Logistic natural growth help needed (1 Viewer)

[cssftw]

Hi guys here's a question involving logistic growth that I can't do.

Q3 part a is what I am having problem with. In order to show that P = that expression, do I just sub it in dP/dt, and eventually I would get kP(1-RP)?

What would I have to write as a concluding statement etc? Can someone pleease offer a complete solution?

Thanks, appreciate the help.

 

[funnytomato]

LHS=dP/dt , so differentiate the expression for P with respect to t, which means you'll need the chain rule
for the RHS , just sub in the expression for P

u may need to change ur expression a bit so that both sides look exactly the same

 

[funnytomato]

if u have difficulty in differentiating

try this easier example first

d/dx (5+3x)^3

for which you'll get 3*(5+3x)^2*3, where the 3 is the derivative of whatever is in side the brackets, note the derivative involves an exponential term in ur question

...

 

[cssftw]

if u have difficulty in differentiating

try this easier example first

d/dx (5+3x)^3

for which you'll get 3*(5+3x)^2*3, where the 3 is the derivative of whatever is in side the brackets, note the derivative involves an exponential term in ur question

...

lol no, I just want to know what I have to do to answer Q3a. E.g. Step 1 --> you do this.... Step 2: You do that etc... If I got that in an exam, what would I have to do to answer the question? Because I don't understand



EDIT --> Just wondering is there anyway to get the value of P from beginning with the expression dP/dt = kP(1-RP)??

Or in order to answer the question, must you substitute the expression for P into dP/dt?

 

[funnytomato]

lol no, I just want to know what I have to do to answer Q3a. E.g. Step 1 --> you do this.... Step 2: You do that etc... If I got that in an exam, what would I have to do to answer the question? Because I don't understand



EDIT --> Just wondering is there anyway to get the value of P from beginning with the expression dP/dt = kP(1-RP)??

Or in order to answer the question, must you substitute the expression for P into dP/dt?

alright

step 1: find dP/dt for LHS, i think that's what u meant by "substitute the expression for P into dP/dt"
step 2: sub P into kP(1-RP) for RHS
step 3: tidy up your answers for LHS and RHS to make the equality apparent

 

[funnytomato]

you don't need to solve the differential equation yourself , in exams you're given the solution and you're expected to verify it(which is much easier than solving it)

i.e. find its derivative , and substitute back

 

[funnytomato]

lol no, I just want to know what I have to do to answer Q3a. E.g. Step 1 --> you do this.... Step 2: You do that etc... If I got that in an exam, what would I have to do to answer the question? Because I don't understand



EDIT --> Just wondering is there anyway to get the value of P from beginning with the expression dP/dt = kP(1-RP)??

Or in order to answer the question, must you substitute the expression for P into dP/dt?

probably i didn't clarify it enough, but that's meant to help you to find dP/dt

 

[cssftw]

Sorry about this, but could someone please mark/check my solution?

Basically I provided a statement in how and why the proof will work.

Then I substituted P = (expression we have to prove) in to dP/dt --> and I repeated this process except I substituted the expression into kP(1-RP) --> these two expressions eventually were in equality.

Thanks.

 

[funnytomato]

that's basically right
but it could be improved a little bit:

in a normal exam, you probably don't need to write down the first few lines of explanation, but it's not wrong either

for the line before LHS=dP/dt, "Now dP/dt=kp(1-rp)" , some ambiguity could occur as this is not shown yet.
what u meant might be "to show dP/dt=kp(1-rp)"

I'd replace that line with "for P=expression"

then:

LHS=... (what you wrote down)

RHS=...(what you wrote down till the 4th last line) = LHS

therefore , P=expression is a solution for dP/dt=kp(1-rp)