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Chapter 3: Problem 19

An amount of invested money is said to draw interest compounded continuously if the amount of money increases at a rate proportional to the amount present. Suppose $$\$ 1000$$ is invested and draws interest compounded continuously, where the annual interest rate is \(6 \%\). (a) How much money will be present 10 years after the original amount was invested? (b) How long will it take the original amount of money to double?

Short Answer

Expert verified
(a) After 10 years, there will be approximately $1,822.12 in the account. (b) It will take approximately 11.55 years for the original amount of money to double.

Step by step solution

01

Convert the percentage interest rate to a decimal

To convert the interest rate from a percentage to a decimal, divide it by 100. \[ r = \frac{6}{100} = 0.06 \]
02

Plug in the values

Now that we have the interest rate as a decimal, we can plug our values into the continuously compounded interest formula: \[ A = P * e^{(r*t)} \] The principal \(P = 1000\), the interest rate \(r = 0.06\), and the time \(t = 10\) years.
03

Solve for A

Plug in the values to the formula and solve for \(A\): \[ A = 1000 * e^{(0.06 * 10)} \] \[ A = 1000 * e^{0.6} \] \[ A \approx \$1,822.12 \] After 10 years, there will be approximately $1,822.12 in the account. (b) Calculate the time needed to double the original investment
04

Set up the equation

We want to find the time \(t\) when the amount \(A\) is double the initial investment. So, we set up the equation: \[(2 * P) = P * e^{(r*t)}\]
05

Plug in the values

We know the principal \(P = 1000\) and the interest rate \(r = 0.06\). Now, plug in these values into the equation: \[ 2 * 1000 = 1000 * e^{(0.06 * t)} \]
06

Solve for t

Divide both sides of the equation by 1000: \[ 2 = e^{(0.06 * t)} \] Now, take the natural logarithm of both sides of the equation: \[ \ln(2) = \ln(e^{(0.06 * t)}) \] Using the properties of logarithms, we can rewrite this as: \[ \ln(2) = 0.06 * t \] Finally, solve for \(t\): \[ t = \frac{\ln(2)}{0.06} \] \[ t \approx 11.55 \] It will take approximately 11.55 years for the original amount of money to double.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Exponential Growth
Exponential growth is when something increases at a rate proportional to its current size or value. This idea is central to understanding continuously compounded interest. Imagine your investment like a snowball rolling downhill. As it goes, it picks up more snow, growing bigger and rolling faster, all because it keeps adding more snow based on how big it's already grown.
That's the power of exponential growth! In the formula for continuously compounded interest, \[ A = P \times e^{(r \times t)} \]the "e" represents this exponential factor. It shows how the investment grows over time.
The rate of growth is tied to the interest rate, "r," making it compound faster as time passes. This continuous compound growth is more aggressive than other methods, like simple interest or even periodically compounded interest like yearly, monthly, or daily.
  • An initial small amount can become much larger with time.
  • The time factor, "t," is crucial because the longer the investment stays untouched, the bigger it gets.
  • Small increases in the interest rate can result in significant growth over long periods.
Financial Mathematics
Financial mathematics plays a key role in understanding investments and interest, especially when talking about the compound aspect. The formula used in our problem dives into a vital part of financial mathematics where we need to manage calculations such as interest rates and time in a clear and logical manner. This ensures investors can make informed decisions.
The principal amount, "P," is the starting amount of money you invest. Knowing how this principal changes helps assess the profitability of investments over time through calculated equations and formulas.
The exponential function in our formula helps predict the future value of investments. This can directly influence financial planning decisions, providing clarity on how investment strategies perform.
  • A solid understanding of financial mathematics allows for effective interest rate negotiations.
  • It can impact personal savings, loans, mortgages, and academic projects.
  • It teaches how compounded interest can affect overall financial growth.
Interest Rate Conversion
Interest rate conversion is necessary to use the percentage values in financial formulas accurately. Typically, the first step involves converting the percentage into a decimal by dividing by 100. This conversion is critical in ensuring you plug the correct values into your interest-related calculations.
For example, if the annual interest rate is 6%, you convert it to a decimal format: \[ r = \frac{6}{100} = 0.06 \]This conversion allows the interest rate to work with equations, ensuring calculations account for them correctly and accurately reflect their impact over time.
  • Conversion helps in maintaining the integrity of mathematical equations.
  • It simplifies utilizing widely recognized financial models and frameworks.
  • Allows for comparison between different interest rates and terms effectively.

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Most popular questions from this chapter

An object weighing \(16 \mathrm{lb}\) is dropped from rest on the surface of a calm lake and thereafter starts to sink. While its weight tends to force it downward, the buoyancy of the object tends to force it back upward. If this buoyancy force is one of \(6 \mathrm{lb}\) and the resistance of the water (in pounds) is numerically equal to twice the square of the velocity (in feet per second), find the formula for the velocity of the sinking object as a function of the time.

The rate at which a certain substance dissolves in water is proportional to the product of the amount undissolved and the difference \(c_{1}-c_{2}\), where \(c_{1}\) is the concentration in the saturated solution and \(c_{2}\) is the concentration in the actual solution. If saturated, \(50 \mathrm{gm}\) of water would dissolve \(20 \mathrm{gm}\) of the substance. If \(10 \mathrm{gm}\) of the substance is placed in \(50 \mathrm{gm}\) of water and half of the substance is then dissolved in \(90 \mathrm{~min}\), how much will be dissolved in \(3 \mathrm{hr}\) ?

A body of mass \(m\) is in rectilinear motion along a horizontal axis. The resultant force acting on the body is given by \(-k x\), where \(k>0\) is a constant of proportionality and \(x\) is the distance along the axis from a fixed point \(\mathrm{O}\). The body has initial velocity \(v=v_{0}\) when \(x=x_{0}\). Apply Newton's second law in the form ( \(3.23\) ) and thus write the differential equation of motion in the form $$ m v \frac{d v}{d x}=-k x . $$ Solve the differential equation, apply the initial condition, and thus express the square of the velocity \(v\) as a function of the distance \(x\). Recalling that \(v=d x / d t\), show that the relation between \(v\) and \(x\) thus obtained is satisfied for all time \(t\) by $$ x=\sqrt{x_{0}^{2}+\frac{m v_{0}^{2}}{k}} \sin \left(\sqrt{\frac{k}{m}} t+\phi\right) $$ where \(\phi\) is a constant.

A ship which weighs 32,000 tons starts from rest under the force of a constant propeller thrust of \(100,000 \mathrm{lb}\). The resistance in pounds is numerically equal to \(8000 v\), where \(v\) is in feet per second. (a) Find the velocity of the ship as a function of the time. (b) Find the limiting velocity (that is, the limit of \(v\) as \(t \rightarrow+\infty\) ). (c) Find how long it takes the ship to attain a velocity of \(80 \%\) of the limiting velocity.

A chemical reaction converts a certain chemical into another chemical, and the rate at which the first chemical is converted is proportional to the amount of this chemical present at any time. At the end of one hour, \(50 \mathrm{gm}\) of the first chemical remain; while at the end of three hours, only \(25 \mathrm{gm}\) remain. (a) How many grams of the first chemical were present initially? (b) How many grams of the first chemical will remain at the end of five hours? (c) In how many hours will only \(2 \mathrm{gm}\) of the first chemical remain?
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