a-sum-of-5000-is-invested-at-an-interest-rate-of-9-per-year-compounded-continuously-a-find-the-val-2

Question

A sum of $\$ 5000$ is invested at an interest rate of $9 \%$ per year, compounded continuously.
a. Find the value $A(t)$ of the investment after $t$ years.
b. Draw a graph of $A(t)$.
c. Use the graph of $A(t)$ to determine when this investment will amount to $\$ 25,000$.

Ankit Gupta · Accepted Answer

and the question we are given. But the principal amount equals $5000. The raid off interest is 9% compounded continuously. When David by the value off 80 after years, that&#x27;s gonna be B E T. That&#x27;s 5000 times evenings, 2.9 B for part, Bigger us to craft on use that year 84 calculator to craft our function. That&#x27;s 3000 e days too. Point Do you know mine? B But we use exit for exactly that independent variable recent does window so that we can see the craft Clearly this is on a brooks. You will know that. Yes, there definitely is on Ah, information. So this is what we get. This is the graph that&#x27;s required and for party here us to find the time at which the amount as $25,000. So we go back to the ground end of I do equals ready 3000 and find the intersection point. So the go through the car window and select intersect option. Find the intersection between these two curved and get it a 17 point in a to 643 We make a note that digs 17.88 years for this amount

Julian Wong · Answer

this problem. We want to come up with the ah function A of t that calculates the compound interest. So, uh, using the same kind of ideas before, instead of calling f ve, we&#x27;re gonna call it a of T. That is the function notation for thea amount of interest in the amount of investment in the account. This is equal to a national value of 5000 times one plus. The rate being told is nine over 100 times ah k, which is the number of compound ng periods for a year. That&#x27;s 12 because it&#x27;s calm pounded monthly. So the power of 12 t and ah, we can simplify that to 5000 times. Ah, nine divided by 1200 plus one would be 1.75 to the power of 12 teeth in part B. We want a graph this ah, this equation dysfunction and then use that function to ah do a cup to do ah estimation for part c So part b, every graft it I&#x27;m just comping pace that here This is the function for part B. We were being told to graph this from 0 to 25 on the X axis. So the red curve is thief function A of t and part. See, ah is asking us to use this graph to determine the minimum number of years for this investment to have a value greater than 20,000. So, you see this blue line here? This is Ah, $20,000 mark and it takes us 15.461 years to get to 20,000. Um, the answer wants us or the question wants us to find the minimum number of whole years, so ah, 15 years is not going to cut. It won&#x27;t get us 20,000. We&#x27;re gonna have to go to 16 years. We gotta round up in this case 16 years. Will, ah allow us to exceed ah investment value of 20,000.

Caleb Miller · Answer

Hey, guys, welcome back. This problem is and interest from $1000 borrowed 8% interests. And there were Newsome, different formulas based on how much the interest is compound. So we know that we&#x27;ll use the formula. A is equal to P Times one over or the end times nt You know that teams air time of year&#x27;s time in years. She&#x27;ll be three years. You know that P is how much money start with or principle in this case at his $1000 here that are is our interest rate. This case is 8% which, if you move that in no decimal, be 0.8 and then we know that end is how many times number of times bring your that your compound ing money times per year. And that is what will change a bunch and all the different parts of this problem. So first we want to see what happens if we compound it annually. So if you do it annually, you know that and will be one time per year. So we&#x27;ll say that a This eagle two 1000 one plus R points your way over end. It won raised to the NT, and it&#x27;s just one. He has three. And if that, you know, value of 1000 200 59 in 71 cents, and that is our answer to part a part B or asked what happens if a compound it quarterly. If its quarterly that be four times a year, we will do the same equation plugging in and as you were before, So we have a 2 1000 one plus 10 over four raised to the fore Times three. And we knew that you get a value of 1200 and 68 24 Next dress. What happens if you do it monthly Its monthly? An end to 12. We&#x27;ll have a 0 to 1000 one plus point. Oh, wait over 12 12 times three. That would come out to 1000 270 point 24 Next rest. It happens if you do it weekly. Do it weekly. There are 52 weeks in a year. We have a 0 to 1000 on plus point away over 52 52 times three. That comes out to $1271 two cents. Now ask what happens if we do it daily? Yes, 365 days in a year. We have a zeal to 1000 one plus point. Oh, wait over 365. Raise to the 365 times three. And that comes out to 1000 270 $1. 20. 22 cents. 20 Then you&#x27;re asked what happens if we do it? Hourly? Do it hourly. That would be 365 days in a year. Times 24 hours in a day Ejected 8000 760 hours in a year. 7000. It&#x27;s our 8760 times get compound. If a is equal to 1000 one plus point, no wait over 8000 160 raised to the 8000 760 times three. That comes out to 1000 200 71.25 Then rest. What happens if you do it continuously? Well, if you do it continuously, that&#x27;s basically saying what happens if an goes to infinity? If n goes to infinity, that are, formula becomes a different simplifies down and it&#x27;s just it a is equal to P. Yeah, the rt fill it in. It would be equal to 1000 e the point. Oh, wait. Times three, we get the exact same value. We had our last part, which is 1271.25 and then finally using our formula A is equal to P E to the rt. You want a graph zero to T years ago. Three zero from T is equal to three different interest rates to on the graph are you 2.6 are 0.8 and our issue 2.1 I want had and I grabbed this on Dismas and that graph can be shown here. I don&#x27;t exercise not cancelled Carry over. This was acceptable to zero Xeloda one exit time. Is he going to I&#x27;m is equal to three in this obviously in years and her wise and dollars you can see how the Green line when are you would appoint one is slightly higher. Our blue line When are you gonna 0.0 waits in the middle and then our lowest is when already lit a 0.6 And that concludes this problem. Thank you for watching

Heather Zimmers · Answer

here we have a compound interest problem, and we&#x27;re going to use our non continuously compound interest formula for the majority of part A. The amount invested is $1000 for a time period of three years, with an interest rate of 8% and that translates into a decimal value of 0.8 And so we&#x27;re going to go through several different amounts of calm poundings per year, starting with annually once per year. We use an equals one, and we get $1259.71. Then, for quarterly, we use an equals four. We substitute that into our formula. We get $1268.24 for monthly and equals 12 and that gives us $1270.24 for weekly and equals 52. That gives us $1271.1. We see the amount of money is growing, but the rate of growth is slowing as we accumulate more calm poundings for 365 compound ings per year, that would be daily. We get $1271.22 an hourly that would be 8760 compound ings per year. We get $1271.25 and for the last part, we&#x27;re going to do continuous compounding, so we use a different formula. The continuous compounding formula is a equals a not e to the R. T. So we still have the same interest rate point await. We still have the same amount of time three years, and we still have the same initial investment, and we end up with $1271.25. Finally, for Part B, we want to graph the continuously compounded function for different amounts of interest rates. So 6%. Let&#x27;s try that again. We have 6% we have 8% and we have 10%. So let&#x27;s grab a graphing calculator. We type those in, and then we&#x27;re going to look at them from time zero to time three so we can go to window and we can change our window dimensions to go from 0 to 3 on the X axis. And then I chose negative 100 to 1600 for my Y axis and then looking at the graphs. The lowest one logically is the one that has a 6% interest rate, and the middle one is the one with 8% interest rate, and the highest one is the one with a 10% interest rate.

Ryan Mcalister · Answer

this question asks us to consider two possible investments. We know we have initial value P equal to $10,000. The two possible investments that we have one of them has a rate of an interest rate of 5% or 0.5 in compounds quarterly or four times a year. The second has an interest rate of 4.5% or 0.45 and compounds 12 times a year monthly. This question at first wants us to write the formulas for each function we know Our non continuously compounding interest function is the amount A equals p times one plus the rate over the number of times that compounds raised to the number of times times time here t will be our independent variable, and they will be your dependent variable. So for the first type of investment, we have a equals 10,000 times, uh, one plus 0.5 over four race to the four times T and the other. We have a equals 10,000 times one plus 0.0 for five over 12 times. 12th race, the 12 times t. So these are two, um, possible investment equations. Let&#x27;s go ahead and put these into a graphing calculator. This is the T 84 and I will be entering this in in y one and y two. The first investment with the higher interest rate we said was 10,000 times one plus, and I put our fraction in parentheses. 0.5 divided by four. All that was raised to four times t in here. T his ex are independent variable. And our second equation was also 10,000 times one plus again. Fraction in, uh, parentheses. 0.0 for five, divided by 12 raised to the 12 times X. And I actually want I think I might have made a type of here. I did 0.5 instead of 0.5 I&#x27;m gonna go in and insert zero here. I think that worked. Yeah. Okay, so now we have our two equations, and let&#x27;s look at the graph. Well, when resumed in like this, we&#x27;re obviously not gonna be able to see anything. So I&#x27;m gonna go to the zoom button and scroll down until I see zoom fit. Zoom fit is gonna, uh, change the zoo so we can see the important stuff we look at zoom fit. We see this. We have our XX is on the bottom. Pretty spread out. Or why access Pretty close. So let&#x27;s go ahead and take a look. We&#x27;ll trace either. This is our first function, the one with the lower interest rate. And you can see there the one with the higher interest rate. Excuse me. You can see his ex increases or what we&#x27;re calling T. It is higher than the one with the lower interest rate. So why one? The one with a 0.5 interest rate compounded quarterly after 1.8 years we see has value of $10,940. Well, the second investment after 1.8 years has a value of $10,846 a little bit lower. So this is ah, how we see the investment to grow. That&#x27;s your final answer.

Yoshihito Takagi · Answer

Hello, guys. So we have this problem here. It&#x27;s a principal problem. Interest problem. Make more sense. Um, so in this interest problem, we have to figure out, um, we have to fill out this chart right here, using this time equation that were given. But what is the time equation for two? Well, um, like, what is the time? I mean, this case So according to the problem at the time, refers to how long it will take for your initial investment to multiply K amount of times. So it&#x27;s hypothetical would be safe to say you invest $1000 right? And you want to find out. How long will it take for this to turn into $2000? Twice as much? Well, you would plug in to Miss Kay and then plugging the so formula. And then you should get this amount of time a size what it refers to so essentially, to fill out this graph, you just have to plug it into a calculator that many times. But this can be a little bit difficult toe plug into your caterpillar, depending what you have, So you might want to rewrite it. So I&#x27;m just gonna do this for the people who who need this to be rewritten to be a bit easier, um, Tiu to simplify it, you could, um, simplify the denominator. So you see how it has a coefficient here? It has a coefficient you can simplify by bringing it as an expert because one of the lager than little states that lug eight of parties say, Miss Love, um, is be times log. Hey, in this case, we could write it as a log e the power of 0.45 All right, And this here is a constant right. So you can just plug this into the calculator, having us the denominator and then plug in different case, and that should be easier. So I&#x27;m gonna plug in some values, and hopefully you should get the same one. So I cook in, huh? Log log. Can I put E to the power of 0.45? And for the expression, I get zero point 01 nine five or I&#x27;m going as far as I can. Um, in case you don&#x27;t have an e, but in which I&#x27;m sure most of you do. But just in case you don&#x27;t. According to my calculator, E is 2.7 18 two eight Casino. We have this constant, and we have the top lock, Kate. So we just have to plug in law, okay? Every time. So I&#x27;m gonna do that log and then the first case, one the bottom we put, um, her number, which is Sierra 0.19 fight for. And we get zero. Which makes sense, because this is saying that how long will it take for your, um, investment to be the same will take no time at all. So this checks out now. Go back to it and plug in the number two. It&#x27;s F one. Um, Now it says three decimal places. I&#x27;m gonna write a three decimal places. It&#x27;s gonna be Yeah, teen 0.4 zero. Bye. Eight. Since the fourth place is an age, that means I can round up the five, so I&#x27;ll make it a six. I&#x27;ll just take that. Get put it in there. Okay? So that should be your second. So this, um, units are in years, so that takes 15 years for this to duck, which is, uh, I guess pretty good&#x27;s a penny. Watch your interesting but, uh, let&#x27;s go for three. So go back looking three. And this time we get 24 point for 17 66 is greater than five second round up to seven and make it an eight. So that&#x27;s, er next value, and it should be increasing because it doesn&#x27;t make sense that you are well, investment would start decreasing for any reason because it&#x27;s not outlined and issue this is it compounded. Continue sleep. So that implies it&#x27;s always increasing. They should never decrease. Good way to double check. Go back plug in four. Now I get 30 0.8 one 166 is greater than five. You can place the one with two. So that&#x27;s and four boot before Yeah, for five. Same thing. Dirty five 0.77 one and two. But two is not greater than five. We just don&#x27;t need it, and it&#x27;s for that now. I hope you see somewhat of a pattern here. It seems to be increasing less and less every time. That might be something relevant in real life problems, so it&#x27;s always good to look at what the trenches. But now all of a sudden, we&#x27;re jumping to 10. So make sure you don&#x27;t accidentally do six. I would probably do that. Accident Me. So now we have 51.177 00 was obviously less than all fire. So you can get just get it. And that&#x27;s definitely bigger number. But also realize that it&#x27;s a bit big jump for K for to be 10 times and take 51 years. Um, just two more times six. Did you 6.5, Kate two 99 is definitely bigger than five. So that comes with three. Did you get that? You&#x27;re different in last one. They&#x27;re ready. And it is 70 five point bye. No, Uh, 47 seven is greater than five. Simon&#x27;s weekend. We write that for as of, uh, there we go, who find we got it done. And that&#x27;s party. So hopefully you see some kind of trend kind, you know, sort of what? The problems asking. And, uh, that&#x27;s party. Now, a new party says that we have to make a table using the the points we have here, So, uh, yeah, I&#x27;m going to moves to this page right here. So we have to think about the values we have when we&#x27;re graphing. So if we go back here, um, we have around 15. 24 30 35 51 66 75. Right. These air pretty wide range with numbers. And for K, we have, um, rather smaller set. So, um, you have to think what is the deep end? It&#x27;s an independent variable. So this time depend on cater to skate depend on time. Well, obviously, time depends on Kate. Depending what the K is your time will be different. Getting what the time is. Your cave may be different, but not in this formula. So let&#x27;s establish that at the beginning. This okay, this look his time. All right. So who so, um, for kay, we go 12 30. I&#x27;m depending on your grid. Um, you might two different intervals between them, but I&#x27;m going to to it in to soil. Let&#x27;s let&#x27;s see what we can do. 2468 10. 12 e. I don&#x27;t think I can do to. So it&#x27;s 25 Then if I might be too much for let&#x27;s do for or should I zero for eight? Well, 16 one tea 24. 28. Yeah. Yeah, I think we&#x27;re going to have to two and maybe extended a bit, too. For six, 8 10 Well, 14 16 18 20 two, 24 26 28 30. Okay, so we have our cake. We have all the way from 0 to 30 times. That&#x27;s good. But the more complicated one it&#x27;s gonna be t We have all these complicated numbers and it&#x27;s gonna be kind of hard to fit them all in. But we have Ciro to 75 so I would think 0 75 we might want to do in terms of five. These are fairly close toe factors of five. So let&#x27;s let&#x27;s see. What if aphids go five, 10 15 Corny. Five, 30 35. 40 Already Fine. D five, 60 65 70. And what did we need to go up to? 75. So let&#x27;s just add that up there. Okay, so that that mostly works hopefully have a need of one. But this will do. Which is the CIA General trends that are going on. So our first plot is fairly easy. It&#x27;s 10 So you go one around here, zero up. So that&#x27;s around there. Go yet one in the K zero in the time and you go back. You go to and around 15. Since we&#x27;re in terms of five, getting pretty much on 15 is is good enough. So too, And 15 2 15 right around there. Then we have three and 24. That&#x27;s just below 25. It&#x27;s a three and 24 run there. Then four this 30 close to 31 so above 30 right above it, though then five is 35 more on 36. So above 35 just a little bit. And then we have 10 and 51. So a little. Oh, both decent tens over here fifties right there. Just a little bit above again. You don&#x27;t have to be exact 20 and 66 maybe 67. So I&#x27;m gonna do it. A good chunk over 65. And that is for those who began 20. So 20 years, all the way over here, A good chunk of her 65 right around there. Let&#x27;s see then last one is 30 with 75 around 30 with 75 way off there. So I hope you notice but it&#x27;s kind of Ah, exponential curve or more of a log, a rhythm curve because its opposite of exponential curve. But it&#x27;s it was first really fast and starts to slow down. If you see that curves like this, if I could drive, Yeah, ignore the curves. But but But I hope you see that trend, um, CIA. So that&#x27;s part B. Um, yes. So it doesn&#x27;t necessarily say to to scrap a line, just a graph, the plots. So I think you&#x27;re supposed to kind of see the on the pattern now the final notice along problem. But we&#x27;re at the final one. It says If Paul infests $1000 it&#x27;s principal is $1000. When will he have his investment double and then also has tripled? So it&#x27;s actually it&#x27;s actually two questions. When does the Doublemint is the triple? So double would mean that cake was too right. Because, you know, two times triple means chemicals. Three. So we&#x27;re just plugging in when Kate goes to and K equals three. So actually, the principal doesn&#x27;t matter. You know why? Because it&#x27;s not even in the equation. So this actually doesn&#x27;t matter what it is It&#x27;s just kind of giving you a real life example off. What, you know, that this could be applied for. So let&#x27;s let&#x27;s plug in the equation so poor back here. So we do, um, the log. But instead of, um, 30 or one this time we&#x27;re going to put just to away Sorry, my bad. They already have it for you. So this wasn&#x27;t always the time. They would have the graph right here. We have it so that when you double already tells you how long it will take and then what was the 2nd 1? It was asking triple. We already have it right here. So we already wrote the answers down, so that was rather sneaky of them, making us learn without us knowing. But I appreciate that. Um, yes. So that&#x27;s that&#x27;s the answer. Um, yeah, these these problems are very long seen Garrity&#x27;s. But at the end of the day, these air these their skills you ought to know. And this, especially this longer than curve is it&#x27;s very important when it comes to thio interest and banking stuff like that. So if you&#x27;re at all interested in these types of curves, you need to know your longer themselves. What works. So I know that was a long one. But I hope you learn from that. And I hope you can do these type of problems in the future.

Problem 39 Medium Difficulty

A sum of $\$ 5000$ is invested at an interest rate of $9 \%$ per year, compounded continuously.
a. Find the value $A(t)$ of the investment after $t$ years.
b. Draw a graph of $A(t)$.
c. Use the graph of $A(t)$ to determine when this investment will amount to $\$ 25,000$.

Answer

(a) $A(t)=5000 e^{0.09 t}$
(b)
(c) After 17.88 years

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(a) $A(t)=5000 e^{0.09 t}$(b)(c) After 17.88 years

College Algebra

Heather Z.

Alayna H.

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Kristen K.

Absolute Value - Example 1

Absolute Value - Example 2

James Stewart, Lothar Redlin, Saleem WatsonCollege Algebra

Heather Z.

Alayna H.

Maria G.

Kristen K.

(a) $A(t)=5000 e^{0.09 t}$
(b)
(c) After 17.88 years

James Stewart, Lothar Redlin, Saleem Watson

College Algebra