Cone shaped tank, derivative

[Jenna97]

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?

 

[HallsofIvy]

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.

It's a little late to be studying isn't it?

A cone-shaped tank (with the tip facing upwards) with a radius of 1 meter

You mean, I presume, that the base is a circle of radius 1.

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).

Do you know the formula for the volume of a cone? It is \(\displaystyle \pi r^2h/3\). The entire tank has r= 1 and h= 3 so has volume \(\displaystyle \pi (1)(3)/3= \pi\) cubic meters. Since the tip is upward, the volume of the water is the volume of the entire cone minus the volume of the cone above the water. When the water height is h m, the height of the cone above the water is 3- h m.

But what is the radius of the "base" (the circular surface of the water)? To get that imagine looking at the cone from the side. From the side we see an isosceles triangle with base 2 (the diameter of the circle) and height 3. The small cone above the water has all the same angles as the entire tank so forms a triangle similar to that first one. Similar triangles have sides in the same proportion. Taking "height over base" for each triangle and calling the radius of the smaller cone r and height y, we have \(\displaystyle \frac{2r}{y}= \frac{2}{3}\) so \(\displaystyle r= \frac{y}{3}[/tex]. The smaller cone, the space in the tank above the water, has volume $$\pi r^2h/3= \pi y^2h/27= \pi(3- h)^2h/27$$.
The actual volume of the water is the entire volume of the tank minus the volume of the region above the water:
$$\pi- \pi(3- h)^2h/27= \pi(1- (3- h)^2/27= \pi\frac{27-(9- 6h+ h^2)h}{27}= \pi\frac{18h+ 6h^2- h^3}{27}$$

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?

You now know that $$V= \pi\frac{18h+ 6h^2- h^3}{27}$$, you know that $$\frac{dV}{dt}= \frac{dV}{dh}\frac{dh}{dt}= \pi\frac{18+ 12h- 3h^2}{27}\frac{dV}{dt}$$.

Set $$\frac{dV}{dt}= -\frac{1}{2}$$, h= 2, and solve for $$\frac{dh}{dt}$$.\)

 

[Prove It]

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 $\displaystyle \begin{align*} \frac{9}{2\,\pi}\,\textrm{m} / \textrm{min} \end{align*}$.