Exponential Decay / Finding Half Life
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- Опубліковано 10 вер 2024
- Thanks to all of you who support me on Patreon. You da real mvps! $1 per month helps!! :) / patrickjmt !! Exponential Decay / Finding Half Life - In this video, I find the half life of a substance that is decreasing annually by 4%.
For more free math videos, visit PatrickJMT.com
Thanks so much, these videos you put up are not in vain. Students enjoy someone who is willing to teach us without pay. What amazes me is the fact that you do this out of your time to help anyone who needs help with math. Thankyou Patrick!!!
*ad* : are the videos from 2006 really heloing?
me: YES
@Deeya Mulchandani same
Very very helpful, thank you! Also, the bracket at 4:00 is a work of calligraphy, bravo! Mine are rubbish.
There are a lot of left handed math tutors on youtube.
Great job of explaining decay. Might call it an isotope to be more correct.
Thanks so much for this! I missed a week of school and came back on a test day. I learned the entirety of the material with just two of your videos. Thanks, man.
boo ya!
i always count the number of left handed people in my classes...
Our teacher showed us a different way, but this is much easier, I hope he accepts this method on the test. Thanks, you rock and keep up the good work!
@yeijie456 the formula you have is for a CONTINUOUS decay rate, mine is for ANNUAL. those two words should be key words to help you decide which formula to use. :) ask your teacher for some clarification if you are not sure!
your method is so much easier than y(t)=y(0)e^kt
that's what i was thinking :o
If I could be granted one wish, I would wish for PatrickJMT to teach at my school because he's a very effective teacher--very easily understood! KUDOS! :)
how do I solve for the annual decay rate if I'm given just only the half life which is 22 years, what do I do or how do I solve it?
Would you mind making a video with one of these problems using the N(t)=N(initial)*e^rt version of the equation? My instructor is picky with how we solve the problems. Great video though! At the least I can use the answer as a reference to verify I'm doing things properly.
So for finding any half-life, couldn't we just start with:
ln (1/2) / ln (1+(constant annual decay rate)) = t
And just plug in the constant annual decay rate?
you are not multiplying in either case. and it does not matter what base you use.
I'm writing a test on Tuesday and you're saving my ass rn with these videos. thanks, man!
@duckhero3 let a = some amount, let y = half of that amount, solve for t.
10/10 your explanation is for some reason much easier to understand than others....
@lmaominh sorry to hear that. my 8th grade science teacher, mr cox (now there is a name you can make fun of) told me i was a nerd cause i knew the answer to one of his questions. lots of weird people out there, teachers included.
Why is it divided by 100 at 3:10 is it cause 100 is the highest number or because it's an easy number to throw out there??? Pls help asap
why do you multiply the equation by LN and not by Log? My teacher taught me that when you have an exponential equation with different bases you can multiply by log, so i don't know when i should multiply by Ln or Log.
@KittyKatCurls this is annual decay rate; you are using the formula for continuous decay rate
THANK YOU PATRICKJMT def showed me a flaw in a project i was about to hand in. Freakin life saver right here!
"I like using numbers because they make more sense to poor Patrick" XD
Thanks you so much dear professor for precise formula.
@Stairs3000 he's the Patrick of math.
thank you for the awesome job u do via the channel. don't wait to much in getting ur phd since having u as a university teacher would be great, because u know how to explain math very well and thats what we need in our schools. thank u n don't stop making this math videos
Half Life 3 never came out
Your video really saved me today with my homework, thank you so much
So can anyone help me with a problem? I take a dosage of something every 24 hours, and the drug has a half life of 20 hours, how would I find out how much is in my system after x hours? I don't know how to handle it because I'm not just taking the initial amount, but I'd be topping up every 24 hours.
thanks so much my math teacher told us to learn and research this ahead so i know what we are going to be talking about tomorrow!
no problem mr. nick-bucket!
i would ask first!
sometimes, it is about the process - not about the actual answer (well of course, the answer is important, but they are trying to teach a process in algebra class).
your instructor may be trying to emphasize something that this procedure leaves out ! (not sure what it would be, but again, i would ask to make sure)
trying to look up half life related tutorials on youtube = searching through endless videos of Half Life (The video game...)
I use the rule that says N(t)=N0 e^rt but the one u used is more clear. Thank you thats really helpful.
you are an excellent teacher but this Half-Life method is seeming to confusing for me. my teacher shows us a chart method whree you have 3 rows in the chart. Number of half lifes/Time elasped/Amt. remaining. Do you know this method?
@aka7minimouse it just sounds like your teacher is trying to make some systematic procedure for people to follow.
@Jmerino7 there is no calculus in this video
@saifirefly well, exponentials would be taught in any algebra class
Well, part of this is right. He forgot that when you use natural logarithms you must write your equation using in the proper form using e. He should have been using log base 10 for the form he used.
What about finding the rate of decay of a half life if, for example, its 400 years
My chemistry teacher makes us use k(rate constant)=.693/t 1/2(half life). So the problem you showed in this video was "what is the half life", but what if it asked for like the remaining on a substance or how old is it? then what formula do you use? My chemistry teacher makes us use this one: Ln(natural log) Nt(remaining) / No(initial mass) = -Kt
I am really confused, my problems that our teacher taught we use y=Ce^xt
Why is it that there is no e in the formulas you used for this video ?
carbon 14 has a half year of 5730years an initial sample has a mass of 70mg, how much is left after 171900 years? how many half lives have gone? what is the mass of the daughter isotope?
I got lost at the natural logarithm part....why did you have to use it and how did you get those numbers? (- .693147 / - .040822)???
My professor is using a different process and I'm lost. She's using Qbase0 e^-kt.
She didn't explain. It's with chapter of Logarthmic Functions if that helps any. I'm assuming in later converts into the natural logarithms you have here. Is there an example you can show me for this?
Question looks like: (Find the associated exponential decay or growth model) Q=1000 when t=0 and half-life =1.
Another Question: Find half-life or doubling time when Q=1000e^0.5t
Please help math genius!!
i don't get it. isn't this too easy to be calculus? it is in your second semester calc playlist but i don't see any derivatives or integrals used to solve these.
it shouldnt be in the calculus playlist
patrickJMT oh okay. I'm just confused because I'm self studying for ap calculus bc and one of the topics is exponential growth and decay
I came here for a calculus problem as well lol
Thank you so much!!! I never understood this in my two years of calc, but this helped so much!
I still don't get it, if half life = the amount of time for a substance to get to half of it's original start, and in this problem it's is decaying a constant anual rate of 4% we want to get find out how many years will it take that 4% to equal 50% or 1/2... Why not just make it 4X = 50 in which x represents the years and 50 is the amount lost after completing one half life (100g/2), then I get 12.5 years as the "half life", what am I doing wrong, I know the formula and how to solve it, I just don't understand why it works, plz help
What if the constant annual rate is a number bigger than 1? Making it a negative result and if I'm correct (unlikely) then I can't use logarithms.
@patrickJMT so does this meam i shouls solve it like 0.5a=ae^8.56t and then tjat would be ln0.5/ln8.56?
My final sample exam has this problem on it:
You have a radioactive material which decayed by 20% in one
hour. By which percentage will the material decay in 2 hours?
How exactly would I go about doing this? Is it as simple as it sounds or is there more to it? I may have figured it out simply with a ratio problem using this formula(excluding the variable A and making y=x^2), getting 11% as the answer, but I do not think that is correct.
@MechInvent my name is not youtube
you saved my life. words cannot express how amazing you are!
Strontium-90 has a halflife of 29 years. If there are 400 grams of strontium-90 initially, how much, to the nearest gram, will remain after 87 years?
+juliya smith hella
thank you! o A o but uhmm caan i ask....do you use LOG with half-life?
the formula my teacher gave us is A=Ao(1/2)^t/h
A= amount left after given time
T=Time
h= half life
Ao=initial amount
o 3 o
Ok, I think I actually get it now. there's an r in both formulas, y=a(1+r)^t and y=ae^(rt), and they're BOTH growth/decay rates....BUT if you put the same number "r" in both formulas, you get (slightly) different functions. So, that's why you look for the word "annual" (or some other fixed unit, like "daily" or "hourly") , or "continuous", before the words "growth/decay rate". "Annual" tells you to use y=ab^t (or y=a(1+r)^t ) "Continuous" tells you to use y=ae^(rt).
hello this vid helped me super mucho but have a conflicting issue on how to connect that into this problem: radioactive iodine decays accorindg to the equation y=ae^-0.0856t where t=days...how should i find the half life here? is the percentage decay 8.56%? nd then go from there likeyour vid?
Is the method pe^rt very good? where p is the number that we start with and r is the rate and t is the years....
sir, can you explain the calculation of rate of decay with half-life given?
what if the t has a fraction in it?
our teacher did`nt show us by log method
this seems easy but we havent studied log yet
she showed us by dividing 100 by 2 then continuing
I'm stuck , why does my math teacher always tell me to do 0.5 = 1e^kt ?
is anyone else having problems with the video? it stops loading in less than the first minute..
Extraordinarily helpful. Thank you!
@VerifiablyNoticible put a math term in the search as well next time : )
I have a question. My pre-calc teacher never really taught us logs and ln so I'm now I calc and lost beyond belief. Your example helped but what else is out there?
This still confuses me as well. Why wouldn't y=Ce^(k
t) and y=Cb^t give you the same equation, just with b=e^k? Are they actual different situations or just different ways of finding the same thing? Please help :)
I take my final today adn this has helped me a lot thanks
So just wondering, what is the difference between this formula that you teach, and A = Pe^rt? Are they the same? Or if not, under what circumstances would you use each equation? Thanks for the help!
Also, when I calculated it, 10+20+40+80+160+320+640=1270 and ur answer is 1280
do I have a mistake in calculating??
thanks pat
The short cut I usually use is T1/2=ln2/decay constant. Gives results in seconds.
Hey Due u are the secret of my success in math
thanks
I’ve been trying to find videos to find what’s wrong with me, so I’m watching the half life calculus, because I am living half the life I should be living
can someone help me? I have to do a word problem that only gives me the remaining % and the time.
"if 82% of a sample remains after 2 hours, find the decay rate."
come up with any amount of the sample you want, I would do 100 to make the math easier, that's your initial amount (a in the formual Patrick uses), then your current amount (y) would be 82% of that so if you use 100 for the initial it's just 82, then you know the t is 2, using the y = a(1+r)^t formula you can then algebraically solve for r.
your a genius dude, you make everything so simple
i spend money to not learn this in college and learn it on youtube for free! Thanks dude!
important!! for growth and decay, the formula my teacher gave me is A(sub t)=A(sub o)e^kt
is there a difference? or is this a different formula alltogether
Half-Life 3 confirmed.
Finnaly a half life comment 💀
what if its decreasing every 7 years, where would plug that in?
Thank you so much! That was very helpful!
Shouldn't it be a radioactive isotope, instead of ion?
you can use either one
In- natural log
log- base log
so you would the reason why he uses In instead of log is because it is easier to work with..so... yh that's the reason!!
Awesome explanation. Thks!
is this possible without log rhythms?
yes
@arcstormtroo natural log. LN, look on the calculator
Man you are leading me through college. Thank you for your videos!
Half life, full consequences
Hey Patrick. Do you have any videos on continuous growth/decay rate. I am learning exponential growth and decay but with the formula, y(t)=y0e^kt (y0 = y sub 0)
Thanks a lot! if only my math teacher were as good as you!!!!
Thanks Bro... Thumbs Up... I Hope You're my Professor in my Mathematics subjects.. tsk..
What costs $27 million per gram and can be
used to treat brain cancer, analyze coal for its sulfur content, and
detect explosives in luggage? The answer is californium-252, a
radioactive isotope so rare that only 8 g of it have been made in
the western world since its discovery by Glenn Seaborg in 1950.
The half-life of the isotope is 2.645 years-long enough for a
useful service life and short enough to have a high radioactivity
per unit mass. One microgram of the isotope releases 170 million
neutrons per second.
a. What is the value of k in the decay equation for this isotope?
b. What is the isotope’s mean life?
wao
im confused to what deos ln mean? sry im only in physical science!
Hi patrickJMT! Why use Natural logarithm instead of Common logarithm? Thanks for the video tho. :)
Thx so much!!!
Thanks Sir!
that help me so much. i was so stuck on my home work. thanks
Is episode 3 hidden in this equation?
Thanks helpful
great video. u helped me alot