Babynini
  • Babynini
Help :/ rate of water and troughs
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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Babynini
  • Babynini
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anonymous
  • anonymous
are you at k12
Babynini
  • Babynini
hm?

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anonymous
  • anonymous
do you no what k12 is
Babynini
  • Babynini
Nope. I'm using derivatives and Area of Trapezoid
anonymous
  • anonymous
oh ok i go to k12 it amzing dude
Babynini
  • Babynini
haha maybe later..I just want to solve this xD
Babynini
  • Babynini
@Empty
Babynini
  • Babynini
@jim_thompson5910 :)
anonymous
  • anonymous
i think you should of pt a 3 to
jim_thompson5910
  • jim_thompson5910
the cross section looks like this |dw:1446600270190:dw|
anonymous
  • anonymous
tell me what you did to find the answer ok
Babynini
  • Babynini
|dw:1446600374531:dw|
anonymous
  • anonymous
ok i see what time is it there
anonymous
  • anonymous
multiple all that and tell me what you get ok deal
jim_thompson5910
  • jim_thompson5910
|dw:1446600456534:dw|
jim_thompson5910
  • jim_thompson5910
|dw:1446600541602:dw|
Babynini
  • Babynini
Right, so I wrote it out like V =(1/2)h(b + B)*500 V = (1/2)h(20+(20+2a)*500 ...
Babynini
  • Babynini
If that makes sense, I think we went about it different ways. Perhaps yours is simpler xD
jim_thompson5910
  • 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
  • Babynini
right right, ok. Continue o.o
jim_thompson5910
  • 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
  • Babynini
Well dh/dt is given in the problem. It's 0.2, no?
jim_thompson5910
  • jim_thompson5910
|dw:1446600771181:dw|
jim_thompson5910
  • jim_thompson5910
well since the dimensions are given in cm, let's convert 0.2 m^3/min to cm^3/min
Babynini
  • Babynini
= 2
jim_thompson5910
  • jim_thompson5910
1 m = 100 cm (1 m)^3 = (100 cm)^3 1 m^3 = 1,000,000 cm^3
Babynini
  • Babynini
ou
jim_thompson5910
  • jim_thompson5910
|dw:1446600941506:dw|
jim_thompson5910
  • jim_thompson5910
|dw:1446600972085:dw|
Babynini
  • Babynini
ahaa
jim_thompson5910
  • jim_thompson5910
What do you get when you simplify that?
Babynini
  • Babynini
How do I multiply the bottom out? o.o
jim_thompson5910
  • jim_thompson5910
|dw:1446601213765:dw|
jim_thompson5910
  • jim_thompson5910
up top you'll have 0.2 times 1,000,000 in the bottom you'll have 1 times 1
Babynini
  • Babynini
200,000
Babynini
  • Babynini
over 1
jim_thompson5910
  • jim_thompson5910
dV/dt = 200,000 cm^3/min
Babynini
  • Babynini
2000 cubic m /min ?
Babynini
  • Babynini
hmm idk if that is correct because all the practice answers and problems have been 1/something
Babynini
  • Babynini
oh wait. then we plug in dv/dt and solve?
jim_thompson5910
  • jim_thompson5910
yes that comes later after we figure out what dV/dt is in general
Babynini
  • Babynini
Dv/dt = 0.2 m3/min
Babynini
  • Babynini
as given in the problem
jim_thompson5910
  • jim_thompson5910
yes correct dV/dt = 0.2 m^3/min = 2,000,000 cm^3/min
jim_thompson5910
  • jim_thompson5910
hopefully you agree that the cross section area of the water is 50h? see this drawing |dw:1446601540474:dw|
jim_thompson5910
  • jim_thompson5910
oh wait nvm I messed up let me fix
jim_thompson5910
  • jim_thompson5910
|dw:1446601584238:dw|
jim_thompson5910
  • jim_thompson5910
|dw:1446601615221:dw|
jim_thompson5910
  • jim_thompson5910
|dw:1446601638707:dw|
jim_thompson5910
  • jim_thompson5910
we have 2 similar triangles |dw:1446601693433:dw|
jim_thompson5910
  • jim_thompson5910
so we can say 30/b = 60/h 30h = 60b h = 2b b = h/2
Babynini
  • Babynini
Yeah
jim_thompson5910
  • jim_thompson5910
so one triangle has an area of A = (base*height)/2 A = (h/2*h)/2 A = (h^2)/4
jim_thompson5910
  • 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
  • Babynini
o.o
jim_thompson5910
  • jim_thompson5910
hopefully you're seeing how I'm getting that?
Babynini
  • Babynini
haha yes it's just a bit much xD why are we looking for the area of the shaded portion?
jim_thompson5910
  • jim_thompson5910
|dw:1446602123570:dw|
jim_thompson5910
  • 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
  • jim_thompson5910
|dw:1446602222756:dw|
Babynini
  • 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 :)
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Babynini
  • Babynini
[sorry, you can keep explaining!]
jim_thompson5910
  • 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
  • Babynini
Hm the thing is the answer keeps coming out wrong. so i'm trying your way now haha but still wrong :/
jim_thompson5910
  • jim_thompson5910
let me try out your way for a sec
Babynini
  • Babynini
kk [note I converted the length to cm]
jim_thompson5910
  • 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
  • Babynini
aaah
Babynini
  • Babynini
so instead it should be?
jim_thompson5910
  • 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
  • Babynini
dh/dt = 200,000/22500
Babynini
  • Babynini
...right? which = 8.889
jim_thompson5910
  • jim_thompson5910
that reduces to dh/dt = 80/9 cm/min
jim_thompson5910
  • jim_thompson5910
put another way, the water rises 80 cm in 9 min when the water is 50 cm high
jim_thompson5910
  • jim_thompson5910
multiply 80/9 by 1/100 to convert to m/min
Babynini
  • Babynini
80/900 ?
jim_thompson5910
  • jim_thompson5910
80/900 then reduces to 4/45
Babynini
  • Babynini
that would be the final answer then?
jim_thompson5910
  • 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
  • jim_thompson5910
yes
Babynini
  • Babynini
tis wrong! gah
jim_thompson5910
  • jim_thompson5910
4/45 is wrong?
jim_thompson5910
  • jim_thompson5910
try the decimal form maybe
Babynini
  • Babynini
yeah. The answer in the practice problem and video-explaining problem are both 1/something
Babynini
  • Babynini
nope nope it wants it in a fraction.
jim_thompson5910
  • jim_thompson5910
oh I see what went wrong
jim_thompson5910
  • 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
  • jim_thompson5910
you forgot to pull the 2 down it should be dV/dt = 10,000*dh/dt + 2*250(50)*dh/dt
Babynini
  • Babynini
hmm
jim_thompson5910
  • jim_thompson5910
since y = x^2 ---> dy/dx = 2x
Babynini
  • Babynini
so then.. 200,000/(10,000+2500)=dh/dt
jim_thompson5910
  • jim_thompson5910
2*250(50) = 25,000 not 2,500
Babynini
  • Babynini
ah yeah sorry .
Babynini
  • Babynini
200,000/35000
Babynini
  • Babynini
aai I can't get the correct answer o.0
jim_thompson5910
  • jim_thompson5910
200,000/35000 = 40/7 now multiply by 1/100
jim_thompson5910
  • jim_thompson5910
after multiplying by 1/100 and reducing, you should get 2/35
Babynini
  • Babynini
that's in m, yeah?
Babynini
  • Babynini
yaay that's correct. u.u
jim_thompson5910
  • 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
  • Babynini
Sorry for all the confusion!
jim_thompson5910
  • jim_thompson5910
that's ok
Babynini
  • Babynini
Right right, I see. I clearly need more practice with these =.= haha thank you so much
jim_thompson5910
  • jim_thompson5910
you're welcome
jim_thompson5910
  • 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
  • Babynini
haha I was going to say xD
Babynini
  • Babynini
no worries

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