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Mathematics 16 Online
OpenStudy (anonymous):

If we have a graph of data which has: Time (t) (hours) verses Ln(Population) and the data goes as follows: Time (hours) - 1 2.5 5 7 12 15 18 Ln(Pop) - (2.89)(3)(3.13)(3.3)(3.61)(3.83)(4.03) Dont mind the brackets^ (just to separate the numbers) The question asks to investigate whether this is an exponential growth pattern =/.. In words how would i explain it is an exponential pattern?

OpenStudy (anonymous):

Try graphing the points and seeing of the resulting curve is in the form of an exponential.

OpenStudy (amistre64):

plot the points and see what they look like on a graph is my solution

OpenStudy (amistre64):

you can see if its linear by taking the slope between a few points and see if they match

OpenStudy (amistre64):

you could also assume it exponential and try to determine a formula for it

OpenStudy (amistre64):

use discrete recursion equations maybe?

OpenStudy (anonymous):

Lagrange polynomials?

OpenStudy (anonymous):

just shoot me now :P

OpenStudy (amistre64):

y = b*a^x and solve for the solutions?

OpenStudy (anonymous):

I've already graphed it and it looks linear, however,the assignment is based on exponentials =/

OpenStudy (amistre64):

yes, it does match linearly quite well; but perhaps we are just to close to it so that it looks linear?

myininaya (myininaya):

m1=(3.83-3.61)/(15-12)=11/150 m2=(3.61-3.3)/(12-7)=.062 not linear

OpenStudy (anonymous):

Jhonte was able to find a linear formula though?

OpenStudy (amistre64):

jhon found a best fit match; not quite the same as 'its linear'

myininaya (myininaya):

are we doing linear regression?

OpenStudy (anonymous):

Yeah thats the best fit.

OpenStudy (amistre64):

the trouble with best fit; is that we learn in calculus that if we look close enough at a curve it becomes straight

OpenStudy (amistre64):

the earthis flat after all :)

OpenStudy (anonymous):

hahah true :P

OpenStudy (anonymous):

So what myininaya post earlier: m1=(3.83-3.61)/(15-12)=11/150 m2=(3.61-3.3)/(12-7)=.062 How does this show it is not linear?

OpenStudy (anonymous):

linear functions have one slope

OpenStudy (amistre64):

linear has the same slopes; those are 2 different slopes

OpenStudy (anonymous):

Oh, okay

OpenStudy (anonymous):

its derivative is a constant, whereas this function's derivate changes signs more than once is not

OpenStudy (anonymous):

it looks like a polynomial to me

OpenStudy (anonymous):

So, if i were to explain this was an exponential growth pattern. I could include why it is not linear and what else could i perhaps add to support this argument?

OpenStudy (amistre64):

donuts; they work good for proving your point :)

OpenStudy (anonymous):

lol

OpenStudy (anonymous):

ahaha

OpenStudy (anonymous):

Thanks to everyone that helped :D <3. Ill just mention why it isn't linear :P!

myininaya (myininaya):

ok good luck

OpenStudy (anonymous):

thank you :)

OpenStudy (anonymous):

Oh damn it! is was meant to ask, could you explain to me what Fruitless said earlier: "its derivative is a constant, whereas this function's derivate changes signs more than once is no"

myininaya (myininaya):

do you know anything about calculus?

OpenStudy (anonymous):

Yeah, but does he mean to derivative of Ln(population)?

OpenStudy (anonymous):

I meant the slopes increase/decrease variably between points, so I am assuming it is a polynomial.

OpenStudy (anonymous):

Not negatively decrease, I mean decrease relative to the previous slope.

myininaya (myininaya):

hey so jhonte, you are going to argue that it is an exponential function right? You could say it is possible since it the function keeps increasing.

OpenStudy (anonymous):

Yeah

OpenStudy (anonymous):

Alright thanks alot :D!

OpenStudy (anonymous):

So, all in all.. An exponential growth pattern increases faster as x increases?

OpenStudy (anonymous):

Not exactly... It is exponential growth as long as the function increases; the slope cannot be less than 0.

OpenStudy (anonymous):

In this case the function increases each time doesn't it?

OpenStudy (anonymous):

Yes. That could be one of your arguments to support that it is an exponential.

OpenStudy (anonymous):

THANKS SO MUCH OMG YOU GUYS ROCK!

OpenStudy (anonymous):

Thanks :)

OpenStudy (anonymous):

One second, sorry for bringing the topic back up, but does the ln of data have to show a linear graph?

OpenStudy (anonymous):

If you are assuming it is exponential, then make a curve "approximation" between each two points.

OpenStudy (anonymous):

I think I've missed an import piece of data in being the Population (millions) 18 20 23 27 37 46 56

OpenStudy (anonymous):

I found the ln(pop) from those

OpenStudy (anonymous):

I'm being told the ln(population) verse time graph must be linear....

OpenStudy (anonymous):

My current argument is: The Ln(Population) verse Time graph at first glance looks linear, however, we are able to prove it is in-fact not linear by finding the gradient at different points and seeing if the gradient is constant. So, to find the gradient we use m= (y2-y1)/(x2-x1) Therefore, we know y = Ln(Population) and x = Time. Now we find the gradient at any random points. m1=((3.83-3.61))/((15-12))=0.73 m2= ((3.61-3.3))/(12-7)=0.62 It becomes evident to as why this is not a linear growth pattern. In-order to be linear the gradient has to be constant throughout. Another key in stating whether a graph is an exponential growth pattern is by indentifying whether the function increases and it’s slope doesn’t equal less than 0. In this case we are able to conclude this is in fact an exponential growth pattern as it’s function does increase throughout.

OpenStudy (anonymous):

I agree with your argument.

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