Find the effect of (-kw) in dW/dt = (m - n - kw) W

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shivastar

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The number of wombat at t=0 is 200. m represents the birthrate and n represents the death rate. Kw is the rate at which food is supplied in relation to the rates.(not sure how exactli it the question is worded but it has to do with how the food supply affects the birth/death rates)

Find effect of (-kw)
Find t when W=10 and W=100

dW/dt= (m-n-kw)W
 
You have Kw, kw, and W. I'm wondering if K = k and if W = w. You'll have to clarify that.

Isn't it seperable?
 
I have a different problem. If w is the amount of food given per wombat, the more food you give, the faster they die off???
That aside I get
dW/dt= (m-n-Kw)W
dW/W=(m-n-Kw)dt
ln(W) = (m-n-Kw)t + C
W= C * e^((m-n-Kw)t)
When t = 0, W=200 so C=200
so we have
W=200*e^((m-n-Kw)t)
 
hEy thanX for da prompT repLy ppl...

w= the number of wombat....
k= the amount of food

dW/dt= (m-n-kw)W
a) wat effect does (-kw) have?
b) find t (datz on its own i fink:S)...
c) Find W when t=10 and t=100

i aint too sure how exactli the question reads... i ll get full ques in 3 daiz...

ThanX again ppl:)
 
ohh sowi dere was more infO*

b) and c).... is when m=0.1 and n=0.06 and k=0.00005
 
Wow! Did you forget how to speak English? Your first post was a good style. Please go back to it.

You still haven't answered my questions about K, k, W, and w.
 
I'll give those who enjoy sneering at me a chance.
I think:
m-n = dW/dt so
dW/dt=(dW/dt-KW)W
dW/dt=W(dW/dt)-KW²
(dW/dt)(1-W)=-KW²
dW/dt=-KW²/(1-W)=KW²/(W-1)
But I can't recall how to integrate it.
 
Where did you get m - n = dW/dt?

dW/dt = KW<sup>2</sup>/(W-1)

(W-1)/W<sup>2</sup> dW = k dt

ln(W) + 1/W = k*t + C
 
Just pulled it out of the air. Birth rate - death rate has to be the change rate of the poulation = dW/dt. I didn't think of moving stuff to the left side though, even after doing it before w equaled W. I knew I wanted a ln(W) but... :oops:
 
Hello everyone... I hope to completely solve this problem :p

This shall take a while (hopefully I can help)

OK we first have established that: dw/dt= (m-n-kw)w

and for intensive purposes I will use these values (as they have been previously mentioned) :

m= 0.1 n= 0.06 (Therefore we can say that m-n = 0.04)

K = 0.00005

INITIAL NUMBER OF WOMBATS = 200

From this new information we can expand dw/dt= (m-n-kw)w

This yields: dw/dt = 0.04w - 0.00005w^2

Now we can re-write this as dw/dt = w/25 - w^2/20000

(as 1/25 is the same as 0.04 and 1/20000 is the same as 0.00005)

So from this we are faced with the problem that dw/dt is in terms of w NOT IN TERMS OF t - Which is what it would need to be to be able to properly differentiate it.

No problem. :p

instead we use subtraction of fractions with different bases (learnt in year 7)

To get (after simplified) dw/dt= (800w-w^2)/20000

now using magic (aka flipping both fractions)

we can say that dt/dw = 20000 / (800w-w^2) (NOTE Dt/Dw!!!)

NOW we can differentiate by taking out the 20000
we can now safly say:

t = ∫[20000/(800w-w^2)]dw (DONT FORGET + C !!!)

THIS IS QUITE HARD but if you break down the 20000/(800w-w^2) using PARTIAL FRACTIONS you get 25/(w-800) - 25/w

Therefore:
t = -25*∫[(1/(w-800)) - 1/w
t = -25*[log(e)(|w-800|) - log(e)(|w|)]

(NOTE THE ABSOLUTE VALUE FUNCTIONS!!!! THEY ARE EXTREMELY IMPORTANT!!!)

t = -25Log(e)[ |w-800| / |w|] + C (simplified version of the above using log laws)

so substituting t = 0 and w = 200 we can find that C = 25log(e)(3)

Therefore t =-25Log(e)[ |w-800| / |w|] + 25log(e)(3)

You can just use this formula to find out predicted values

in which case when t = 10 you get TWO ANSWERS (you have to interpret them)

Taking note of the absolute functions you get TWO equations in terms of w (really quite hard to re-arrange) :

W = [800e^(t/25)]/[e^(t/25)-3]

and

W = [800e^(t/25)]/[e^(t/25)+3] (THIS IS THE CORRECT ONE)

you decipher that the second is correct by observing the graph - If you aren't doing extremely advanced maths I recommend that you use a calculator - One of the graphs will be negative for all W values THIS IS OBVIOUSLY NOT THE RIGHT ONE !! as you cannot have a negative population.

W = [800e^(t/25)]/[e^(t/25)+3]

NOW USING THIS ONE WE CAN SUBSTITUTE VALUES IN!!!

for example:

t = 10 yields w ≈ 265.96 (ROUND DOWN to 265!!)
t = 100 yields w ≈ 758.332 (ROUND DOWN to 758!!!)

Well that was indeed a marathon. Thank you for the question. I enjoyed it.

:p

Peace out.
 
That's admirable, but why did you dig up a two-year old problem?.
 
Well now that one is out of the way.

An extension for this problem is :

Dw/dt = (m-n-kw)w - D (where D is the ammount of deaths incurred by Cars)

Derive an expression.

Hint: Use the descriminent

This one got me good.


Good luck
 
"That's admirable, but why did you dig up a two-year old problem?."

Why thank you. You are too kind.

However I have much recently come into contact with almost exactly the same problem.
 
Well we now have dw/dt = (m-n-kw)w - D

So if we expand this equation we get:

-kw^2 + (m-n)w - D

And by using the descriminent (b^2 - 4ac)

and concidering the case where dw/dt = 0

we have:

(m-n)^2 -4(-k)(-D)=0
=(m-n)^2 = 4kD

D=[(m-n)^2]/4k

:p - Im not sure if this is right... Please correct...

So this equation can calculate how many wombats can die by miscellaneous methods without effecting the population.

(I know it's strange me answering the questions that I posed)

-Leonator
 
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