Question

how do you do this? http://puu.sh/7wWHq.png

Answer 1

Do you know DEs to begin with?

Answer 2

what do you mean?

Answer 3

differential equations? yes

Answer 4

can you walk me through this?

Answer 5

start with a picture.

Answer 6

@inkyvoyd Carry on; I'm out.

Answer 7

note that you have .5 lbs of salt per second coming in, and you have .05Q(t) lbs of salt per second coming out

Answer 8

does it make sense why?

Answer 9

@KNorne7592 , please send me a response so I know you're still here :)

Answer 10

I'm here

Answer 11

so what would the de look like?

Answer 12

does it make sense why you have .5lbs of salt per second coming in, and .05Q(t) lbs of salt per second going out?

Answer 13

dy/dx = .05 y?

Answer 14

so you have two separate des?

Answer 15

No.

Answer 16

lol, you told us you wanted a step by step walkthrough, so I'm giving you one. I know you're eager to get to the DE so you can solve it, but let's face it, wolfram alpha can solve the DE any day, but it's the word problems and applications of the DE that gets us.

Answer 17

you're right

Answer 18

let's start over again. The best way to solve a problem like this is to draw a picture so you have an understanding of what's going on.

Answer 19

I've drawn a diagram where the arrows denote flow rates of quantity of salt input and output per second.

Answer 20

note that I got .5 lb gal/sec from the fact that the concentration of salt is .1 lb/gal, and that there are 5 gallons input per second.

Note that I got 5/100*Q(t) gal/sec from the fact that there is Q(t) gallons of salt in the tank at any time t, but only 5 out of 100 of those gallons are drained.

Answer 21

@KNorne7592 does it make sense what I'm doing here? What does it seem like I've drawn to you?

Answer 22

I'm unsure about the output equation

Answer 23

What do you mean, the output equation?

Answer 24

the water coming out

Answer 25

okay, first off, I just want to clarify. Everything I've worked with up til now has been terms and expressions - there have been no equations yet.
Anyways, to calculate the amount of salt that's going OUT of the tank, what you have to do is figure out the amount of water that is going out of the tank per second, and the salt concentration of that water. Does that make sense?

Answer 26

I understand

Answer 27

so what is the amount of water (solution) coming out of the tank per second?

Answer 28

I meant your ratio of salt water concentration coming out of the tank

Answer 29

5gal/sec

Answer 30

okay. So what is the concentration of said solution?
use the definition (by weight) of concentration to do it: concentration = weight of solute/volume of solution

Answer 31

Q(t) x 100?

Answer 32

Q(t) represents the weight of salt in the tank, and 100 represents the volume of water. you should divide the two numbers, not multiply them, because concentration is in units lb/gal.

Make sense?

Answer 33

@KNorne7592 , hello?

Answer 34

hmm

Answer 35

so why do you have 5/100 in your picture?

Answer 36

I'm sorry I'm so slow x[

Answer 37

That's what I'm trying to answer lol.
we already figured out what (and why) the weight of salt going in is .5lb/sec, and we seek to figure out what (and why) the weight of salt going out is.

What you have to do to understand this is to use the definitions of concentration, solution/solute, and the function given to you.

Answer 38

the reason I have 5/100 in my picture is because Q(t)/100 is the concentration of the salt water, but there are only 5 gallons of the stuff going out per second.
Note that concentration*volume=weight (the units cancel if you're confused), so you have to multiple Q(t)/100 by 5.

Answer 39

so qt is lbs?

Answer 40

yes.

Answer 41

ok so that's the amount in the tank at any given time

Answer 42

yes.

Answer 43

in lbs

Answer 44

so does my explanation make sense?

Answer 45

oh I see now

Answer 46

so after multiplying, what do you end up with

Answer 47

I get lb/s

Answer 48

yup.

Answer 49

and what is that equal to?

Answer 50

5/100*Q(t)

Answer 51

I mean what do you do next?

Answer 52

well, I've just described the rate of change of salt in the tank.

Answer 53

ooooh

Answer 54

the first rate, .5 lb/sec, is an increase of .5 lb/sec of salt in the tank, correct?

Answer 55

it is the rate, or dQ/dt

Answer 56

I see I see

Answer 57

yes!

Answer 58

so dQ/dt=the rate at which the salt is going in - the rate at which the salt is going out, correct?

Answer 59

yes

Answer 60

so given what we just figured out, what is dQ/dt equal to?

Answer 61

y/20

Answer 62

?

Answer 63

so the rate is how much salt is moving in and out of the tank?

Answer 64

Q(t)/20

Answer 65

how much salt is moving in the tank per second minus how much salt is moving out of the tank per second

Answer 66

we figured out that there are .5 lbs/sec of salt moving in, and .05Q(t) lbs/sec of salt moving out.

Answer 67

but that's all compounded for in Q(t)/20, right?

Answer 68

oh..

Answer 69

so it's .5 - qt/20

Answer 70

exactly. what's that equal to?

Answer 71

dqt/dt

Answer 72

and separate and solve?

Answer 73

wow that logic

Answer 74

well, that's not a seperable DE I don't think.
we just arrived at Q'(t)=.5-Q(t)/20

rearranging gives us -Q(t)/20-Q'(t)=.5

this isn't actually a seperable differential equation, but it IS solvable.

Answer 75

to solve it you need to understand how to solve linear first order ODEs - at this point, if you want to learn how to do it, it'd do you good to go to http://tutorial.math.lamar.edu/Classes/DE/Linear.aspx

It's a lot of work, but all valuable things are...

Answer 76

are you sure this is correct?

Answer 77

this problem was meant to be done without a calculator

Answer 78

You don't need a calculator to do this problem, but you need an understanding of how to solve a first order linear ordinary differential equation

Answer 79

I don't think for this problem I would need that level of maths

Answer 80

I know I sent you a wall of text, but THAT is exactly HOW to solve the first order linear differential equation we have arrived at.

Answer 81

pushing it through wolframalpha even gives nothing back

Answer 82

I can walk you through the solution... but I doubt you'll really understand it unless you go read paul's notes

Answer 83

does it converge to a number after time goes to infinity?

Answer 84

of course it does. You can try it in real life, just build a rather large tank and pump solution in and solution out

Answer 85

in fact, we can give a heuristic estimate: the concentration of solute in the tank will never be higher than .1 lbs/gal because that is the maximum amount of concentration the solution is being pumped in

Answer 86

I mean what do you get when you work this out aha