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Chapter 6: Problem 33

Prove that the doubling time for an exponentially increasing quantity is constant for all time.

Short Answer

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Question: Prove that the doubling time for an exponentially increasing quantity is constant for all time. Short Answer: The doubling time for an exponentially increasing quantity can be shown to be constant by solving the exponential function f(t) = a * b^(t/c) for the initial doubling time and comparing it to the equation obtained when time is doubled. As demonstrated in the step-by-step solution, the equation simplifies to show that the quantity indeed doubles in a constant time, proving that the doubling time is constant for all time.

Step by step solution

01

Definition of an exponential function

An exponential function is a function of the form f(t) = a * b^(t/c), where "a" is the initial quantity, "b" is the growth factor, "t" is the time, and "c" is a scaling factor. The growth rate is determined by the value of "b".
02

Calculate the initial doubling time

To find the doubling time, we can divide the initial quantity by the growth factor, and then multiply by the scaling factor. We set f(t) = 2a, since this is what we want to double. So, we have 2a = a * b^(t/c). The "a" terms cancel out, which leaves: 2 = b^(t/c)
03

Solve for the doubling time

To solve for the doubling time, we take the natural logarithm of both sides of the equation: ln(2) = ln(b^(t/c)) Using the logarithm properties, we can rewrite this equation as: t/c * ln(b) = ln(2) Now, let's isolate t: t = c * ln(2) / ln(b)
04

Double the time

In order to show that the doubling time is constant, we must now double the time and calculate the new quantity. Let's set t2 = 2t and find f(t2): f(t2) = a * b^(t2/c) f(t2) = a * b^(2t/c)
05

Using our initial solution

Now, using the solution for the initial doubling time, we can substitute it into the second equation: f(t2) = a * b^(2c * ln(2) / ln(b)) This will simplify to: f(t2) = a * (b^(c * ln(2) / ln(b)))^2
06

Check for doubling

Now we need to prove that f(t2) = 4a, which would show that we indeed doubled the quantity in a constant time: f(t2) = a * (b^(c * ln(2) / ln(b)))^2 = 4a a * (b^(c * ln(2) / ln(b)))^2 = 4a Since the initial doubling time equation resulted in f(t) = 2a, the equation for doubling time (t2) holds true, proving that the doubling time for an exponentially increasing quantity is constant for all time.

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Most popular questions from this chapter

Compute the following derivatives using the method of your choice. $$\frac{d}{d x}\left(x^{\tan x}\right)$$

Determine whether the following statements are true and give an explanation or counterexample. a. \(\frac{d}{d x}(\sinh \ln 3)=\frac{\cosh \ln 3}{3}\) b. \(\frac{d}{d x}(\sinh x)=\cosh x\) and \(\frac{d}{d x}(\cosh x)=-\sinh x\) c. Differentiating the velocity equation for an ocean wave \(v=\sqrt{\frac{g \lambda}{2 \pi} \tanh \left(\frac{2 \pi d}{\lambda}\right)}\) results in the acceleration of the wave. d. \(\ln (1+\sqrt{2})=-\ln (-1+\sqrt{2})\) e. \(\int_{0}^{1} \frac{d x}{4-x^{2}}=\frac{1}{2}\left(\operatorname{coth}^{-1} \frac{1}{2}-\operatorname{coth}^{-1} 0\right)\)

Calculate the work required to stretch the following springs 0.5 m from their equilibrium positions. Assume Hooke's law is obeyed. a. A spring that requires a force of \(50 \mathrm{N}\) to be stretched \(0.2 \mathrm{m}\) from its equilibrium position. b. A spring that requires \(50 \mathrm{J}\) of work to be stretched \(0.2 \mathrm{m}\) from its equilibrium position.

a. Confirm that the linear approximation to \(f(x)=\tanh x\) at \(a=0\) is \(L(x)=x\) b. Recall that the velocity of a surface wave on the ocean is \(v=\sqrt{\frac{g \lambda}{2 \pi} \tanh \left(\frac{2 \pi d}{\lambda}\right)} .\) In fluid dynamics, shallow water refers to water where the depth-to-wavelength ratio \(d / \lambda<0.05 .\) Use your answer to part (a) to explain why the shallow water velocity equation is \(v=\sqrt{g d}\) c. Use the shallow-water velocity equation to explain why waves tend to slow down as they approach the shore.

A rigid body with a mass of \(2 \mathrm{kg}\) moves along a line due to a force that produces a position function \(x(t)=4 t^{2},\) where \(x\) is measured in meters and \(t\) is measured in seconds. Find the work done during the first 5 s in two ways. a. Note that \(x^{\prime \prime}(t)=8 ;\) then use Newton's second law \(\left(F=m a=m x^{\prime \prime}(t)\right)\) to evaluate the work integral \(W=\int_{x_{0}}^{x_{f}} F(x) d x,\) where \(x_{0}\) and \(x_{f}\) are the initial and final positions, respectively. b. Change variables in the work integral and integrate with respect to \(t .\) Be sure your answer agrees with part (a).
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