Question

1. water is being drained from a container which has the shape of an inverted right circular cone. the container has a radius of 7 inches at the top and a height of 9 inches. at the instant when the water in the container is 8 inches deep, the surface level is falling at a rate of 1.4 $\frac{in}{s}$. find the rate at which the water is being drained from the container at the same instant. exact answers only (i.e.) make sure your answer has $pi$ in it! hint: draw a picture to help you get started, and remember similar triangles.

Explanation:

Step1: Establish ratio of radius to height

For the large cone, the ratio of radius to height is $\frac{r}{h}=\frac{7}{9}$. For the cone - shaped water - volume at any time, the ratio of the radius of the water - surface $r$ to the depth of water $h$ is the same due to similar triangles. So $r = \frac{7}{9}h$.

Step2: Write volume formula for water

The volume $V$ of a cone is $V=\frac{1}{3}\pi r^{2}h$. Substitute $r=\frac{7}{9}h$ into the volume formula: $V=\frac{1}{3}\pi(\frac{7}{9}h)^{2}h=\frac{49}{243}\pi h^{3}$.

Step3: Differentiate volume with respect to time

Differentiate $V$ with respect to time $t$ using the chain - rule. $\frac{dV}{dt}=\frac{49}{81}\pi h^{2}\frac{dh}{dt}$.

Step4: Substitute given values

We are given that $h = 8$ inches and $\frac{dh}{dt}=- 1.4$ in/s (negative because the height is decreasing). Substitute these values into the $\frac{dV}{dt}$ formula: $\frac{dV}{dt}=\frac{49}{81}\pi(8)^{2}\times(-1.4)$.
Calculate $\frac{dV}{dt}=\frac{49}{81}\pi\times64\times(-1.4)=-\frac{49\times64\times1.4}{81}\pi=-\frac{49\times64\times14}{810}\pi=-\frac{49\times64\times7}{405}\pi=-\frac{21952}{405}\pi$ in³/s.

Answer:

$-\frac{21952}{405}\pi$ in³/s