# Thread: related rates and upside down cones!

1. ## related rates and upside down cones!

So, my textbook only covers implicit diff. as it relates to ellipses and circles. Needless to say, it all went over my head. Now I have a related rates question where I absolutely need to use implicit diff. Here goes:

A water tank has the shape of an inverted cone with a base radius of 4 meteres and a height of 6 meteres. If water is being pumped into the tank at a rate of 3 m^3/min, find the rate at which the water level is rising when the tank is 4 meters deep. Specify the units of your answer.

So I have two triangles, one with measurements given, and one with only height (because radius at 4m deep is currently unknown)

I also have the formula V = 1/3 pi*r^2*h

Now, to set up a proportion to find r?

r 4
-- = --
h 6

6r = 4h, in terms of r

r = 4h/6, substitute the known h

r = 8/3m

If I did everything right, I now know the radius of the cone when height = 4m. However, I'm sure I need to implicitly differentiate the volume formula and I'm not sure how.

2. you need to get a volume formula in terms of a single variable.

$\displaystyle \frac{r}{h} = \frac{2}{3}$

$\displaystyle r = \frac{2h}{3}$

$\displaystyle V = \frac{\pi}{3}\left(\frac{2h}{3}\right)^2 h$

$\displaystyle V = \frac{4\pi}{27} h^3$

take the time derivative of the above volume function ... sub in your given values for $\displaystyle \frac{dV}{dt}$ and $\displaystyle h$ ... calculate the value of $\displaystyle \frac{dh}{dt}$.

3. Originally Posted by angrynapkin
So, my textbook only covers implicit diff. as it relates to ellipses and circles. Needless to say, it all went over my head. Now I have a related rates question where I absolutely need to use implicit diff. Here goes:

A water tank has the shape of an inverted cone with a base radius of 4 meteres and a height of 6 meteres. If water is being pumped into the tank at a rate of 3 m^3/min, find the rate at which the water level is rising when the tank is 4 meters deep. Specify the units of your answer.

So I have two triangles, one with measurements given, and one with only height (because radius at 4m deep is currently unknown)

I also have the formula V = 1/3 pi*r^2*h

Now, to set up a proportion to find r?

r 4
-- = --
h 6

6r = 4h, in terms of r

r = 4h/6, substitute the known h

r = 8/3m

If I did everything right, I now know the radius of the cone when height = 4m. However, I'm sure I need to implicitly differentiate the volume formula and I'm not sure how.
In the problem, we are told that $\displaystyle \frac{\,dV}{\,dt}=3~\frac{m^3}{min}$. You are correct in saying that a proportion is need to get radius in terms of height: it should be $\displaystyle \frac{r}{h}=\frac{4}{6}\implies r=\frac{2}{3}h$

Since the volume of a cone is $\displaystyle V=\tfrac{1}{3}\pi r^2 h$, we now see that $\displaystyle V=\tfrac{1}{3}\pi\left(\tfrac{2}{3}h\right)^2h\imp lies V=\tfrac{4}{27}\pi h^3$

Now [implicitly] differentiate both sides with respect to t:

$\displaystyle V=\tfrac{4}{27}\pi h^3\implies\frac{\,dV}{\,dt}=\tfrac{4}{9}\pi h^2\frac{\,dh}{\,dt}$

We can now plug in all the known values and solve for $\displaystyle \frac{\,dh}{\,dt}$.

Can you try to take it from here?

--Chris

4. I believe so.

Plug in 3m^3/min and 4

Right side evaluates to 64pi/9

Move 64/9 to left side

27/64

move pi to left side

27/64pi m^3/min.