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Characteristics of the exponential function, y = A 0 e kt

An exponential function with the form y = A 0 e k t has the following characteristics:

  • one-to-one function
  • horizontal asymptote: y = 0
  • domain: ( ,   )
  • range: ( 0 , )
  • x intercept: none
  • y-intercept: ( 0 , A 0 )
  • increasing if k > 0 (see [link] )
  • decreasing if k < 0 (see [link] )
Two graphs of y=(A_0)(e^(kt)) with the asymptote at y=0. The first graph is of when k>0 and with the labeled points (1/k, (A_0)e), (0, A_0), and (-1/k, (A_0)/e). The second graph is of when k<0 and with the labeled points (-1/k, (A_0)e), (0, A_0), and (1/k, (A_0)/e).
An exponential function models exponential growth when k > 0 and exponential decay when k < 0.

Graphing exponential growth

A population of bacteria doubles every hour. If the culture started with 10 bacteria, graph the population as a function of time.

When an amount grows at a fixed percent per unit time, the growth is exponential. To find A 0 we use the fact that A 0 is the amount at time zero, so A 0 = 10. To find k , use the fact that after one hour ( t = 1 ) the population doubles from 10 to 20. The formula is derived as follows

  20 = 10 e k 1     2 = e k Divide by 10 ln 2 = k Take the natural logarithm

so k = ln ( 2 ) . Thus the equation we want to graph is y = 10 e ( ln 2 ) t = 10 ( e ln 2 ) t = 10 · 2 t . The graph is shown in [link] .

A graph starting at ten on the y-axis and rising rapidly to the right.
The graph of y = 10 e ( ln 2 ) t
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Half-life

We now turn to exponential decay . One of the common terms associated with exponential decay, as stated above, is half-life , the length of time it takes an exponentially decaying quantity to decrease to half its original amount. Every radioactive isotope has a half-life, and the process describing the exponential decay of an isotope is called radioactive decay.

To find the half-life of a function describing exponential decay, solve the following equation:

1 2 A 0 = A o e k t

We find that the half-life depends only on the constant k and not on the starting quantity A 0 .

The formula is derived as follows

1 2 A 0 = A o e k t 1 2 = e k t Divide by  A 0 . ln ( 1 2 ) = k t Take the natural log . ln ( 2 ) = k t Apply laws of logarithms . ln ( 2 ) k = t Divide by  k .

Since t , the time, is positive, k must, as expected, be negative. This gives us the half-life formula

t = ln ( 2 ) k

Given the half-life, find the decay rate.

  1. Write A = A o e k t .
  2. Replace A by 1 2 A 0 and replace t by the given half-life.
  3. Solve to find k . Express k as an exact value (do not round).

Note: It is also possible to find the decay rate using k = ln ( 2 ) t .

Finding the function that describes radioactive decay

The half-life of carbon-14 is 5,730 years. Express the amount of carbon-14 remaining as a function of time, t .

This formula is derived as follows.

         A = A 0 e k t The continuous growth formula . 0.5 A 0 = A 0 e k 5730 Substitute the half-life for  t  and  0.5 A 0  for  f ( t ) .        0.5 = e 5730 k Divide by  A 0 . ln ( 0.5 ) = 5730 k Take the natural log of both sides .           k = ln ( 0.5 ) 5730 Divide by the coefficient of  k .          A = A 0 e ( ln ( 0.5 ) 5730 ) t Substitute for  r  in the continuous growth formula .

The function that describes this continuous decay is f ( t ) = A 0 e ( ln ( 0.5 ) 5730 ) t . We observe that the coefficient of t , ln ( 0.5 ) 5730 1.2097 × 10 −4 is negative, as expected in the case of exponential decay.

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The half-life of plutonium-244 is 80,000,000 years. Find function gives the amount of carbon-14 remaining as a function of time, measured in years.

f ( t ) = A 0 e 0.0000000087 t

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Radiocarbon dating

The formula for radioactive decay is important in radiocarbon dating , which is used to calculate the approximate date a plant or animal died. Radiocarbon dating was discovered in 1949 by Willard Libby, who won a Nobel Prize for his discovery. It compares the difference between the ratio of two isotopes of carbon in an organic artifact or fossil to the ratio of those two isotopes in the air. It is believed to be accurate to within about 1% error for plants or animals that died within the last 60,000 years.

Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
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Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
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Source:  OpenStax, Precalculus. OpenStax CNX. Jan 19, 2016 Download for free at https://legacy.cnx.org/content/col11667/1.6
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