# Relative rates, cone filling with water

• Feb 8th 2013, 09:13 AM
togo
Relative rates, cone filling with water
I think there is a problem in the d/dt part
Problem:
A cone 4 m high, base diameter of 6 m, water is entering at 3 m^3 per minute
when water is at a height of 3 m, what is the rate of rise?

1/3pi(r^2)h = V
1/3pi(3^2)(3) = 28.274 m^3 at level of 3 m
dh/dt = ?
dv/dt= 2m^3/min
f'ghi + fg'hi + fgh'i + fghi' = dv/dt
0 + 0 + 1/3pi(2r)(dh/dt) + 1/3pi(r^2)(1)
working this out gets me a rate of
1.18 m^3/min

but I feel there is a problem in the fghi part
• Feb 8th 2013, 09:58 AM
Soroban
Re: Relative rates, cone filling with water
Hello, togo!

Quote:

A cone 4 m high, base diameter of 6 m.
Water is entering at 3 m^3 per minute
When water is at a height of 3 m, what is the rate of rise?

I assume it is an inverted cone.
Code:

                  3     - *-------+-------*     :  \      |      /     :  \-----+-----/     :    \    |  r /     4    \  |  /     :      \  |h /     :      \ | /     :        \|/     -        *
The volume of a cone is: .$\displaystyle V \:=\:\frac{\pi}{3}r^2h$ .[1]

From similar triangles we have: .$\displaystyle \frac{r}{h} = \frac{3}{4} \quad\Rightarrow\quad r = \frac{3}{4}h$

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

Differentiate with respect to time: .$\displaystyle \frac{dV}{dt} \:=\:\frac{9\pi}{16}h^2\,\frac{dh}{dt}$

We are given: .$\displaystyle \frac{dV}{dt} = 3,\;h = 3.$

Therefore: .$\displaystyle 3 \:=\:\frac{9\pi}{16}(3^2)\frac{dh}{dt} \quad\Rightarrow\quad \frac{dh}{dt} \:=\:\frac{16}{27\pi}\text{ m/min}$
• Feb 8th 2013, 10:04 AM
togo
Re: Relative rates, cone filling with water
thanks, however I was really hoping someone could point out where my method went wrong, because I am most comfortable with that one
• Feb 8th 2013, 10:32 AM
togo
Re: Relative rates, cone filling with water
I'm confused with pi. Google says the derivative of pi is zero. However in some of the answers in this book, the derivative of pi ends up being pi.
• Feb 8th 2013, 02:18 PM
Prove It
Re: Relative rates, cone filling with water
Quote:

Originally Posted by togo
I'm confused with pi. Google says the derivative of pi is zero. However in some of the answers in this book, the derivative of pi ends up being pi.

\displaystyle \displaystyle \begin{align*} \pi \end{align*} is a constant, so its derivative is 0. HOWEVER, in this case you have \displaystyle \displaystyle \begin{align*} \pi \end{align*} times a FUNCTION and the derivative of a function times a constant is the same as the constant times the function. I.e. \displaystyle \displaystyle \begin{align*} \frac{d}{dx} \left[ k\, f(x) \right] = k \, \frac{d}{dx} \left[ f(x) \right] \end{align*}.
• Feb 9th 2013, 10:32 AM
togo
Re: Relative rates, cone filling with water
however can someone explain how to find derivative of this:
V = pi(r^2)h

f = pi
g = r^2
h = h

f' = pi
g' = 2r
h' = 1

so f'gh + fg'h + fgh' =
pi(r^2)h + pi(2r)h + pi(r^2)(1)

so is this correct?? thanks!!

also edit: say related rates question includes:
Volume remains constant
How fast does surface area change when radius 10 cm and height 8 cm

so then:
pi(r^2)h + pi(2r)h + pi(r^2)(1) =
pi(r^2)h + pi(2r)dr/dt(h) + pi(r^2)dh/dt = V

correct?
• Feb 9th 2013, 11:52 AM
HallsofIvy
Re: Relative rates, cone filling with water
Quote:

Originally Posted by togo
thanks, however I was really hoping someone could point out where my method went wrong, because I am most comfortable with that one

You went wrong by treating r as if it were a constant. As soroban told you that is not true- as h increases, r must also increase