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

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The discussion focuses on the concept of doubling time in functions, specifically comparing exponential and linear functions. It clarifies that only exponential functions have a constant doubling time, while the doubling time for a linear function can be expressed as t0 + b/a. A participant initially misinterprets the problem but is guided to understand that the doubling time should be calculated as the time elapsed after t0. The solution involves solving for the time t when the function doubles, leading to the correct interpretation of the problem. This highlights the importance of understanding the context of doubling time in different types of 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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Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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