Exponential and Logarithmic Functions
Exponential
Explanation
Exponential function and logarithmic functions are used for for equations that either, increase or decrease exponentially, or the rate at which they increase or decrease is increasing or decreasing exponentially. Some notable examples for this could be in populations that grow exponentially, bank interest, or atomic half-lifes from decay. The basic laws to explain the derivative process is to use the chain rule. For example when finding the derivative of ex one takes the outside derivative and then the inside derivative which would result in ex. The main exponential derivative rule is the second of the listed rules. Its the natural log of the base multiplied to the original exponential function. |
Exponents Derivatives
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Examples
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Half Life
Then to find the rate of the change, we can use the exponential rules. However we see that there is also some chain rule involved, (go to the chain rule page for some clarification on that process)
Lets Identify our inside and outside functions fist
f(t)=g(h(x))
g(t)= 10(1/2)h(x)
h(t)=t/2.7
g'(t)= 10(ln(1/2))*(1/2)h(x)
h'(t)=1/2.7
F'(t)=10(ln(1/2))*(1/2)(t/2.7)*1/2.7
At what rate is the quantity of Au-198 in the patients body decreasing at 30 minutes after injection? 6 days after injection? And how much is left at those times?
30 Minutes
1 day = 24 hours = 1440 minutes. 30 minutes = 30/1440= 0.02083333333 days
f(x)=10(1/2)(t/2.7)
f(x)=10(1/2)((0.02083333333)/2.7)
f(x)=10(1/2)(0.007716049383)
f(x)=10*(0.9946659191)
f(x)=9.946659191
At 30 minutes, there are still 9.94665919 µgrams of Au-198
f'(x)=10(ln(1/2))*(1/2)(t/2.7)*1/2.7
f'(30m)=10(ln(1/2))*(1/2)(0.02083333333/2.7)*1/2.7
f'(30m)=10(ln(1/2))*(1/2)(0.007716049383)*1/2.7
f'(30m)=10(-0.6931471806)*(1/2)(0.007716049383)*1/2.7
f'(30m)=-6.931471806*0.9946659191*0.3703703704
f'(30m)=-2.5535180649
At 30 minutes past injection, the amount of Au-198 present in the area will be decreasing at a rate of 2.5535180649µgrams per day.
6 Days
f(x)=10(1/2)(t/2.7)
f(x)=10(1/2)(6/2.7)
f(x)=10(1/2)2.2222222222
f(x)=10*(0.2143109957)
f(x)=2.143109957
At 6 days, there are only 2.143109957 µgrams of Au-198
f'(x)=10(ln(1/2))*(1/2)(t/2.7)*1/2.7
f'(x)=10(ln(1/2))*(1/2)(6/2.7)*1/2.7
f'(x)=10(ln(1/2))*(1/2)(2.2222222222)*1/2.7
f'(x)=10(-0.6931471806)*(1/2)(2.2222222222)*1/2.7
f'(x)=-6.931471806*(1/2)(2.2222222222)*1/2.7
f'(x)=-6.931471806*0.2143109957*0.3703703704
f'(x)=-0.5501817128
At 6 days past injection, the amount of Au-198 present in the area will be decreasing at a rate of 0.5501817128 µgrams per day.
Logarithmic
Explanation
Exponential function and logarithmic functions are used for for equations that either, increase or decrease exponentially, or the rate at which they increase or decrease is increasing or decreasing exponentially. Notably it is the inverse of a exponential function. |
Logarithmic Derivatives
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Examples
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Populations of Paris, France
The thick line is the actual data and official projected data (www.un.org), the dotted line shows the formula r(x) = ln(x)+7.3
The units on the y-axes are millions of people. The x axes increases by 5 years for each tick mark, 0 being 1950.
The derivative of r(x)= ln(x)+7.3 is 1⁄x
How fast is the population projected to be growing in 2015? 2050? How fast was it growing in 1960? Do you think these numbers accurate?
2015:
2015 -1950 = 65/5 =13
r'(x)=1⁄x
r'(13)=1⁄(13)
r'(13)=0.076923076923
In 2015 the population of paris was growing at a rate of 0.076,923,076,923 million people per five years.
2050:
2050 -1950 = 100/5 =20
r'(x)=1⁄x
r'(20)=1⁄(20)
r'(20) = 0.05
In 2050 the population of paris was growing at a rate of 0.05 people per five years.
1960:
1960 -1950 = 10/5 =2
r'(x)=1⁄x
r'(20)=1⁄(2)
r'(20)=.5
In 1960 the population of paris was growing at a rate of .5 million people per five years.
The rate of growth for 2015 and 2060 seem pretty accurate. The rate of growth for 1960 does not. As visible in the graph above, the formula r(x)= ln(x)+7.3 only starts modeling the pattern of Paris' growth around 1975.