Help :/ rate of water and troughs

- Babynini

Help :/ rate of water and troughs

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- Babynini

##### 1 Attachment

- anonymous

are you at k12

- Babynini

hm?

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## More answers

- anonymous

do you no what k12 is

- Babynini

Nope. I'm using derivatives and Area of Trapezoid

- anonymous

oh ok i go to k12 it amzing dude

- Babynini

haha maybe later..I just want to solve this xD

- Babynini

@Empty

- Babynini

@jim_thompson5910 :)

- anonymous

i think you should of pt a 3 to

- jim_thompson5910

the cross section looks like this
|dw:1446600270190:dw|

- anonymous

tell me what you did to find the answer ok

- Babynini

|dw:1446600374531:dw|

- anonymous

ok i see what time is it there

- anonymous

multiple all that and tell me what you get ok deal

- jim_thompson5910

|dw:1446600456534:dw|

- jim_thompson5910

|dw:1446600541602:dw|

- Babynini

Right, so I wrote it out like
V =(1/2)h(b + B)*500
V = (1/2)h(20+(20+2a)*500
...

- Babynini

If that makes sense, I think we went about it different ways. Perhaps yours is simpler xD

- jim_thompson5910

I would use h instead of a (see drawing) because the volume will depend on the height and they ask how fast the water is rising at a given height
ie they want to know dh/dt at a certain value of h

- Babynini

right right, ok. Continue o.o

- jim_thompson5910

I guess you could use 'a' but you'd have to use that to figure out what dh/dt and h is

- Babynini

Well dh/dt is given in the problem. It's 0.2, no?

- jim_thompson5910

|dw:1446600771181:dw|

- jim_thompson5910

well since the dimensions are given in cm, let's convert 0.2 m^3/min to cm^3/min

- Babynini

= 2

- jim_thompson5910

1 m = 100 cm
(1 m)^3 = (100 cm)^3
1 m^3 = 1,000,000 cm^3

- Babynini

ou

- jim_thompson5910

|dw:1446600941506:dw|

- jim_thompson5910

|dw:1446600972085:dw|

- Babynini

ahaa

- jim_thompson5910

What do you get when you simplify that?

- Babynini

How do I multiply the bottom out? o.o

- jim_thompson5910

|dw:1446601213765:dw|

- jim_thompson5910

up top you'll have 0.2 times 1,000,000
in the bottom you'll have 1 times 1

- Babynini

200,000

- Babynini

over 1

- jim_thompson5910

dV/dt = 200,000 cm^3/min

- Babynini

2000 cubic m /min ?

- Babynini

hmm idk if that is correct because all the practice answers and problems have been 1/something

- Babynini

oh wait. then we plug in dv/dt and solve?

- jim_thompson5910

yes that comes later after we figure out what dV/dt is in general

- Babynini

Dv/dt = 0.2 m3/min

- Babynini

as given in the problem

- jim_thompson5910

yes correct
dV/dt = 0.2 m^3/min = 2,000,000 cm^3/min

- jim_thompson5910

hopefully you agree that the cross section area of the water is 50h? see this drawing
|dw:1446601540474:dw|

- jim_thompson5910

oh wait nvm I messed up
let me fix

- jim_thompson5910

|dw:1446601584238:dw|

- jim_thompson5910

|dw:1446601615221:dw|

- jim_thompson5910

|dw:1446601638707:dw|

- jim_thompson5910

we have 2 similar triangles
|dw:1446601693433:dw|

- jim_thompson5910

so we can say
30/b = 60/h
30h = 60b
h = 2b
b = h/2

- Babynini

Yeah

- jim_thompson5910

so one triangle has an area of
A = (base*height)/2
A = (h/2*h)/2
A = (h^2)/4

- jim_thompson5910

|dw:1446601867362:dw|
the total shaded area of the cross section of the water is
(area of triangle) + (area of rectangle) + (area of triangle) = (h^2)/4 + 20h + (h^2/4) = (h^2)/2 + 20h

- Babynini

o.o

- jim_thompson5910

hopefully you're seeing how I'm getting that?

- Babynini

haha yes it's just a bit much xD
why are we looking for the area of the shaded portion?

- jim_thompson5910

|dw:1446602123570:dw|

- jim_thompson5910

because we ultimately want the volume of the water in the trough
volume of water = (area of cross section) * (length of trough)
|dw:1446602185851:dw|

- jim_thompson5910

|dw:1446602222756:dw|

- Babynini

K so this is sort of how I was explaining it earlier. It's a completely different approach which is why I think i'm a little confused by yours. Though i'm starting to get it :)

##### 1 Attachment

- Babynini

[sorry, you can keep explaining!]

- jim_thompson5910

Your work looks great. You're going a slightly different way, but it will lead to the same answer if you followed my steps. So either route works.

- Babynini

Hm the thing is the answer keeps coming out wrong.
so i'm trying your way now haha but still wrong :/

- jim_thompson5910

let me try out your way for a sec

- Babynini

kk
[note I converted the length to cm]

- jim_thompson5910

ok I agree up til the point where you said
dV/dt = 10000*dh/dt+250(50)*dh/dt
keep in mind that dV/dt = 200,000 cm^3/min
but in your steps you plugged in dV/dt = 0.2 m^3/min

- Babynini

aaah

- Babynini

so instead it should be?

- jim_thompson5910

so you should have
200,000 = 10000*dh/dt+250(50)*dh/dt
solve for dh/dt
whatever you get, it will be in cm/min
you'll have to convert to m/min

- Babynini

dh/dt = 200,000/22500

- Babynini

...right?
which = 8.889

- jim_thompson5910

that reduces to dh/dt = 80/9 cm/min

- jim_thompson5910

put another way, the water rises 80 cm in 9 min when the water is 50 cm high

- jim_thompson5910

multiply 80/9 by 1/100 to convert to m/min

- Babynini

80/900 ?

- jim_thompson5910

80/900 then reduces to 4/45

- Babynini

that would be the final answer then?

- jim_thompson5910

4/45 = 0.0889 approx
so the water rises 4 m in 45 min at the height h = 50 cm
this is an instantaneous velocity at h = 50 cm

- jim_thompson5910

yes

- Babynini

tis wrong! gah

- jim_thompson5910

4/45 is wrong?

- jim_thompson5910

try the decimal form maybe

- Babynini

yeah.
The answer in the practice problem and video-explaining problem are both 1/something

- Babynini

nope nope it wants it in a fraction.

- jim_thompson5910

oh I see what went wrong

- jim_thompson5910

you have this
V = 10,000h + 250h^2
dV/dt = 10,000*dh/dt + 250(50)*dh/dt
which is very close

- jim_thompson5910

you forgot to pull the 2 down
it should be
dV/dt = 10,000*dh/dt + 2*250(50)*dh/dt

- Babynini

hmm

- jim_thompson5910

since y = x^2 ---> dy/dx = 2x

- Babynini

so then..
200,000/(10,000+2500)=dh/dt

- jim_thompson5910

2*250(50) = 25,000
not 2,500

- Babynini

ah yeah sorry .

- Babynini

200,000/35000

- Babynini

aai I can't get the correct answer o.0

- jim_thompson5910

200,000/35000 = 40/7
now multiply by 1/100

- jim_thompson5910

after multiplying by 1/100 and reducing, you should get 2/35

- Babynini

that's in m, yeah?

- Babynini

yaay that's correct. u.u

- jim_thompson5910

\[\Large \frac{dh}{dt} = \frac{40}{7}\frac{\text{m}}{\text{min}}\]
\[\Large \frac{dh}{dt} = \frac{2}{35}\frac{\text{cm}}{\text{min}}\]

- Babynini

Sorry for all the confusion!

- jim_thompson5910

that's ok

- Babynini

Right right, I see. I clearly need more practice with these =.= haha thank you so much

- jim_thompson5910

you're welcome

- jim_thompson5910

oh wait I had those backwards
\[\Large \frac{dh}{dt} = \frac{40}{7}\frac{\text{cm}}{\text{min}}\]
\[\Large \frac{dh}{dt} = \frac{2}{35}\frac{\text{m}}{\text{min}}\]
my bad

- Babynini

haha I was going to say xD

- Babynini

no worries

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