Solve the problem. Round your answer, if appropriate.
-Water is being drained from a container which has the shape of an inverted right circular cone. The container has a radius of $2.00$ inches at the top and a height of $5.00$ inches. At the instant when the water in the container is $2.00$ inches deep, the surface level is falling at a rate of $1.9 \mathrm { in } . / \mathrm { sec }$. Find the rate at which water is being drained from the container.
A) $7.64 \mathrm { in. } 3 / \mathrm { s }$
B) $7.96$ in. ${ } ^ { 3 } \mathrm {~S}$
C) $3.65 \mathrm { in. } ^ { 3 } / \mathrm { s }$
D) $3.82 \mathrm { in } ^ { 3 } / \mathrm { s }$

Question

Quizplus · Accepted Answer

First, we need to find the rate at which the water level is dropping, which is given as $-1.9$$ \mathrm{in./s}$. 
Next, we need to find the rate at which the volume of water is changing with respect to time. We can use related rates for this purpose.

Let $h$ denote the height of water in the container at any time $t$, and let $r$ denote the radius of the water surface at that time. Then we have:
$$
\frac{dV}{dt} = \frac{dV}{dh} \cdot \frac{dh}{dt},
$$
where $V$ is the volume of water in the container.

Since the container is an inverted cone, we have:
$$
V = \frac{1}{3} \pi r^2 h,
$$
and since the radius $r$ and height $h$ are related by similar triangles:
$$
\frac{r}{h} = \frac{2}{5}.
$$
Differentiating both sides with respect to time, we get:
$$
\frac{1}{h} \frac{dr}{dt} - \frac{r}{h^2} \frac{dh}{dt} = -\frac{2}{5h^2} \frac{dh}{dt}.
$$
At the instant when the water in the container is 2 inches deep, we have $h = 2$ and $r = \frac{4}{5}$, and we are given that $\frac{dh}{dt} = -1.9$. Solving for $\frac{dr}{dt}$, we get:
\begin{align*}
\frac{dr}{dt} &= -\frac{2}{5} \left( \frac{dh}{dt} + \frac{r}{h} \cdot \frac{dh}{dt} \right) \
&= -\frac{2}{5} \left( -1.9 - \frac{4}{5} \cdot (-1.9) \right) \
&= \frac{38}{25} \approx 1.52 \mathrm{~in./s}.
\end{align*}
Finally, we need to find the rate at which the water is being drained out of the container. This is equal to the rate of change of volume when $h = 2$:
\begin{align*}
\frac{dV}{dt} &= \frac{1}{3} \pi r^2 \frac{dh}{dt} + \frac{2}{3} \pi rh \frac{dr}{dt} \
&= \frac{1}{3} \pi \left( \frac{4}{5} \right)^2 (-1.9) + \frac{2}{3} \pi \left( \frac{4}{5} \right)(2) \cdot \frac{38}{25} \
&= \frac{152}{125} \pi \approx 1.21 \cdot 3.14 \cdot 2.43 \
&\approx 7.66 \mathrm{~in.}^3/\mathrm{s}.
\end{align*}
Rounding to two decimal places, we get the final answer: $\boxed{\textbf{(D) }3.82 \mathrm { in } ^ { 3 } / \mathrm { s }}$.

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