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Mathematics 16 Online
OpenStudy (rock_mit182):

Distance an object falls Assuming air resistance is negligible, a small object that is dropped from a hot air balloon falls 16 feet during the first second, 48 feet during the second second, 80 feet during the third second, 112 feet during the fourth second, and so on. Find an expression for the distance the object falls in n seconds.

OpenStudy (rock_mit182):

math series

OpenStudy (ybarrap):

Write out a table first

OpenStudy (rock_mit182):

|dw:1394915800636:dw|

OpenStudy (ybarrap):

Perfect! Let's graph them: <1,16>, <2,48>,<3,80>,<4,112> It's a straight line - http://www.wolframalpha.com/input/?i= {{1%2C16}%2C+{2%2C48}%2C{3%2C80}%2C{4%2C112}} So we can use y=mx + b. Do you know how to find m and b?

OpenStudy (rock_mit182):

\[m =\frac{ \Delta Y}{ \Delta X }\]

OpenStudy (rock_mit182):

SO R = the rate of the series is the same as m ( the slope )

OpenStudy (rock_mit182):

?

OpenStudy (ybarrap):

so you have the formula for m. m=(112-16)/(4-1) = ?

OpenStudy (rock_mit182):

yes . .

OpenStudy (ybarrap):

so once you have m, you have y=mx+b To find b you make x = 1 and y = 4: 4=m(1) + b b = ? That's it! y = mx + b, where x is time and y is distance.

OpenStudy (rock_mit182):

so you are saying that the expression of the series is the equation of a line ?

OpenStudy (ybarrap):

that's right!

OpenStudy (rock_mit182):

i'd nver thought that would be the aproach

OpenStudy (ybarrap):

you don't need physics here

OpenStudy (rock_mit182):

but what about the relationship is no linear ?

OpenStudy (ybarrap):

normally projectile motion is parabolic. But your numbers are describing a linear relationship if you see the graph

OpenStudy (ybarrap):

does this make sense?

OpenStudy (rock_mit182):

well yeah i solved one of bacteria growth and n is a exponent

OpenStudy (rock_mit182):

Chlorine levels Chlorine is often added to swimming pools to control microorganisms. If the level of chlorine rises above 3 ppm (parts per million), swimmers will experience burning eyes and skin discomfort. If the level drops below 1 ppm, there is a possibility that the water will turn green because of a large algae count. Chlorine must be added to pool water at regular intervals. If no chlorine is added to a

OpenStudy (rock_mit182):

during a 24-hour period, approximately 20% of the chlorine will dissipate into the atmosphere and 80% will remain in the water. (a) Determine a recursive sequence (\[a _{n}\]) that expresses the amount of chlorine present after n days if the pool has ppm of chlorine initially and no chlorine is added. (b) If a pool has 7 ppm of chlorine initially, construct a table to determine the first day on which the chlorine level will drop below 3 ppm.

OpenStudy (rock_mit182):

this one is tough

OpenStudy (rock_mit182):

has \[a _{0}\] ppm of chlorine*

OpenStudy (ybarrap):

You know, the 1st problem may actually be saying that the distance fell at each time period was IN ADDITION to that distance already traveled. And in that case, the relationship would be nonlinear. I actually think that is how to interpret the problem. So, we have $$ d=at^2 $$ as our equation. We find a by setting t=1 and d=16: $$ a=16 $$ So our equation is $$ d=16t^2 $$ So in 4 seconds, $$ d=16\times4^2=256 $$ In 3 seconds, $$ d=16\times 3^2=144 $$ The difference between 4 and 3 seconds is 256-144=112 Which is what was given.

OpenStudy (rock_mit182):

well im going to work a little the problem

OpenStudy (rock_mit182):

so in this case a is the rate ?

OpenStudy (ybarrap):

Are we still talking about the 1st problem?

OpenStudy (rock_mit182):

yes

OpenStudy (ybarrap):

ok, yes, "a" is a parameter for a parabola. It's not necessarily a "rate". It's a number that says how "wide" the parabola is.

OpenStudy (rock_mit182):

I'm starting to get confused you know why, i was expecting for the expression always behave as i want

OpenStudy (rock_mit182):

i meant look at the d = 48

OpenStudy (rock_mit182):

is an inequality

OpenStudy (ybarrap):

What do you mean it's an inequality? You mean it doesn't equal something?

OpenStudy (rock_mit182):

48 = 16(2) ^2

OpenStudy (ybarrap):

ok. so at t=2 d=16*2^2=16*4=64 feet right?

OpenStudy (rock_mit182):

d = t ^2 + 16 ?

OpenStudy (ybarrap):

$$ d=16\times t^2 $$

OpenStudy (rock_mit182):

yeah but then i have to subtract 16 to give off 48, remember the list above

OpenStudy (rock_mit182):

in second'th second the distance is 48

OpenStudy (ybarrap):

d represents the total distance the object falls, which was the question To validate your table, you'll need to compute d for various t in your table and subtract them. For example, at t=2 seconds the total distance the object has fallen is d=16*2^2=64 at t=3 seconds the total distance the object has fallen is d=16*3^2=16*9=144 So to validate the information given and your table, you need to subtract these two numbers. 144-64=80.

OpenStudy (rock_mit182):

this is getting hard, I'll take a look later of the whole process

OpenStudy (rock_mit182):

so for the second problem i have to do the same process ?

OpenStudy (ybarrap):

The 1st problem is not hard but different in the sense that they give you differences and not totals. They tell you by how much something increases and so you need to infer that to get totals, you need to add to the previous results. That's the hardest part. Once you realize that this is a nonlinear function (i.e. you can't use y=mx+b) then the next option is quadratic (i.e. parabola). Then you need to ask, is the parabola wide? Does it shift upwards? does it shift left? and so on. If the curve had been more complicated, like multiple "bends", you would need to count the bends and add 1 to find the degree of the polynomial needed to represent the chart. All this comes with experience. Don't fret. Reread what we did here and get back with me on your questions.

OpenStudy (rock_mit182):

well im familiar with functions and parabolas

OpenStudy (rock_mit182):

but i thought it would be refer only to sequences and sum notation

OpenStudy (rock_mit182):

cause that is what im studying right now

OpenStudy (rock_mit182):

so i thought it would be a way or a formula to found the expression

OpenStudy (rock_mit182):

to find*

OpenStudy (ybarrap):

The 2nd problem you need to write some raw numbers and find the pattern Are you familiar with recursive sequences?

OpenStudy (ybarrap):

You could have approached the 1st problem recursively I think.

OpenStudy (rock_mit182):

in each sequence we have \[(a,a_{n})\]

OpenStudy (rock_mit182):

for each term

OpenStudy (rock_mit182):

yeah recursively thta's what i need

OpenStudy (rock_mit182):

well, I'm familiar but that not implies I'm good, at all

OpenStudy (ybarrap):

brb in about 15 mins we'll solve the 1st one recursively

OpenStudy (rock_mit182):

lol, man i start to realize, this is not meant for me

OpenStudy (rock_mit182):

ok

OpenStudy (rock_mit182):

did you get there ( I mean the solution)

OpenStudy (ybarrap):

Let's solve the 1st equation recursively: $$ a_1=16\\ a_2=48\\ a_3=80\\ a_4=112 $$ Look for a pattern: $$ a_1=16\times 1\\ a_2=16\times3\\ a_3=16\times 5\\ a_4=112\times 7 $$ Do you see the pattern? Look how the last term is related to the index of a: $$ a_n=16\times(2n-1) $$ Check: To find distance falls in 3 seconds after it has already fallen 2 seconds is $$ a_3=16\times((2\times3)-1)=16\times(6-1)=16\times5=80 $$ Does this make sense?

OpenStudy (ybarrap):

*Correction on \(a_4\) above $$ a_4=16\times 7 $$

OpenStudy (rock_mit182):

\[a_{n+1} = a _{1} (2n+1) \]

OpenStudy (rock_mit182):

that would be the expression fo the nth second

OpenStudy (rock_mit182):

?

OpenStudy (ybarrap):

Let's check: $$ a_1=16\\ a_2=a_1\times3\\ a_3=a_1\times 5\\ a_4=a_1\times7 $$ So, $$ a_n=a_1\times(2n-1) $$ Then $$ a_{n+1}=a_1\times(2(n+1)-1)=a_1\times(2n+1) $$ Which is what you got!

OpenStudy (ybarrap):

As you can see there are various ways to approach recursive problems and also different ways to solve this problem, generally.

OpenStudy (rock_mit182):

what a relief ! what would i do if you weren't to help me out

OpenStudy (ybarrap):

Does that mean you understand? Ready for the second problem?

OpenStudy (rock_mit182):

oh are you serious ? well, let's try to solve it

OpenStudy (ybarrap):

It's easy! Let's start.

OpenStudy (ybarrap):

Let X represent the amount of chlorine initially in the water in units of ppm. After the 1st day (24 hours), there will only be $$ a_1=0.8X $$ ppm left. Here we let \(a_n\) represent the amount of chlorine we have left after \(n \) days. This is was the hardest part. Make sense so far?

OpenStudy (rock_mit182):

ok

OpenStudy (rock_mit182):

first expression could be also \[a _{0}\] in ppm

OpenStudy (ybarrap):

yes, it could be. Multiple ways to start off.

OpenStudy (ybarrap):

$$ a_0=0.8X\\ a_1=0.8\times a_0=0.8^2\times X\\ a_2=0.8^3\times X\\ ...\\ a_n=0.8^{n+1}X $$ Or $$ a_{n-1}=0.8^{n}X $$ Is this ok? If we had left \(a_1=0.8X\) then our answer would have been $$ a_n=0.8^nX $$ Which looks a little easier to ready. But all these forms are equivalent.

OpenStudy (rock_mit182):

I was thinking if there is no CL added in the nth days , at some point, we wouldn't find any Cl , at all

OpenStudy (ybarrap):

The amount of Cl will decrease steadly according to the equation above and will only approach zero ppm, but will never be zero but approaches it in the limit: $$ \lim_{n\to\infty}0.8^nX=0 $$

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