If we compare linear and exponential growth, we will see that over time, *any* exponential growth will surpass *any* linear growth, no matter how steep it is.
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- I don't understand why Sal didn't express the rates of growth in terms of functions. I tried taking that approach to solve a practice exercise on Kahn Academy, but my results don't match up with the ones on the website.
Here is the practice exercise:
Sheila is a wildlife biologist. Her daily task is to count the number of wild turkeys and white-tail deer in a large game reserve.
- Sheila counts 12 wild turkeys by the first hour after sunrise, and the cumulative number of turkeys she has counted increases by approximately 40 percent each hour.
Sheila counts 18 white-tail deer by the first hour after sunrise, and she counts 10 deer each hour after that.
In which hour after sunrise will Sheila's cumulative count of the turkeys first exceed the cumulative count of the deer?
I will represent the exponential growth in turkeys with f(x) = 12 times 1.4^x
I will represent the linear growth in deer with f(x) = (18)x + 10
Using those functions, I calculate that by 2 hours after sunrise, there will be 23.52 turkeys (since 12 times 1.4^2 = 23.52) and 46 deer (since 18 times 2 + 10 = 46). However, when I ask Khan Academy for hints on this practice problem, I am told that there are 17 turkeys and 28 deer by 2 hours after sunrise.
I don't understand why my functions did not produce the correct results. I also don't understand why the use of functions is not encouraged to get results.(13 votes)
- David - Per your request, here's my take.
Your functions are assuming that the clock starts after she does here initial count. You can't make that assumption. You need to account for the 1st hour as noted in the other responses the you received.
There is 12 turkeys that increase by 40% after the end of the 1st your.
Let x = times. Your exponent needs to be x-1 so that the 1st hour is not increased by 40%. You want only hours 2 and above to increase by 40%. Thus, the function should be:
T(x) = 12*1.4^(x-1)
T(1) = 12*1.4^(1-1) = 12*1.4^0 = 12*1 = 12
T(2) = 12*1.4(2-1) = 12*1.4 = 16.8, or when rounded 17 turkeys.
There is 18 deer that increase by 10 each hour after the end of the 1st your. The 18 is your starting point. The 10 is your slope. But, like with the other function, you need to adjust time to account for the 1st hour.
Let x = times. You multiply 10 by x-1 so that the 1st hour does not increase by 10. You want only hours 2 and above to increase by 10. Thus, the function should be:
D(x) = 18+10(x-1)
D(1) = 18+10(1-1) = 18+10(0) = 18+0 = 18
D(2) = 18+10(2-1) = 18+10(1) = 18+10 = 28
Hope this helps.(26 votes)
- how do i solve this?
A(t) = 10,000 + 5,000t
B(t) = 500 * 2^t
500 * 2^t > 10,000 + 5,000t / :500
= (500 * 2^t) / 500 > (10,000 + 5,000t ) / 500
= 2^t > 20 + 10t
from here i am stuck and i am not sure how to solve it.
please help.(13 votes)
- This lesson doesn't expect you to know how to solve that algebraically. I don't know how to do it either. A(t) is a linear equation. B(t) is an exponential equation. The answer you want is the value of t that makes the two equations equal (which corresponds to the intersection of the two graphs). How do you solve for t when one of them is the exponent and one is not? I don't know. If you Google "finding intersection of linear and exponential functions," you will find numerical methods (not algebraic) for solving this kind of problem to an arbitrary precision. I'm just reviewing Algebra so I'm not watching many videos, but if he didn't, Sal should have mentioned this problem (because any thoughtful student of math would wonder the same thing as you).(10 votes)
- Is there an equation that can be used to solve this quicker, or do you have to draw the table?(8 votes)
- I know this is kind of late but I have a method that worked for me, however it involved a calculator. I used the geometric and arithmetic explicit equations and just plugged in the values from the problem into the graphing calculator. Then I looked into the table on the graphing calculator and found my answer.(3 votes)
- Is there a faster way to solve a problem like this?(5 votes)
- Samuel you could could graph it and record the intercept of the two lines but this way is simpler because in order to graph the equation you would have to make a table anyway.
Hope I helped,
Xavier Jalem(2 votes)
- So I was thinking instead of going through each x-y pairs to check when the excess happens, how about writing the inequality:
Let m = # of months
10,000 + 5000*m < 500 * 2^m
how can I solve for m here? Do I not have enough math tools to solve this inequality? If not, can you show me how to solve, perhaps using system of equations instead?(3 votes)
The last equation is the furthest I have been able to simplify with both your equation and someone else's. I'm not sure if with all the math tools learned from this point in math and prior it'd be possible to further simplify this inequality, but I am curious about your thought on using system of equations. What 2 (or more) equations are you suggesting we use? (I have seen this idea come up a lot so I'm quite curious)(4 votes)
- Is it possible to do this with an equation instead of graphing it each time? Thanks :)(4 votes)
- Is it cheating If I use these videos to get answers for the practices?(1 vote)
- If you're using the video to learn how to answer your questions, then no that's not cheating. However, it's cheating if you use the video to ask people to do your homework for you.(6 votes)
- [Instructor] Company A is offering $10,000 for the first month and will increase the amount each month by $5,000. Company B is offering $500 for the first month and will double their payment each month. For which monthly payment will Company B's payment first exceed Company A's payment? So pause this video and try to work that out. All right, let's work this out together. So let me set up a little bit of a table. So this is going to be, the first column I'm gonna have is month. The first column is month. The second column is how much Company A is going to pay. And then third column, let's think about how much Company B is going to pay. Well, they tell us a few things. They say Company A is offering $10,000 for the first month. So in month one, Company A is offering $10,000, we'll assume, well, I'll just write the dollars there. And then Company B is offering $500 for the first month. $500 for the first month. But then they tell us Company A is offering, will increase the amount, each month by $5,000. So month two will be 5,000 more, we'll get to $15,000. Month three we'll get to $20,000. Four, we'll get to $25,000. Five, I think you get the point, we'll get to $30,000. Six, we'll get to $35,000. Seven, we'll get to $40,000. Let me scroll down a little bit. Month eight, I'll stop there. Month eight, we will get to $45,000. Let me extend these lines a little bit. Now let's think about what's gonna happen with Company B. Company B is offering 500 for the first month, but will double their payment each month. So the second month is gonna be double that. So that's going to be $1,000. Then we're gonna double that again, $2,000. Then we're gonna double that again, $4,000. Double that again, $16,000. Double that again, $32,000. Double that, oh, I skipped one. I went from 4,000 to 16,000. 4,000, $8,000. Then we double it again, $16,000. Again, 32, I sound like my two year old again. All right, $32,000. Then we get to $64,000. And at that point something interesting happens. It's actually good that I went to the eighth month because every month before the eighth month, Company A's payment was higher until that eighth month. In that eighth month, Company B is going to pay more. So first we can just answer their question. For which monthly payment will Company B's payment first exceed Company A's payment? Well, that is month eight, month eight. And there's a broader lesson going on here. You might recognize that the rate at which Company A's payment is increasing is linear. Every month it increases by the same amount. So plus 5,000, plus 5,000. It increases by 5,000 the same amount. Company B is increasing exponentially. It's increasing by the same factor every time, so we're multiplying by the same value every time. We're multiplying by two, we're multiplying by two, multiplying by two. And so there's actually a very interesting thing here that you can actually make the general statement that an exponential function will, we'll want something that is exponentially increasing, will eventually always surpass something that is linearly increasing. And it doesn't matter what the initial situation is, and it also doesn't even matter that rate of exponential increase, it will eventually always pass up something that's increasing linearly. And you can think about that visually if you like. If I were to draw a visual function, a linear function, so this is x-axis, this is the y-axis, a linear function, well, this is gonna be described by a line. So it could look something like this. A linear function is always gonna be a line of some slope. And an exponential function, even though it might start a little bit slower, it's eventually going to pass up the linear function. And this is going to be the case even if the linear function has a pretty high slope or a pretty high starting point, if it's something like that, and even if the exponential function is starting pretty slow, it will eventually, and even if it's compounding or growing relatively slow but exponentially, you know if it's going 2% or 3%, it still will eventually pass up the linear function which is pretty cool.