- Interpreting change in exponential models: with manipulation
- Interpret change in exponential models: with manipulation
- Interpreting change in exponential models: changing units
- Interpret change in exponential models: changing units
- Exponential models: FAQ
Sal analyzes the rate of change of various exponential models for different time units by manipulating the functions that model the situations.
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- I noticed that about half way through the video, when the table is filled out, the initial value of 315 was written as 3.5. I suspect this was a mistake and the value Sal meant to say was 315 the whole time?(42 votes)
- If you are watching in fullscreen, you can exit fullscreen and see that there is a correction box at the bottom right-hand corner that lets you know he messed up. Happens all the time. EDIT: Just realized this is 7 years late...(10 votes)
- Given the phrasing of the question, wouldn't the answer be 1.01? I'm looking to see if this question comes up in the quiz, and if we are marked wrong because of putting in too many decimal places. Sal didn't prep very well for this one.(22 votes)
- What I did was said that if in 10 years it went up 6%, then yearly that means it goes up 0.6%, which is close to the final answer of 0.58%. Was what I did acceptable?(10 votes)
- No, because it's not a linear function. The error in reasoning can be seen more clearly if we use a more exaggerated example. You can't say that if in one year you get 100 percent increase (or double the original), then in 10 years you'll get 1000 percent, or 10 times as much. Try doubling a number 10 times and see the difference.(13 votes)
Isn’t it 315?(7 votes)
- Using this logic, can we solve compound interest problems involving fractional periods?(4 votes)
- Yes. You can manipulate the exponent t to however your situation is. For example, if it happens every year, you can put (t/365) for the exponent. Likewise, if it's every month, you can put (t/12) — however, be wary of how the question uses the variable.
Hopefully that helps !(4 votes)
- Where did he get the 3.5?(3 votes)
- So, when I tried to model the function in terms of years instead of decades, intuitively I thought that it should look something like this:
a_years(t) = 315 * (1.06)^((1/10)t)
because in order for a_years to output an equivalent answer to a_decades, you should need to input 10 times more input into a_years. 1 year is a tenth of a decade, after all.
but Sal's model is slightly different
a_years(t) = 315 * (1.0058)^t
The first model is correct, right?(1 vote)
- So how would I do this for days? For example if the problem was
Mweek^(t)= 15 x (0.79)^t(3 votes)
- You could replace the week variable with a day variable, and then change the formula to divide it by seven. The model you are creating represents growth per week, so I'm pretty sure you have to convert the days into weeks somehow. There may be another more complicated solution, but if you're okay with changing the way that the formula looks, this should work. Hope this helps!(1 vote)
- [Voiceover] The amount of carbon dioxide, CO2, in the atmosphere increases rapidly as we continue to rely on fossil fuels. The relationship between the elapsed time, t, in decades, let me highlight that, because that's not a typical unit, but, in decades, since CO2 levels were first measured, and the total amount of CO2 in the atmosphere, so the amount of CO2, A, of sub decade of (t), in parts per million, is modeled by the following function. So, the amount of CO2 as a function of how many decades have passed is going to be this, so t is in decades in this model right over here. Complete the following sentence about the yearly rate of change, the yearly rate of change in the amount of CO2 in the atmosphere. Round your answer to two decimal places. Every year, the amount of CO2 in the atmosphere increases by a factor of? If they said every decade, well, this would be pretty straightforward. Every decade, you increase t by one, and so you're going to multiply by 1.06 again. So every decade, you increase by a factor of 1.06. But what about every year? Now, I always find it helpful to make a bit of a table, just so we can really digest things properly. So I'll say t, and I'll say A of t. So, when t is equal to zero, so at the beginning of our study, well, 1.06 to the zeroth power is just going to be one, you have 315 parts per million. So, what's a year later? So, a year later is going to be a tenth of a decade. Remember, t is in decades. So a year later is 0.1 of a decade. So, 0.1 of a decade later, what is going to be the amount of carbon we have? Well, it's going to be 315 times 1.06 to the 0.1 power. And what is that going to be? Let's see, if we... so 1.06 to the, so to the 0.1 power, I didn't have to actually use the parentheses there, is equal to 1.0058, I'll just take it with that. 1.0058. So this is the same thing as 3.5 times 1.0058. And I should say approximately equal to, I did a little bit of rounding there. So, after another year, so now now we're at t equals 0.2, we're at two-tenths of a decade. Where're we going to be? We're going to be at 3.5 times 1.06 to the 0.2, which is the same thing as 3.5 times 1.06 to the 0.1, and then that raised to the second power. So we're going to multiply by this 1.06 to the one-tenth power again, or we're going to multiply by 1.0058 a second time. Another way to think about it, if we wanted to reformulate this model in terms of years, so per year of t, it's going to be 315 and now our common ratio wouldn't be 1.06, it would be 1.06 to the .1 power, or 1.0058, and then we would raise that. Now t would be in years now. Here, it is in decades. And I can say approximately, since this is rounded a little bit. And so, every year, the amount of CO2 in the atmosphere increases by a factor of, I could say 1.06 to the 0.1 power, but if I'm rounding my answer to two decimal places, the, well, we're going to increase by 1.0058, in fact, they should, the increase is by a factor of, they should, I'm guessing they want more than two decimal places. Well, anyway, this is arguably, this right over here is five significant digits. But anyway, I'll leave it there.