Jenna97 said:
Desperate times call for desperate measures.
I hope someone can show me how to do this.
I don't want to offend anyone, but the truth is i have no work to show.
I have exam on monday and i know a task like this will be given, exactly the same just different numbers.
I have no vision on studying math but i need to pass this class to study biology. Again hope to not offend anyone and that someone can show me how to do this.
A cone-shaped tank (with the tip facing upwards) with a radius of 1 meter
and height 3 m is emptied of water. Show that when the water height is h meters, where
0 ≤ h ≤ 3, the volume of water in the tank is given by
v(h)=pi(h-(h^2/3)+(h^3/27)
When the water level in the tank is 2 meters, the tank is emptied with a speed of 1/2
cubic meters per minute. How fast is the water depth declining at this point?
I'm sure you know that the volume of a right cone is $\displaystyle \begin{align*} V = \frac{1}{3}\,\pi\,R^2\,H \end{align*}$. So that means, when the tank is full, the volume is
$\displaystyle \begin{align*} V_{\textrm{Full}} &= \frac{1}{3}\,\pi\,\left( 1 \right) ^2 \left( 3 \right) \\ &= \pi \end{align*}$
As the tank is emptied, the empty part of the tank also takes the shape of a cone. When the height of the water is "h" metres, the the height of the "empty cone" is $\displaystyle \begin{align*} 3 - h \end{align*}$. The "empty cone" is similar to the conical tank, where the radius is 1/3 of the height, and thus $\displaystyle \begin{align*} r = \frac{1}{3}\,\left( 3 - h \right) \end{align*}$. Thus the "empty volume" is
$\displaystyle \begin{align*} V_{\textrm{Empty}} &= \frac{1}{3}\,\pi\,\left[ \frac{1}{3}\,\left( 3 - h \right) \right] ^2 \, \left( 3 - h \right) \\ &= \frac{1}{3}\,\pi\,\frac{1}{9}\,\left( 3 - h \right) ^2 \,\left( 3 - h \right) \\ &= \frac{1}{27}\,\pi\,\left( 27 - 27\,h + 9\,h^2 - h^3 \right) \\ &= \pi - \pi\,h + \frac{1}{3}\,\pi\,h^2 - \frac{1}{27}\,\pi\,h^3 \end{align*}$
and therefore the volume of the water is what's left when the "empty volume" is subtracted from the volume of the tank...
$\displaystyle \begin{align*} V_{\textrm{Water}} &= V_{\textrm{Full}} - V_{\textrm{Empty}} \\ &= \pi - \left( \pi - \pi\,h + \frac{1}{3}\,\pi\,h^2 - \frac{1}{27}\,\pi\,h^3 \right) \\ &= \pi\,\left( h - \frac{1}{3}\,h^2 + \frac{1}{27}\,h^3 \right) \end{align*}$Now as you are told that when the height is 2m, the tank empties with a speed of $\displaystyle \begin{align*} \frac{1}{2}\,\textrm{m}^3 / \textrm{min} \end{align*}$, so $\displaystyle \begin{align*} \frac{\mathrm{d}V}{\mathrm{d}t} = \frac{1}{2} \end{align*}$. You are asked to find how fast the water depth is declining at this point, so to find $\displaystyle \begin{align*} \frac{\mathrm{d}h}{\mathrm{d}t} \end{align*}$ when $\displaystyle \begin{align*} h = 2 \end{align*}$.
Here you will need to use the Chain Rule:
$\displaystyle \begin{align*} \frac{\mathrm{d}V}{\mathrm{d}h}\cdot \frac{\mathrm{d}h}{\mathrm{d}t} &= \frac{\mathrm{d}V}{\mathrm{d}t} \\ \frac{\mathrm{d}h}{\mathrm{d}t} &= \frac{\frac{\mathrm{d}V}{\mathrm{d}t}}{\frac{\mathrm{d}V}{\mathrm{d}h}} \end{align*}$
Now as $\displaystyle \begin{align*} V = \pi\,\left( h - \frac{1}{3}\,h^2 + \frac{1}{27}\,h^3 \right) \end{align*}$ that means $\displaystyle \begin{align*} \frac{\mathrm{d}V}{\mathrm{d}h} = \pi \,\left( 1 - \frac{2}{3}\,h + \frac{1}{9}\,h^2 \right) \end{align*}$ and when $\displaystyle \begin{align*} h = 2 \end{align*}$ we have
$\displaystyle \begin{align*} \frac{\mathrm{d}V}{\mathrm{d}h} &= \pi\,\left( 1 - \frac{2}{3} \cdot 2 + \frac{1}{9}\cdot 2^2 \right) \\ &= \pi\,\left( 1 - \frac{4}{3} + \frac{4}{9} \right) \\ &= \pi\,\left( \frac{9}{9} - \frac{12}{9} + \frac{4}{9} \right) \\ &= \frac{\pi}{9} \end{align*}$
and thus when $\displaystyle \begin{align*} h = 2 \end{align*}$
$\displaystyle \begin{align*} \frac{\mathrm{d}h}{\mathrm{d}t} &= \frac{\frac{1}{2}}{\frac{\pi}{9}} \\ &= \frac{9}{2\,\pi} \end{align*}$
So finally at the point where $\displaystyle \begin{align*} h = 2\,\textrm{m} \end{align*}$ the depth is declining at a rate of $\displaystyle \begin{align*} \frac{9}{2\,\pi}\,\textrm{m} / \textrm{min} \end{align*}$.
