word problem :(

[rolav]

Okay, I have no idea how to do this problem. Please help!

A container, in the shape of an inverted right circular cone, has a radius of 8 inches at the top and a height of 10
inches. At the instant when the water in the container is 8 inches deep, the surface level is falling at the rate of
-1.3 in./s. Find the rate at which water is being drained

Thanks in advance!

 

[soroban]

Hello, rolav!

A container, in the shape of an inverted right circular cone,
has a radius of 8 inches at the top and a height of 10 inches.
At the instant when the water in the container is 8 inches deep,
the surface level is falling at the rate of -1.3 in/sec.
Find the rate at which water is being drained.

Here is the side view of the conical container.

      : - 8 - : - 8 - :
    - *-------+-------*
    :  \      |      /
    :   \     |  r  /
    :    \----+----/
   10     \:::|:::/
    :      \::|h:/
    :       \:|:/
    :        \|/
    -         *

The water in the container is also a cone
. . with radius $\displaystyle r$ and height $\displaystyle h.$

The volume of water is: .$\displaystyle V \:=\:\frac{\pi}{3}r^2h$ .[1]

From similar right triangles: .$\displaystyle \frac{r}{h} \,=\,\frac{8}{10} \quad\Rightarrow\quad r \:=\:\frac{4}{5}h$

Substitute into [1]: .$\displaystyle V \:=\:\frac{\pi}{3}\left(\frac{4}{5}h\right)^2h \quad\Rightarrow\quad V \:=\:\frac{16\pi}{75}h^3$

Differentiate with respect to time: .$\displaystyle \dfrac{dV}{dt} \:=\:\dfrac{16\pi}{25}h^2\!\cdot\!\dfrac{dh}{dt}$

We are told: .$\displaystyle h = 8,\:\frac{dh}{dt} = \text{-}1.3$

Therefore: .$\displaystyle \dfrac{dV}{dt} \:=\:\dfrac{16\pi}{25}(8^2)(\text{-}1.3) \:=\:-53.248\pi \text{ in}^3\text{/sec}$

 

[rolav]

Hello, rolav!


Here is the side view of the conical container.

      : - 8 - : - 8 - :
    - *-------+-------*
    :  \      |      /
    :   \     |  r  /
    :    \----+----/
   10     \:::|:::/
    :      \::|h:/
    :       \:|:/
    :        \|/
    -         *

The water in the container is also a cone
. . with radius $\displaystyle r$ and height $\displaystyle h.$

The volume of water is: .$\displaystyle V \:=\:\frac{\pi}{3}r^2h$ .[1]

From similar right triangles: .$\displaystyle \frac{r}{h} \,=\,\frac{8}{10} \quad\Rightarrow\quad r \:=\:\frac{4}{5}h$

Substitute into [1]: .$\displaystyle V \:=\:\frac{\pi}{3}\left(\frac{4}{5}h\right)^2h \quad\Rightarrow\quad V \:=\:\frac{16\pi}{75}h^3$

Differentiate with respect to time: .$\displaystyle \dfrac{dV}{dt} \:=\:\dfrac{16\pi}{25}h^2\!\cdot\!\dfrac{dh}{dt}$

We are told: .$\displaystyle h = 8,\:\frac{dh}{dt} = \text{-}1.3$

Therefore: .$\displaystyle \dfrac{dV}{dt} \:=\:\dfrac{16\pi}{25}(8^2)(\text{-}1.3) \:=\:-53.248\pi \text{ in}^3\text{/sec}$

Thank you, good sir! But one thing, wouldn't it be v= 1/3 pi r^2 h instead of $\displaystyle V \:=\:\frac{\pi}{3}r^2h$. Or am I just silly and that means the same thing?

 

[mmm4444bot]

wouldn't it be v= 1/3 pi r^2 h instead of $\displaystyle V \:=\:$$\displaystyle \dfrac{\pi}{3}\;$$\displaystyle r^2\;h$. Or am I just silly

Egads! Something is silly; I don't know whether it's you, on-line courses, secondary schools, or something else, but a calculus student ought to be solid on this maneuver:


$\displaystyle \frac{1}{n} \cdot c \;=\; \frac{1}{n} \cdot \frac{c}{1} \;=\; \frac{c}{n}$


In other words, both of the red expressions in your quote above are equal :!:

Hope you learned something, from the earlier camera-ready copy. :?