related rates- trash compactor and depth of water

[math]

The sides walls of a trash compactor are moving apart at rate of 0.3 m/s. The volume of rancid water at bottom of compactor is constant at 20 m^3. When the side walls are 5 m long and the end walls are 2 m long, at what rate is the depth of the rancid water changing?

My attempt:

They give me that da/dt = 0.1, db/dt = -0.3, and V = 20. They ask me to find dh/dt when a = 5 and b = 2.

The volume is V = ab^2, so:

. . .20 = ab^2

. . .0 = 2ab (db/dt) + b^2 (da/dt)

. . .0 = b [2a (db/dt) + b (da/dt)]

. . .2a (db/dt) = -b (da/dt)

I don't think I'm doing this right. I'm especially confused about the depth of water changing part.

 

[wjm11]

trash compactor. the sides walls are moving apart at rate of .3 m/s. volume of rancid water at bottom of compactor is constant at 20 m^3 . when side walls are 5 m long and end walls are 2 m long, at what rate is depth of water changing?

my attempt:
da/dt = .1 db/dt= -.3 V=20 Find dh/dt a=5 b=2

Your problem statement is confusing. Since you said that the side walls are moving and that they are 5 feet long, I will assume that a set of five foot wide walls are moving away from each other and that the gap between them is two feet at the moment requested. That gap is changing with time.

Let h be the height of the water and dh/dt be the rate of change of that height.

You have defined “a=5” so that is the width of the moving walls and is a constant. da/dt is therefore equal to zero.

For the opening gap, you’ve set “b=2.” The value of b is changing with time and db/dt = .3 m/s.

The volume of water is 20m^3, and V = abh = 5*2*h, or 20m^3 = (10m^2)h, and h = 2m. In general, therefore,

h = V/ab = 20m^3/(5m)(b) = (4m^2)(b^(-1))
dh/dt = -4m^2(b^(-2))db/dt = [(-4m^2)/((2m)^2)] * (.3m/s) = -.3m/s

 

[skeeter]

The water in the compactor has a constant volume 20 m³ and maintains the shape of a rectangular prism. The side walls change in length, and I assume that the end wall length remains constant.

V = (length of end walls)(length of side walls)(height)

20 = 2sh

10 = sh

d/dt(10 = sh)

0 = s(dh/dt) + h(ds/dt)

dh/dt = -h(ds/dt)/s

when the sides walls have a length of 5 m, the height is 2 m ... you were also given the rate of change of the side walls, moving apart (increasing) at a rate of 0.3 m/s

dh/dt = -2(0.3)/5 = -0.12 m/s ... the height of water in the compactor is decreasing at a rate of 12 cm/s

 

[wjm11]

These two solutions represent two different sets of assumptions, i.e., which wall dimensions are changing. Choose appropriately.

Hope that helps.

 

[skeeter]

Upon refelection, I see your point. The side walls are not changing in length since they move apart.

The distance between the side walls (the length of the end walls) is changing.

Ignore my previous post ... it is incorrect.

mea culpa.

 

[wjm11]

Skeeter, your solution could be the correct one; I found the problem statement a bit vague myself.

 

[soroban]

Re: related rates- trash compactor

Hello, everyone!

I have theory about this problem . . .
Both the side walls and the end walls are changing lengths . . .
. . . and we must be very careful about the rate of change!

There are typos . . . and "math" hasn't been back to clarify the problem.

Trash compactor
The side walls are moving apart at rate of 0.3 m/s.
Volume of rancid water at bottom of compactor is constant at 20 m³.

When side walls are 5 m long and end walls are 2 m long,
at what rate is depth of water changing?

My attempt:
da/dt = 0.1 . What is this?
db/dt= -0.3 . Negative? .The sides walls are moving apart.
V=20
Find dh/dt when a=5 b=2

Here's what I think is happening:
. . The side walls are moving apart at 0.3 m/s.
. . The end walls are moving together at 0.1 m/s.

Here's the sneaky part . . .
Look at the diagram of the bottom of the trash compactor

              S↑
      * - - - - - - - *
      |               |
    E |               | E
    → |               | ←
      * - - - - - - - *
              S↓

If the side walls are moving apart at 0.3 m/s,
. . then the end walls are getting longer at 0.3 m/s.
Let \(\displaystyle E\) = length of an end wall. .Then: \(\displaystyle \:\frac{dE}{dt}\,=\,0.3\)

If the end walls are moving together at 0.1 m/s,
. . then the sides walls are getting shorter at 0.1 m/s.
Let \(\displaystyle S\) = length of a side wall. .Then: \(\displaystyle \:\frac{dS}{dt}\,=\,-0.1\)

Let \(\displaystyle D\) = depth of the water.


The volume of the water is: \(\displaystyle \:V \:=\:S\cdot E\cdot D\)

Differentiate with respect to time: \(\displaystyle \L\:\frac{dV}{dt}\:=\:S\cdot E\cdot\frac{dD}{dt}\,+\,S\cdot D\cdot\frac{dE}{dt}\,+\,E\cdot D\cdot\frac{dS}{dt}\)

Since the volume is constant (20): \(\displaystyle \,\frac{dV}{dt}\,=\,0\)
. . So we have: \(\displaystyle \L\:S\cdot E\cdot\frac{dD}{dt}\,+\,S\cdot D\cdot\frac{dE}{dt}\,+\,E\cdot D\cdot\frac{dS}{dt}\:=\:0\)

Hence: \(\displaystyle \L\:\frac{dD}{dt} \;=\;-\frac{1}{S\cdot E}\left[S\cdot D\cdot\frac{dE}{dt}\,+\,E\cdot D\cdot\frac{dS}{dt}\right]\)


If \(\displaystyle S\,=\,5\) and \(\displaystyle E\,=\,2\), then \(\displaystyle D\,=\,2\)
. . and we have: \(\displaystyle \,\frac{dE}{dt}\,=\,0.3,\;\frac{dS}{dt}\,=\,-0.1\)

Substitute: \(\displaystyle \L\:\frac{dD}{dt} \;= \;-\frac{1}{(5)(2)}\left[(5)(2)(0.3)\,+\,(2)(2)(-0.1)\right] \;=\;-0.26\)


Therefore, the level of the water is falling at \(\displaystyle 0.26\text{ m/s}\)

 

[math]

thanks for all the replies. i'm really sorry i just looked back at the problem and left important things out!

The side walls of the compactor are moving apart at the rate of .1 m/s ; the end walls are moving together at .3 m/s . The volume of the rancid water at the bottom of the compactor is constant at 20 cubic meters. When the side walls are 5 meters long and the end walls are 2 meters long, at what rate is the depth of the water changing?

 

[math]

i know everyone who worked on this is probably annoyed .. sorry again!