# 6.1 Exponential functions  (Page 8/16)

 Page 8 / 16

Use a calculator to find $\text{\hspace{0.17em}}{e}^{-0.5}.\text{\hspace{0.17em}}$ Round to five decimal places.

${e}^{-0.5}\approx 0.60653$

## Investigating continuous growth

So far we have worked with rational bases for exponential functions. For most real-world phenomena, however, e is used as the base for exponential functions. Exponential models that use $\text{\hspace{0.17em}}e\text{\hspace{0.17em}}$ as the base are called continuous growth or decay models . We see these models in finance, computer science, and most of the sciences, such as physics, toxicology, and fluid dynamics.

## The continuous growth/decay formula

For all real numbers $\text{\hspace{0.17em}}t,$ and all positive numbers $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}r,$ continuous growth or decay is represented by the formula

$A\left(t\right)=a{e}^{rt}$

where

• $a\text{\hspace{0.17em}}$ is the initial value,
• $r\text{\hspace{0.17em}}$ is the continuous growth rate per unit time,
• and $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ is the elapsed time.

If $\text{\hspace{0.17em}}r>0\text{\hspace{0.17em}}$ , then the formula represents continuous growth. If $\text{\hspace{0.17em}}r<0\text{\hspace{0.17em}}$ , then the formula represents continuous decay.

For business applications, the continuous growth formula is called the continuous compounding formula and takes the form

$A\left(t\right)=P{e}^{rt}$

where

• $P\text{\hspace{0.17em}}$ is the principal or the initial invested,
• $r\text{\hspace{0.17em}}$ is the growth or interest rate per unit time,
• and $t\text{\hspace{0.17em}}$ is the period or term of the investment.

Given the initial value, rate of growth or decay, and time $\text{\hspace{0.17em}}t,$ solve a continuous growth or decay function.

1. Use the information in the problem to determine $\text{\hspace{0.17em}}a$ , the initial value of the function.
2. Use the information in the problem to determine the growth rate $\text{\hspace{0.17em}}r.$
1. If the problem refers to continuous growth, then $\text{\hspace{0.17em}}r>0.$
2. If the problem refers to continuous decay, then $\text{\hspace{0.17em}}r<0.$
3. Use the information in the problem to determine the time $\text{\hspace{0.17em}}t.$
4. Substitute the given information into the continuous growth formula and solve for $\text{\hspace{0.17em}}A\left(t\right).$

## Calculating continuous growth

A person invested $1,000 in an account earning a nominal 10% per year compounded continuously. How much was in the account at the end of one year? Since the account is growing in value, this is a continuous compounding problem with growth rate $\text{\hspace{0.17em}}r=0.10.\text{\hspace{0.17em}}$ The initial investment was$1,000, so $\text{\hspace{0.17em}}P=1000.\text{\hspace{0.17em}}$ We use the continuous compounding formula to find the value after $\text{\hspace{0.17em}}t=1\text{\hspace{0.17em}}$ year:

The account is worth $1,105.17 after one year. A person invests$100,000 at a nominal 12% interest per year compounded continuously. What will be the value of the investment in 30 years?

\$3,659,823.44

## Calculating continuous decay

Radon-222 decays at a continuous rate of 17.3% per day. How much will 100 mg of Radon-222 decay to in 3 days?

Since the substance is decaying, the rate, $\text{\hspace{0.17em}}17.3%$ , is negative. So, The initial amount of radon-222 was $\text{\hspace{0.17em}}100\text{\hspace{0.17em}}$ mg, so $\text{\hspace{0.17em}}a=100.\text{\hspace{0.17em}}$ We use the continuous decay formula to find the value after $\text{\hspace{0.17em}}t=3\text{\hspace{0.17em}}$ days:

So 59.5115 mg of radon-222 will remain.

Using the data in [link] , how much radon-222 will remain after one year?

3.77E-26 (This is calculator notation for the number written as $\text{\hspace{0.17em}}3.77×{10}^{-26}\text{\hspace{0.17em}}$ in scientific notation. While the output of an exponential function is never zero, this number is so close to zero that for all practical purposes we can accept zero as the answer.)

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