Search Menu

Contents

Problems for "Exponential Growth and Decay"

Problems for "Exponential Growth and Decay"

Problems for "Exponential Growth and Decay"

Problems for "Exponential Growth and Decay"

Problems for "Exponential Growth and Decay"

Problems for "Exponential Growth and Decay"

Problem : If P = 300e 2t , at what time will P=600?


600   = 300e 2t  
2   = e 2t  

use natural log to solve for t


ln(2)   = 2t  
t   = .347  

Problem : A strain of bacteria multiply in such a fashion that they double in number every 4 hours. Find an expression that describes this kind of growth.

A function that grows or decays by a fixed percentage over time exhibits growth of the form

B(t) = B 0 e kt    

After 4 hours, B(t) = 2B 0 , so 2B 0 = B 0 e k(4) , 2 = e 4k Use natural log to solve for k


ln(2)   = 4k  
k   = .173  
B(t)   = B 0 e .173t  

Problem : Find a function that meets the following:

= 0.693y    

Any function for which

= ky    

must be of the form y = e kt so y = e 0.693t satisfies the condition above. To verify this, see that

e 0.693t = 0.693 e 0.693t    

or 0.693y

Problem : The half-life of a substance is the time that it takes for the mass of that substance to decay to 50% of its original value. If a certain substance has a half-life of 30 minutes, what is an equation that describes its decay?

This is another case of exponential decay, because a certain percentage of the substance decays at regular time intervals. So, the general form is

C(t) = C 0 e kt    

At t = 30 , 1/2 of the original remains, so


C 0   = C 0 e k(30)  
  = e 30k  
ln   = 30k  
k   - 0.0231  
C(t)   = C 0 e -0.0231t  

Problem : After 40 minutes, only 34% of a radioactive compound is remaining. What is the expression describing its decay?


C(t)   = C 0 e kt  
.34C 0   = C 0 e k(40)  
.34   = e 40k  
ln.34   = 40k  
k   - 0.027  
C(t)   = C 0 e -0.027t