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(a) If 3000 is invested al $ 5\% $ interest, find the value of the investment at the end of 5 years if the interest is compounded (i) annually, (ii) semiannually, (iii) monthly, (iv) weekly, (v) daily, and (vi) continuously.(b) If $ A(t) $ is the amount of the investment at time $ t $ for the case of continuous compounding, write a differential equation and an initial condition satisfied by $ A(t). $
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03:08
Wen Zheng
Calculus 1 / AB
Chapter 3
Differentiation Rules
Section 8
Exponential Growth and Decay
Derivatives
Differentiation
Missouri State University
Campbell University
University of Nottingham
Lectures
04:40
In mathematics, a derivative is a measure of how a function changes as its input changes. Loosely speaking, a derivative can be thought of as how much one quantity is changing in response to changes in some other quantity; for example, the derivative of the position of a moving object with respect to time is the object's velocity. The concept of a derivative developed as a way to measure the steepness of a curve; the concept was ultimately generalized and now "derivative" is often used to refer to the relationship between two variables, independent and dependent, and to various related notions, such as the differential.
44:57
In mathematics, a differentiation rule is a rule for computing the derivative of a function in one variable. Many differentiation rules can be expressed as a product rule.
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here we have a compound interest problem and we'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'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'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's 3000 times one plus 0.0 5/52 raised to the power 52 times five, and that'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'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'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'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'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's multiply the 0.5 and the 3000 together, and we get 1 50 e to the 0.5 t. So that's our differential equation for this situation. And for the initial condition. That would be what'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's the initial condition.
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