Relative rates, cone filling with water

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February 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
February 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: . $V \:=\:\frac{\pi}{3}r^2h$ .[1]

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

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

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

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

Therefore: . $3 \:=\:\frac{9\pi}{16}(3^2)\frac{dh}{dt} \quad\Rightarrow\quad \frac{dh}{dt} \:=\:\frac{16}{27\pi}\text{ m/min}$
February 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
February 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.
February 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.

$\displaystyle \begin{align*} \pi \end{align*}$ is a constant, so its derivative is 0. HOWEVER, in this case you have $\displaystyle \begin{align*} \pi \end{align*}$ times a FUNCTION and the derivative of a function times a constant is the same as the constant times the function. I.e. $\displaystyle \begin{align*} \frac{d}{dx} \left[ k\, f(x) \right] = k \, \frac{d}{dx} \left[ f(x) \right] \end{align*}$ .
February 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
Radius decreases 2cm/min
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?
February 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