When the interest on an investment is compounded continuously, the investment gr

Question

When the interest on an investment is compounded continuously, the investment grows at a rate that is proportional to the amount in the account, so that if the amount present is P, then dP dt where P is in dollars, t is in years, and k is a constant. If $120,000 is invested (when t = 0) and the amount in the account after 13 years is $229,865, find the function that gives the value of the investment as a function of t. (Round your value of k to two decimal places.) What is the interest rate on this investment? (Round your answer to the nearest whole number.)

Explanation / Answer

dP/P = k*dt

Integrating both sides :
ln(P) = kt + C

P = e^(kt + C)

P = Ce^(kt)

When t =0, P = 120000,
so 120000 = ce^(0)

C = 120000

So,
P = 120000e^(kt)

When t = 13, P = 229865 :
229865 = 120000e^(13k)

229865/120000 = e^(13k)

13k = ln(229865/120000)

k = 0.05

So, we have
P = 120000e^(0.05t) -----> ANS

Interest rate :
Equating 1 + r = e^0.05, we get
r = 0.0512710963760241

As in
5.12%

So, rounded to the nearest whole number it is
5% ------> ANS

-----------------------------------------------------------------

So, answers are :
P = 120000e^(0.05t)
5%

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.