Algebra 2 (Eureka Math/EngageNY)
- Interpreting change in exponential models
- Interpret change in exponential models
- Interpreting time in exponential models
- Interpret time in exponential models
- Constructing exponential models
- Constructing exponential models: half life
- Constructing exponential models: percent change
- Construct exponential models
- 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
Sal analyzes the rate of change of various exponential models, where the function that models the situation needs some manipulation.
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- i watched this so many times and still dont get none of it(70 votes)
- you see 1.35^t/6+5=1.35^5times 1.35^t/6, this is logical, because 1.35^5 times 1.35^t/6=1.35^t/6+5
and (1.35^1/6）^t is also logical, bacause he moved the t outside the parentheses, and so the answer is the same, for example if t=5, 5/6= 0.83, and 1/6 times 5/1=5/6,so this is also logical,
but I don't really understand the last step(2 votes)
- At3:51, I repeated the last part many times yet I still don't get it, I just don't understand how 5% is the answer(12 votes)
- The common ratio is 1.051, which means that every time t is increased by 1, M(t) is multiplied by 1.051. Since 1.051 is larger than 0, M(t) will be increasing. 1.051 = 105.1%, and since anything multiplied by 100% is itself, M(t) will increase by 5.1% (105.1-100) every day. Hope this helps!(37 votes)
- can someone please explain this video in simpler terms? like a step-by-step process? thank you.(5 votes)
- So basically what we want to find in the video is the common ratio for 1 day but we are given the common ratio for t/5 +6 days. By using the properties of exponents. we already know, Sal simplifies it in a way such that the power of the common ratio is 1. Hope you find this helpful.(16 votes)
- Does this mean that the initial mass of the sunfish is 1.35^5 = 4.48...?(5 votes)
- Anyone who doesn't understand this lesson needs to go back to Unit 6 and fully master Lesson 4 of Unit 6 (may seem frustrating at first, but you will get the hang of it).(6 votes)
- at this point the hardest part isnt the math but just reading the problem and answering in the correct units lol(6 votes)
- Can someone give me a few examples of how to solve problems like this? Im kinda-very stuck on it...
- Here is an example taken from the exercise following this video:
A sample of an unknown chemical element naturally loses its mass over time.
The relationship between the elapsed time t, in days, since the mass of the sample was initially measured, and its mass, M(t), in grams, is modeled by the following function:
The sample loses 1/3 of its mass every ____ days.
Let's look at the different parts of the equation.
900 grams is the sample's initial mass.
Multiplying that by (8/27)^t means that by the end of each day only 8/27 of the amount that was there at the beginning of that day is left. In other words, the sample loses 19/27 of its mass over the course of one day. This is more than 1/3, so it must be that the sample loses 1/3 of its mass in less than one day.
For the purpose of this exercise, we can mostly ignore the original mass, 900.
If we can change the common ratio from 8/27 to 2/3 somehow, then we will be able to find out how long it takes for the sample to lose 1/3 of its mass.
This means that after about 1/3 of a day 2/3 of the mass that was there at the beginning of the day remains. In other words, the sample loses 1/3 of its mass approximately every 0.33 days.
Hopefully that was helpful. If there is anything unclear or anything you don't understand about this, then you can ask about it.(7 votes)
- Could someone please explain why my strategy didn't work? I made a table with values for t (0 days, 6 days, 12 days) which made it clear that every six days we multiply the mass by 1.35, which is the same thing as adding 35%. Then I thought that if we add 35% every six days, then if we divide 35 by 6 we'll get the percent we add every day. But then I got approximately 5.8% as a result instead of 5.1%(2 votes)
- Your strategy assumes we add the same amount every day, which we don't (that would be linear growth, not exponential). You computed the average amount added per day over those six days, but because the growth is exponential, we have less growth in the beginning and more growth later.(2 votes)
- How and why does the expression have an initial value of 1.35^5? If the common ratio of the growth factor would be 1.051^t and it is growing since birth, where did the initial value come from?(2 votes)
- The expression above is said to model its mass "since an ocean sunfish is born". Apparently, the fish is born with some positive mass, which is the value of the expression at the time t = 0, i.e. 1.35^5 kilos (or grams, or some other unit of weight). Representing an initial value under a common exponent with a growth factor is just a trick to make the expression more compact (and, actually, more sophisticated). This is generally doable, as any positive number (initial value) may be represented as some exponent of any other positive number (growth factor). In this particular case, the initial value is first represented as a 5th power of the growth factor and then they two are blended under the single exponent of (t/6+5).(2 votes)
- [Voiceover] Ocean sunfishes are well-known for rapidly gaining a lot of weight on a diet based on jellyfish. The relationship between the elapsed time, t, in days, since an ocean sunfish is born, and its mass, M of t, in milligrams, is modeled by the following function. All right. Complete the following sentence about the daily percent change in the mass of the sunfish. Every day, there is a blank percent addition or removal from the mass of the sunfish. So one thing that we know almost from the get-go, we know that the sunfish gains weight, and we also see that as t grows, as t grows, the exponent here is going to grow. And if you grow an exponent on something that is larger than one, M of t is going to grow. So I already know it's going to be about addition to the mass of the sunfish. But let's think about how much is added every day. Well, let's think about it. Well let's see if we can rewrite this. I'm going to just focus on the right-hand side of this expression. So 1.35 to the t/6 plus five. That's the same thing as 1.35 to the fifth power, times 1.35 to the t/6 power. And that's going to be equal to 1.35 to the fifth power, times 1.35, and I can separate this t/6 as 1/6 times t. So 1.35 to the 1/6 power, and then that being raised to the t-th power. So let's think about it. Every day, as t increases by one, now we can say that we're gonna take the previous day's mass, and multiply it by this common ratio. The common ratio here isn't the way I've written it. Isn't 1.35. It's 1.35 to the 1/6 power. Let me draw a little table here to make that really, really clear. And all of that algebraic manipulation I just did is just so I could simplify this, so I have some common ratio to the t-th power. So t and M of t. So based on how I've just written it, when t is zero, well if t is zero, this is one, so then we just have our initial mass, it's going to be 1.35 to the fifth power. And then when t is equal to one, it's going to be our initial mass, 1.35 to the fifth power times our common ratio, times 1.35 to the 1/6 power. When t equals two, we're just gonna multiply what we had at t equals one, and we're just gonna multiply that times 1.35 to the 1/6 again. And so, every day, we are growing by our common ratio, 1.35 to the 1/6 power. Actually, let me get a calculator out. We're allowed to use calculators in this exercise. So 1.35 to the, open parentheses, one divided by six, close parentheses, power, is equal to, I'll say 1.051, approximately. So this is approximately 1.051. So we can say this is approximately 1.35 times 1.051 to the t-th power. So every day, we are growing by a factor of 1.051. Well growing by a factor of 1.051 means that you are adding a little bit more than 5%. You're adding 0.51 every day of your mass, so you're adding 5.1%. And if you're rounding to the nearest percent, we would say there's a 5% addition to the mass of the sunfish every day.