#rate of change
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i'd like to know what formula to use for this question, also please ping me once you respond, thank you :)
Rate of change is the amount of change per some amount of time, i.e., just the slope
No formula needed from t=8 to 10, because the graph is linear. Just find the slope
"delta y/delta x", "rise over run", etc.
Yeah
wow i really made things complicated for myself
thank you so much
so the answer should be 1
No, try again
As an exercise, try finding the average rate of change between t=4 and t=10. And what about between t=0 and t=10?
formula should be y1-y2/x1-x2, correct?
Yes
the answer is 2/2
No
What is 5-2
No
What is 5-2
I guess you're using the formula?
What is 5-2
Now try this.
-10/-6
but if you simplify it, it should be -5/-3
and two negatives make positive
so 5/3
Between t=4 and 10?
@lament knot
Remember slope formula from algebra 1?
(y2-y1)/(x2-x1)
That's all it is
So go to the interval (meaning what number on the x axis) it's asking for
Plot the beginning interval, to the end interval
Those are your two points
Plug into slope formula
Solve
for the graph or the excersise i was given?
The interval it's asking for
An interval is just a piece of the x axis aka seconds in this instance
ah so this isnt what i was supposed to do?
sorry
x is horizontal y is verticle
thank you
That's wrong
first interval is right, second is wrong
Try again,
Remember x goes first, y comes second
nah you're good
(8,2) (10,5)
Now what was the slope formula we talked abt
Almost
ah
uh
y2-y1/x2-x1
yes that's correct
i thought you can do both way
Nah
i remember being told that
I mean maybe? But I don't recommend it
ohhh
idk where 3 came form
Meaning 8 comes first, 2nd is 10
okay so 2-5/8-10
Still, no
Remember 2nd is subtracted first
y2-y1
Not y1-y2
1st pair being (8,2) 2nd pair being (10,5)
i have to wirte it down on paper
That's fine
I'd recommend it when starting algebra
Dw tho in a couple more of years of this you'll just start recognizing the slope in like 10 seconds of staring at 2 points
im in grade 12 đ
Tf
i just have horrible memory
Aw hell
I JUST NEED A REFRESHER
Calc
Huh this seems easy
and this unit goes back to grade 9
the unit before was worse
Basically I took algebra 1, algebra 2, geometry, and precalc
and its just one quesiton form my assigment so its easy
sadly precalc only reviewed like
6 weeks of algebra 2
And we got so much shit to cover
are you american?
Yup
Bro you need all the way up to algebra 2 to graduate
Basically algebra 1 is abt lines and graphing and all that
we just have grade 9 and 10 math, functions, advanced functions, data management and calc
aperently theres another one too
is this grade 9 math
Yeah Algebra is grade 9 math
And what we're doing rn isn't even algebra tbh
It's more like intermediate
oh okay
Did you write the problem down btw?
yes i did and i still got 2-5/8-10
i feel bad, after advanced functiosn im done math (until uni)
This isnât wrong
ikm confused then
For the slope between t=8 and 10
Why doesn't he just do it y2-y1/x2/x1
she đ
Itâs the same
đ
nyways yes what's the answer once you do it like that đ
Still sad you didn't plug it in My way-Frank Sinatra âąïž
She already did the problem given in the hw
đ
would you like me to do it yoru way
It makes no difference
The important thing here isnât memorizing a formula the right way. You need to understand the concept of slope, regardless of which formula you use
since then you don't gotta cancel it
I mean the formula is p simple
I'd be more worried abt
Point slope
Slope intercept
Standard/General form
okok i understand it i jsut kidn of have probnlems of when i use a formula
how ocme
also would it be okay if i ask for more help
They're confusing imo
They all sound the same
and ofc
I'm not even a helper
Urgle is
If you ever appreciate someone helping
You can nominate them in a channel lol
Those are just more formulas to memorize, and the more important thing is understanding what pieces of information distinguish a line
OH and they almost always intertwine within questions
Sometimes you gotta go from one to the other to the standard (something like taht I lowkey forgot it's been a long long while)
are they all plugging in
đ
wait
i think i did slope intercept
i cant remember id have to se a picture
anyway im droppign another question in if that is okay
for this i a(t)=100 (1.04)^t as the formula so i basicalluy wrote 100(1.04)^18
is that correct
i got my final answer to be 202.5816
uh oh
This is asking for instantaneous rate of change, so you need the derivative of A at t=18
scared me
đ
okay, so basically just plugging it in
But it says âestimateâ. Try to find the slope between A(17.9) and A(18.1)
For the exact answer, after you differentiate A with respect to t
oh so just at 18, right?
This is the amount of money. The problem is asking for the rate of change of the money
No, you need two points to find a slope
i have to graph it?
Which is why you can try two points very close to A
No, just find the difference between A(17.9) and A(18.1) and divide it by 0.2
how do i find the difference between A(17.9) and A(18.1)
Find A(17.9). Next, calculate A(18.1). Finally, take the difference between those two values
No problem, take it step by step. What is A(17.9)?
into the formula
Yes, plug t=17.9 into A(t)
a(17) is her age, right?
No, t is her age. A(t) is the money she has at that age

oh ok so what i do is i make a seprate equation for both a(17.8) and one for a(18.1)

Yes, find those two values separately
No, you want the exponent to be 17.9. Because thatâs what t is
so a(18)=100(1.04)^17.9
A(17.9), not 18
a(17.9)=100(1.04)^17.9
Yes. After you find that value, do the same with 18.1
oky, i will solve that right now
i got 201.7886
now ill start 18.1
a(18.1)=100(1.04)^18.1 = 203.3777
203.3777-201.7886=?
Yep. Going back to the formula, this is like the âdelta yâ
Ok, now divide that by 0.2. Remember, thatâs like the âdelta xâ
7.9455
Ok, so thatâs our estimate
is that the full quesiton
Yeah, itâs only asking for an estimate. It doesnât specify the level of precision we need, so I guess 7.9455 thousand dollars per year is a decent estimate
Would you like to see how close the estimate was to the exact rate of change?
yes please
ohh okay okay
First we need to find Aâ(t). Using the rule giving the derivative of an exponential function, we get Aâ(t) = 100 * 1.04^t * ln(1.04)
what is ln
Natural log; in calculus this is log base e
i dont think my calculator has a log
This rounds to 3.9221 * 1.04^t. Plugging in 18, we get 7.9454
So our estimate was pretty close
only one number off
Yep
am i keeping you up?
Not really, but I will sleep soon
please do, thank you so much for the help i really appericate this
Np, if u want to learn more about rate of change/calculus, I recommend khan academy
okay, ill try that out
also
does it help for university math
or does it stop at highschool
Calculus is âuniversity mathâ, but I think it also does offer some diff eq and linear algebra. Not sure tho
But it is pretty high quality teaching imo
I wouldnât worry about what grade the material is considered to be tbh, itâs not important at all
If u find it interesting, u definitely should study it
well i might nto until university since i dont think i need grade 12 calculous
wow i never knew that really
Ah I see. In ur case I think it wouldnât hurt at all to begin studying calc, but it would likely take quite a lot of effort
Itâs fine to wait until uni
yeah it woule be best for me
to study earlier