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EXERCISE 1.What constant interest rate is required if an initial deposit placed into an account that accrues interest compounded continuously is to double its value in five years?/

EXERCISE 1.What constant interest rate is required if an initial deposit placed into an account that accrues interest compounded continuously is to double its value in five years?/

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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pls answer the exercise below and ill give you sample problems on guide you on how to do it (THE ANSWER SHOULD 13.86%) ASAP!!!

$1000000- S2000/year eeyr)-1) S= $1 million Compound Interest k +(8000 0.1 Sample Problems: t= 40 years -k k 1. Ayoung person with no initial capital invests k dollars per yeor at an annual rate of retum r. Assume 01*01ea)0 = 8000e(a.1)() Substitute the given in part (c) to the answer in part (a) to solve for the value of r s- - 1) that investments are made continuously and that the return is compounded continuously. k(-1+ e010) = (0.1)(8000)e(0.1)) a. Determine the sum) accumulated at any time !. $2000/year (0.1)(8000)e01)0) k = (-1+ e01(3)) $1000000 = (e4oyr) - 1) b. Ifr=7.5%, determine kso that $1 million will be available for retirement in 40 years. r 0.0977 or 9.77% + Answer k- $3086.64/year - Answer c. IF*= $ 2000 / year, determine the return rate r that must be obtained to have $1 million available in 40 years. 2. A certain college graduate borrows $8000 to buy a car. The lender charges interest at an annual rate of 10%. Assuming that interest is compounded continuously and that the borrower makes payments continuously at a constant annual rate k, determine the poyment rote k that is required to pay off the loan in 3 years. Also determine how much interest is paid during the 3-year period. The interest is the total amount paid (product of the total payment k and the number Solution: of payments n) decreased by the amount borrowed: A= rS +k a. dt ds (S3086.64) |(3 years) – $8000 year The rate of change of the sum is equal to the percentage of the sum at that time plus the Given: Let B represents the balance at time t. I = $1259.92 - Answer amount the person has added on his own. ds At time t=0, B = 8000 r = 10% = 0.10 - rS = k dt Required: Rate poyments (k) in order to have B = 0 int=3 years EXERCISE Solve the differential equation using the method for solving the lineor equation. Solution: The balance increases by the interest (product of the current balance and interest rate) and Note: The dependent variable in this equation is S decreases by the payment: dB = rB - k dt 1.What constant interest rate is required if an initial deposit placed into an account that accrues interest compounded continuously is to double its value in five years? P(t) = - Qt) = k Solve for the integrating factor: dB - rB = -k dt I= ej P)dt = es -ndt I= e-rt Solve the differential equation using the method for solving the linear equation. Solve for S: Se-rt = | ke-rtdt + C Note: The dependent variable in this equation is B Se-rt =. -rert dt +C P(t) = - Q(t) = -k Solve for the integrating factor: Se-rt = - -ke-rt +C I= ej P()dt = ej -nit I=e-rt -k S=*+ Cert + Cert + equation 1 Solve for B: Bet- -ke"rt dt + C Initial Conditions: when t = 0, S= o (he has no money invested at t=0) 0 =*+ Ce'(0) =#S-re"dt +C -redt +C k ke-rt --k k S =+ert Re-rt - Substitute C into the equation 1 k S =- (ert - 1) B==+ Cert - equation 1 + Answer Initial Conditions: when t = 0, B = 8000 8000 ==+ Ce (0) b. Given:r= 7.5% S = $1 million 8000 --= C 1= 40 years k B = =+(8000 -)ert Substitute Cinto the equation 1 - equation 2 Substitute the given in part (b) to the answer in part (a) to solve for the value of k To solve for k, substitute r= 0.1, and the initial condition that at t = 3, B = 0 into the equation 2. s=e" – (et - 1) Sr k = (ert - 1) ($1000000)(0.075) k = (e(0,075)(40) – 1) k- $3930/year - Answer c. Given: k= $2000/year
Transcribed Image Text:$1000000- S2000/year eeyr)-1) S= $1 million Compound Interest k +(8000 0.1 Sample Problems: t= 40 years -k k 1. Ayoung person with no initial capital invests k dollars per yeor at an annual rate of retum r. Assume 01*01ea)0 = 8000e(a.1)() Substitute the given in part (c) to the answer in part (a) to solve for the value of r s- - 1) that investments are made continuously and that the return is compounded continuously. k(-1+ e010) = (0.1)(8000)e(0.1)) a. Determine the sum) accumulated at any time !. $2000/year (0.1)(8000)e01)0) k = (-1+ e01(3)) $1000000 = (e4oyr) - 1) b. Ifr=7.5%, determine kso that $1 million will be available for retirement in 40 years. r 0.0977 or 9.77% + Answer k- $3086.64/year - Answer c. IF*= $ 2000 / year, determine the return rate r that must be obtained to have $1 million available in 40 years. 2. A certain college graduate borrows $8000 to buy a car. The lender charges interest at an annual rate of 10%. Assuming that interest is compounded continuously and that the borrower makes payments continuously at a constant annual rate k, determine the poyment rote k that is required to pay off the loan in 3 years. Also determine how much interest is paid during the 3-year period. The interest is the total amount paid (product of the total payment k and the number Solution: of payments n) decreased by the amount borrowed: A= rS +k a. dt ds (S3086.64) |(3 years) – $8000 year The rate of change of the sum is equal to the percentage of the sum at that time plus the Given: Let B represents the balance at time t. I = $1259.92 - Answer amount the person has added on his own. ds At time t=0, B = 8000 r = 10% = 0.10 - rS = k dt Required: Rate poyments (k) in order to have B = 0 int=3 years EXERCISE Solve the differential equation using the method for solving the lineor equation. Solution: The balance increases by the interest (product of the current balance and interest rate) and Note: The dependent variable in this equation is S decreases by the payment: dB = rB - k dt 1.What constant interest rate is required if an initial deposit placed into an account that accrues interest compounded continuously is to double its value in five years? P(t) = - Qt) = k Solve for the integrating factor: dB - rB = -k dt I= ej P)dt = es -ndt I= e-rt Solve the differential equation using the method for solving the linear equation. Solve for S: Se-rt = | ke-rtdt + C Note: The dependent variable in this equation is B Se-rt =. -rert dt +C P(t) = - Q(t) = -k Solve for the integrating factor: Se-rt = - -ke-rt +C I= ej P()dt = ej -nit I=e-rt -k S=*+ Cert + Cert + equation 1 Solve for B: Bet- -ke"rt dt + C Initial Conditions: when t = 0, S= o (he has no money invested at t=0) 0 =*+ Ce'(0) =#S-re"dt +C -redt +C k ke-rt --k k S =+ert Re-rt - Substitute C into the equation 1 k S =- (ert - 1) B==+ Cert - equation 1 + Answer Initial Conditions: when t = 0, B = 8000 8000 ==+ Ce (0) b. Given:r= 7.5% S = $1 million 8000 --= C 1= 40 years k B = =+(8000 -)ert Substitute Cinto the equation 1 - equation 2 Substitute the given in part (b) to the answer in part (a) to solve for the value of k To solve for k, substitute r= 0.1, and the initial condition that at t = 3, B = 0 into the equation 2. s=e" – (et - 1) Sr k = (ert - 1) ($1000000)(0.075) k = (e(0,075)(40) – 1) k- $3930/year - Answer c. Given: k= $2000/year
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