Constant Doubling Time Functions: Exponential vs. Linear | Homework Help

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SUMMARY

The discussion centers on the concept of doubling time in functions, specifically contrasting exponential functions of the form P0ekt with linear functions defined as f(t) = a(t) + b. It is established that the only functions exhibiting a constant doubling time are exponential functions with k > 0. The doubling time for a linear function at time t0 is calculated as t0 + b/a, emphasizing the need to interpret doubling time as the interval t - t0 for clarity in calculations.

PREREQUISITES
  • Understanding of exponential functions, specifically P0ekt with k > 0
  • Familiarity with linear functions, particularly f(t) = a(t) + b
  • Basic algebraic manipulation skills
  • Knowledge of the concept of doubling time in mathematical functions
NEXT STEPS
  • Study the properties of exponential growth functions
  • Explore the implications of linear versus exponential growth in real-world scenarios
  • Learn about the mathematical derivation of doubling time for various function types
  • Investigate applications of doubling time in fields such as finance and population studies
USEFUL FOR

Students studying mathematics, particularly those focusing on functions and growth rates, as well as educators seeking to clarify the differences between linear and exponential functions.

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Homework Statement


The only functions with a constant doubling time are the exponential functions P0ekt with k > 0. Show that the doubling time of linear function f(t) = a(t) + b at time t0 is t0 + b/a.

Homework Equations


n/a

The Attempt at a Solution



With initial time t0, P = at0 + b

At some time t, P is doubled: at + b = 2P

t = (2P - b)/a

Plug in P and simplify:

t = (2at0 + b)/a
t = 2t0 + b/a

What am I doing wrong?
 
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physicsernaw said:

Homework Statement


The only functions with a constant doubling time are the exponential functions P0ekt with k > 0. Show that the doubling time of linear function f(t) = a(t) + b at time t0 is t0 + b/a.

Homework Equations


n/a

The Attempt at a Solution



With initial time t0, P = at0 + b

At some time t, P is doubled: at + b = 2P

t = (2P - b)/a

Plug in P and simplify:

t = (2at0 + b)/a
t = 2t0 + b/a

What am I doing wrong?

You haven't done anything wrong. But I think they mean the doubling time to be t-t0. That's the length of time after t0 that the function will double. Solve for that.
 
Dick said:
You haven't done anything wrong. But I think they mean the doubling time to be t-t0. That's the length of time after t0 that the function will double. Solve for that.

Ahhh, thank you so much.
 

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