a-if-1000-is-borrowed-at-8-interest-find-the-amounts-due-at-the-end-of-3-years-if-the-interest-is--4

Question

(a) If $\$ 1000$ is borrowed at 8$\%$ interest, find the amounts
due at the end of 3 years if the interest is compounded
(i) annually, (ii) quarterly, (iii) monthly, (iv) weekly,
(v) daily, (vi) hourly, and (vii) continuously.
(b) Suppose $\$ 1000$ is borrowed and the interest is compounded continuously. If $A(t)$ is the amount due after
$t$ years, where $0 \leqslant t \leqslant 3,$ graph $A(t)$ for each of the
interest rates $6 \%, 8 \%,$ and 10$\%$ on a common screen.

Caleb Miller · Accepted 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.

Ankit Gupta · 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.

Ankit Gupta · Answer

Nation E. That is the initial investment at a function of time. It is 1500 times to raise the ball T divide by eight Party of discussion. Find a Mountain Amount account After five years of when time is five years, it&#x27;s just 1500 times to listen or five behind by eight. So what you with that? If you use the camp later. Five days to the part eight. We should write it as two recent for 55 Divide Way eight years, 0.625 1500 times to rate about 2.6 to faith and to raise the power wind 6 to 5. We have 1.5 for two times, 1500 is 2313.316 and it is dollars. Next part of the solution. He after Dainius 1500 times to raise to power. 10. Divide by eight and to raise the bar. 10. Divide by eight. We should right then divided by eight is 1.25 to raise two. Part 1.25 It&#x27;s 2.378 times. 1500 that is 3567 in dollar. It is 35 67 6 to 1 box CEO of this question Is it 18 is the Reuben Itwas initial time. So time is zero that is stirring time 1500 times to raise to power the door divide way it that is $1500. So that is initial investment and e off eight. It is 1500 things to raise the bar. Eight. Divide weight, that is, Teddy told him. Dolar it is ticked at investment doubles after eight years.

Heather Zimmers · Answer

here we have a compound interest problem and we&#x27;re going to do part a mostly with the non continuous compounding formula. And so we have, in an amount invested $3000 an interest rate of 5%. So we use the decimal point verify and a time of five years. So to find the final amount, if interest is calm, pounded annually, which would be an equals 11 time per year, we have a equals 3000 times one plus 0.0 5/1 raised to the power one times five. And we put that in a calculator and we get rounded to the nearest penny. $3828.84. Now we move on to the second part of part A, and we&#x27;re going to calculate it for semi annually, which would be two times a year so and equals two. So a equals 3000 times one plus 0.0 5/2 raised to the two times five, which would be 10 10 compound ings in the five year period. And we put that in a calculator and rounding to the nearest penny we get $3840.25 for the third part of part. They were going to calculate for monthly compounding, so that would be 12 times of compounding per year. So we have a equals 3000 times one plus 0.0 5/12 raised to the power 12 times five, which would be 60 compound ing&#x27;s and rounded to the nearest penny. We get $3850 and 20 and ah, seven cents seven cents. So we could see the amounts are going up a little bit each time as the number of compound ings increases for the fourth part were compounding weekly. There are 52 weeks in a year, so an equals 52. So we have a gal&#x27;s 3000 times one plus 0.0 5/52 raised to the power 52 times five, and that&#x27;s going to give us $3851.61. And for the fifth part of part, a daily compounding will use n equals 365. So we have a equals 3000 times one plus 0.0 5/3 65 raised to the power 3 65 times five. Put that into a calculator and we get $3852.1. And for part six, we&#x27;re going to switch to continuously compounded interest. So we have a different model for this. We have a equals a not times e to the R. T. So that&#x27;s going to be 3000 times E to the 0.5 times five, and we end up with $3852.8 not a lot more than the previous one. But it&#x27;s the highest you can get for continuously compounded interest compared to any of the other calm poundings now for Part B were to find the differential equation that fits the situation. So because we have this basic exponential growth type model here, it fits. The differential equation rate of change of A is proportional to a so rate of change of a equals K times A. Now we know the value of K will be our rate. And in our equation we had a equals 3000 times E to the 0.5 t, so the rate was your a 0.0 fight. So that&#x27;s R K. So now we have de a d t equals 0.5 times a, and we can go ahead and substitute our equation. And for a and we have 0.5 times 3000 each of the 0.5 t. All right, let&#x27;s multiply the 0.5 and the 3000 together, and we get 1 50 e to the 0.5 t. So that&#x27;s our differential equation for this situation. And for the initial condition. That would be what&#x27;s the amount when the time is zero and we know the amount invested was $3000. So at time zero the amount is 3000. That&#x27;s the initial condition.

Problem

If $\$ 3000$ is invested at 5$\%$ interest, find …

Problem 18 Medium Difficulty

Answer

a)
i.$$
\$ 1259.71
$$
ii. $$
\$ 1268.24
$$
iii. $$
\$ 1270.24
$$
iv. $$
\$ 1271.02
$$
v. $$
\$ 1271.22
$$
vi. $$
\$ 1271.25
$$
vii. $$
\$ 1271.25
$$
b) See video for graph

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Essential Calculus Early Transcendentals

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a)i.$$\$ 1259.71$$ii. $$\$ 1268.24$$iii. $$\$ 1270.24$$iv. $$\$ 1271.02$$v. $$\$ 1271.22$$vi. $$\$ 1271.25$$vii. $$\$ 1271.25$$b) See video for graph

Essential Calculus Early Transcendentals

Catherine R.

Anna Marie V.

Kayleah T.

Samuel H.

Precalculus Review - Intro

Functions on the Real Line - Overview

James StewartEssential Calculus Early Transcendentals

Catherine R.

Anna Marie V.

Kayleah T.

Samuel H.

a)
i.$$
\$ 1259.71
$$
ii. $$
\$ 1268.24
$$
iii. $$
\$ 1270.24
$$
iv. $$
\$ 1271.02
$$
v. $$
\$ 1271.22
$$
vi. $$
\$ 1271.25
$$
vii. $$
\$ 1271.25
$$
b) See video for graph

James Stewart

Essential Calculus Early Transcendentals