightarrow (5 6. 2 5 - H ^ {2}) H - 6 0. 7 5 > 0$$

Then $1.105 < H < 6.887$

As $n < H$ then $n \leq 6$.

Maximum of the volume of the bigger cone we can find at $H$ where $\frac{\partial V_b}{\partial H} = 0 \Rightarrow L^2 - 3H^2 = 0$

Then $H = \sqrt{\frac{L^2}{3}} = \sqrt{18.75} \cong 4.33$

and $R = \sqrt{L^2 - H^2} \cong 6.124$, $\frac{R}{H} = 1.414$

Maximum of the volume of the bigger cone is $V_{b} = \frac{1}{3}\pi (3H^{2} - H^{2})H = \frac{2}{3}\pi H^{3} \cong 54.12\pi >\frac{81\pi}{2}$

Thus minimum of the volume of the smaller cone must be $V_{s} > 0$

So $n > 0$

Thus $n = 1,2,3,4,5,6$

**Answer:** The integer solution to the height of the smaller cone can be any integer number from 1 to 6: $n = 1,2,3,4,5,6$.

3. If the slant height of the cone is 16 inches and the total area is 120 pi square inches, find the height of the cone.

**Solution:**

As $A = \pi R^2 + \pi RL$

and $R^2 + H^2 = L^2$

we can write down

$$\begin{array}{l} A = \pi (L^2 - H^2 + \sqrt{L^2 - H^2} L) \Rightarrow \left(\frac{A}{\pi} - L^2 + H^2\right)^2 = L^2 (L^2 - H^2) \\ \Rightarrow (120 - 256 + H^2)^2 = 256 (256 - H^2) \Rightarrow (H^2 - 136)^2 = 65536 - 256 H^2) \\ \Rightarrow H^4 - 16 H^2 - 47040 = 0 \Rightarrow H^2 \cong 225 \Rightarrow H \cong 15 \\ \end{array}$$

**Answer:** The height of the cone is about 15 inches

4. What is the height if a right circular cone having a slant height of 6 square root of 10 units and a base radius of 6 units?

**Solution:**

As $R^2 + H^2 = L^2$

we obtain $H = \sqrt{L^2 - R^2}$

Thus $H = \sqrt{\left(6\sqrt{10}\right)^2 - 6^2} = \sqrt{360 - 36} = 18$ units.

**Answer:** The height of the right circular cone is 18 units.

5. Find the ratio of the slant height to the radius of a right circular cone in which the volume and lateral area are numerically equal. Assume the altitude of the cone to be 9 units.

**Solution:**

As $V = \frac{1}{3} \pi R^2 H$, $A_{lat} = \pi RL$ and $V = A_{lat}$

we obtain $\pi R^2 H = 3 \pi RL$

Thus $\frac{L}{R} = \frac{H}{3} = 3$ units

**Answer:** The ratio of the slant height to the radius is 3 units.