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Chapter 6: Problem 45

The same exponential growth function can be written in the forms \(y(t)=y_{0} e^{k t}, y(t)=y_{0}(1+r)^{t}\) and \(y(t)=y_{0} 2^{t / T_{2}} .\) Write \(k\) as a function of \(r, r\) as a function of \(T_{2}\), and \(T_{2}\) as a function of \(k\).

Short Answer

Expert verified
Question: Identify the relationships between the parameters k, r, and T鈧� for the given exponential growth functions. Answer: The relationships between the parameters are: 1. k = ln(1+r) 2. r = e^(ln(2)/T鈧�) - 1 3. T鈧� = ln(2)/k

Step by step solution

01

Write \(k\) as a function of \(r\).

Equating the first two forms of the exponential growth function, we have: \(y_{0}e^{kt} = y_{0}(1+r)^t\). Divide both sides by \(y_0\): \(e^{kt} = (1+r)^t\). Now, take the natural logarithm of both sides: \(kt = t\ln(1+r)\). Divide both sides by \(t\): \(k = \ln(1+r)\). This gives us \(k\) as a function of \(r\): \(k = \ln(1+r)\).
02

Write \(r\) as a function of \(T_{2}\).

Equating the second and third forms of the exponential growth function, we have: \(y_{0}(1+r)^t = y_{0}2^{t/T_{2}}\). Divide both sides by \(y_0\): \((1+r)^t = 2^{t/T_{2}}\). Now, take the natural logarithm of both sides: \(t\ln(1+r) = \frac{t}{T_{2}}\ln(2)\). Divide both sides by \(t\): \(\ln(1+r) = \frac{\ln(2)}{T_{2}}\). To solve for \(r\), use the exponential function: \(1+r = e^{\frac{\ln(2)}{T_{2}}}\). Subtract 1 from both sides: \(r = e^{\frac{\ln(2)}{T_{2}}} - 1\). This gives us \(r\) as a function of \(T_{2}\): \(r = e^{\frac{\ln(2)}{T_{2}}} - 1\).
03

Write \(T_{2}\) as a function of \(k\).

Equating the first and third forms of the exponential growth function, we have: \(y_{0}e^{kt} = y_{0}2^{t/T_{2}}\). Divide both sides by \(y_0\): \(e^{kt} = 2^{t/T_{2}}\). Now, take the natural logarithm of both sides: \(kt = \frac{t}{T_{2}}\ln(2)\). Divide both sides by \(t\): \(k = \frac{\ln(2)}{T_{2}}\). To solve for \(T_{2}\), divide both sides by \(\ln(2)\): \(T_{2} = \frac{\ln(2)}{k}\). This gives us \(T_{2}\) as a function of \(k\): \(T_{2} = \frac{\ln(2)}{k}\). In summary, the relationships between the parameters are: 1. \(k = \ln(1+r)\). 2. \(r = e^{\frac{\ln(2)}{T_{2}}} - 1\). 3. \(T_{2} = \frac{\ln(2)}{k}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Natural Logarithm
Understanding the natural logarithm is crucial when dealing with exponential growth functions. It is denoted by \(\ln\) and is a special logarithm where the base is the mathematical constant \(e\), approximately equal to 2.71828. This number \(e\) arises naturally in mathematics and is the base of the natural logarithm because of its unique properties in calculus, particularly concerning growth processes and interest calculations.

The natural logarithm has the defining property that \(\ln(e) = 1\), and conversely, that \(e\) raised to the power of a natural logarithm of a number \(x\) gives the number itself: \(e^{\ln(x)} = x\). This interplay between exponential and logarithmic functions allows us to write exponential equations in logarithmic form, thus simplifying the process of solving them, as seen in our exercise.

  • In the exercise, we took the natural logarithm of both sides to isolate the variable.
  • We used the property that the logarithm of a number to a power is the power times the logarithm of the number: \(\ln(a^b) = b\cdot\ln(a)\).
  • This allowed us to solve for the parameters of interest: \(k\), \(r\), and \(T_{2}\).
Exponential Equations
Exponential equations are equations in which variables occur as exponents. They take the general form \(y = ab^{x}\), where \(a\) is a constant, \(b\) is the base, and \(x\) is the exponent. In our exercise, we've dealt with the classic form of exponential growth functions, which model situations where growth rate is proportional to the size of the function at any point in time.

These equations can often be transformed into a more manageable form by using logarithms, particularly the natural logarithm. Here's how we applied this approach:
  • We started with the equation \(y(t) = y_{0}e^{kt}\) and used the natural logarithm to compare it to other forms of the same function.
  • By applying \(\ln\) to both sides, we could linearize the equation, thereby isolating the growth rate constant \(k\) or the rate \(r\).
  • This manipulation often simplifies the process, making it easier to solve for the unknown variables.
Exponential and Logarithmic Functions
Exponential and logarithmic functions are mathematical inverses of one another, which means that they can undo each other's operations. Exponential functions deal with situations where a quantity grows at a rate proportional to its current value, which leads to a rapid increase of the quantity over time. Logarithmic functions, on the other hand, help us determine the 'time' it takes for the exponential function to reach a certain level.

In the context of our exponential growth problem:
  • We recognized the connections between different forms of the exponential function and successfully transitioned from one form to another using the properties of these functions.
  • When we encounter \(y_{0}(1+r)^t\), a compound interest formula, we can relate it to the other forms to extract more information and solve for variables.
  • The reciprocal relationship between exponential and logarithmic functions became the key tool in deriving expressions for \(k\), \(r\), and \(T_{2}\) with respect to each other.

Understanding these concepts enables students to tackle a wide range of problems involving exponential growth and decay, as well as to comprehend complex financial calculations and population dynamics.

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Most popular questions from this chapter

Determine whether the following statements are true and give an explanation or counterexample. a. A pyramid is a solid of revolution. b. The volume of a hemisphere can be computed using the disk method. c. Let \(R_{1}\) be the region bounded by \(y=\cos x\) and the \(x\) -axis on \([-\pi / 2, \pi / 2] .\) Let \(R_{2}\) be the region bounded by \(y=\sin x\) and the \(x\) -axis on \([0, \pi] .\) The volumes of the solids generated when \(R_{1}\) and \(R_{2}\) are revolved about the \(x\) -axis are equal.

For each region \(R\), find the horizontal line \(y=k\) that divides \(R\) into two subregions of equal area. \(R\) is the region bounded by \(y=1-|x-1|\) and the \(x\) -axis.

Archimedes' principle says that the buoyant force exerted on an object that is (partially or totally) submerged in water is equal to the weight of the water displaced by the object (see figure). Let \(\rho_{w}=1 \mathrm{g} / \mathrm{cm}^{3}=1000 \mathrm{kg} / \mathrm{m}^{3}\) be the density of water and let \(\rho\) be the density of an object in water. Let \(f=\rho / \rho_{w}\). If \(01,\) then the object sinks. Consider a cubical box with sides \(2 \mathrm{m}\) long floating in water with one-half of its volume submerged \(\left(\rho=\rho_{w} / 2\right) .\) Find the force required to fully submerge the box (so its top surface is at the water level).

Theo and Sasha start at the same place on a straight road, riding bikes with the following velocities (measured in \(\mathrm{mi} / \mathrm{hr}\) ). Assume \(t\) is measured in hours. Theo: \(v_{T}(t)=10,\) for \(t \geq 0\) Sasha: \(v_{S}(t)=15 t,\) for \(0 \leq t \leq 1\) and \(v_{S}(t)=15,\) for \(t>1\) a. Graph the velocity functions for both riders. b. If the riders ride for 1 hr, who rides farther? Interpret your answer geometrically using the graphs of part (a). c. If the riders ride for 2 hr, who rides farther? Interpret your answer geometrically using the graphs of part (a). d. Which rider arrives first at the \(10-, 15-\), and 20 -mile markers of the race? Interpret your answer geometrically using the graphs of part (a). e. Suppose Sasha gives Theo a head start of \(0.2 \mathrm{mi}\) and the riders ride for 20 mi. Who wins the race? f. Suppose Sasha gives Theo a head start of \(0.2 \mathrm{hr}\) and the riders ride for 20 mi. Who wins the race?

Suppose the acceleration of an object moving along a line is given by \(a(t)=-k v(t),\) where \(k\) is a positive constant and \(v\) is the object's velocity. Assume that the initial velocity and position are given by \(v(0)=10\) and \(s(0)=0,\) respectively. a. Use \(a(t)=v^{\prime}(t)\) to find the velocity of the object as a function of time. b. Use \(v(t)=s^{\prime}(t)\) to find the position of the object as a function of time. c. Use the fact that \(d v / d t=(d v / d s)(d s / d t)\) (by the Chain Rule) to find the velocity as a function of position.
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