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Chapter 2: Problem 34

As \(x \rightarrow-\infty, e^{6 x} \rightarrow 0\) and \(y \rightarrow 2 x+3 .\) Now write \(\left(1+c e^{6 x}\right) /\left(1-\alpha e^{6 x}\right)\) as \(\left(e^{-6 x}+c\right) /\left(e^{-6 x}-c\right) .\) Then, as \(x \rightarrow \infty, e^{-6 x} \rightarrow 0\) and \(y \rightarrow 2 x-3\).

Short Answer

Expert verified
As \(x \to -\infty\), \(y \to 2x+3\). As \(x \to \infty\), \(y \to 2x-3\).

Step by step solution

01

Understanding the Behavior at Negative Infinity

As we consider the behavior of the exponential function, notice that as \(x \rightarrow -\infty\), the expression \(e^{6x}\) approaches 0. In this scenario, the term \(y\) simplifies to \(2x + 3\). We need to find the corresponding behavior when expressed with \(e^{-6x}\).
02

Manipulating the Expression

We are given the expression \(\frac{1+ce^{6x}}{1-\alpha e^{6x}}\). To transform it, let's multiply both numerator and denominator by \(e^{-6x}\). This multiplication is essentially dividing each term by \(e^{6x}\):
03

Multiplying by \(e^{-6x}\)

Multiply the terms: \(\left(1 + ce^{6x}\right) e^{-6x} = e^{-6x} + c\) and \(\left(1 - \alpha e^{6x}\right)e^{-6x} = e^{-6x} - \alpha\). Thus, we rewrite the expression as \(\frac{e^{-6x} + c}{e^{-6x} - \alpha}\).
04

Understanding the Behavior at Positive Infinity

As \(x \rightarrow \infty\), note that \(e^{-6x}\) approaches 0. In this altered expression, \(y\) simplifies to \(2x - 3\). This is because the exponential term vanishes, reflecting a change in the offset of the linear term.

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Key Concepts

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

Limits and Behavior at Infinity
In mathematics, a **limit** describes the behavior of a function as the input approaches a particular value. When we examine functions at infinity, we're assessing how a function behaves as the variable grows infinitely large or small. Consider the exponential function
  • As \( x \to -\infty \), the expression \( e^{6x} \to 0 \). This means that the value of the function diminishes towards zero, effectively disappearing as we move left on the number line.
  • Conversely, as \( x \to \infty \), the expression \( e^{-6x} \to 0 \). Here, the exponential decay results in the function diminishing from zero as we move rightward on the number line.
In practice, these limits help determine the resulting behavior of the function or expression as the input becomes exceedingly large or small. This is crucial when simplifying expressions to find approximate values or behaviors without the overshadowing influence of remaining components.
Expression Manipulation
**Expression manipulation** involves creatively altering mathematical expressions to reveal or simplify their structure. In our scenario, we start with the expression \( \frac{1+c e^{6x}}{1-\alpha e^{6x}} \).
Notice that multiplying the entire expression by \( e^{-6x} \) is a strategic move. It effectively changes the basis of the expression:
  • For the numerator: \( (1+ce^{6x})e^{-6x} = e^{-6x} + c \).
  • For the denominator: \( (1-\alpha e^{6x})e^{-6x} = e^{-6x} - \alpha \).
Multiplying by \( e^{-6x} \) aligns the exponential powers, simplifying the transformation. This approach stems from wanting to analyze the behavior at positive infinity, making it more intuitive. Manipulation aids in restructuring problems to highlight aspects like limits or simplification, essential strategies for algebraic problem-solving.
Exponential Decay
**Exponential decay** is a process where a quantity decreases over time, proportionally to its current value. In our exercise, this concept can be observed in
  • As \( x \to \infty \), \( e^{-6x} \to 0 \) depicts decay since the term diminishes quickly.
This behavior is inherent in many real-world scenarios, such as radioactive decay, depreciation, and cooling laws.
The characteristic feature of exponential decay is its rapid drop-off鈥攐nce a base element decreases, it's multiplied by a constant factor each time period, further reducing its overall presence. This type of decay is indicative of inverse relationships, where large movements in \( x \) result in small resulting values for the associated function. When expressed in a modified function, this allows for simpler calculation by negating terms that become negligible beyond a certain boundary.
Understanding exponential decay is crucial in many fields like physics, finance, and population studies, and helps predict how quantities change over time.

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Most popular questions from this chapter

From \(d A / d t=10-A / 100\) we obtain \(A=1000+c e^{-t / 100} .\) If \(A(0)=0\) then \(c=-1000\) and \(A(t)=\) \(1000-1000 e^{-t / 100}\).

From \(\frac{1}{1-2 y} d y=d t\) we obtain \(-\frac{1}{2} \ln |1-2 y|=t+c\) or \(1-2 y=c_{1} e^{-2 t} .\) Using \(y(0)=5 / 2\) we find \(c_{1}=-4\). The solution of the initial-value problem is \(1-2 y=-4 e^{-2 t}\) or \(y=2 e^{-2 t}+\frac{1}{2}\).

The left-hand derivative of the function at \(x=1\) is \(1 / e\) and the right- hand derivative at \(x=1\) is \(1-1 / e\). Thus, \(y\) is not differentiable at \(x=1\).

(a) Initially the tank contains 300 gallons of solution. since brine is pumped in at a rate of 3 gal/min and the mixture is pumped out at a rate of 2 gal/min, the net change is an increase of 1 gal/min. Thus, in 100 minutes the tank will contain its capacity of 400 gallons. (b) The differential equation describing the amount of salt in the tank is \(A^{\prime}(t)=6-2 A /(300+t)\) with solution $$A(t)=600+2 t-\left(4.95 \times 10^{7}\right)(300+t)^{-2}, \quad 0 \leq t \leq 100$$ as noted in the discussion following Example 5 in the text. Thus, the amount of salt in the tank when it overflows is $$A(100)=800-\left(4.95 \times 10^{7}\right)(400)^{-2}=490.625 \mathrm{lbs}$$. (c) When the tank is overflowing the amount of salt in the tank is governed by the differential equation $$\begin{aligned} \frac{d A}{d t} &=(3 \mathrm{gal} / \mathrm{min})(2 \mathrm{lb} / \mathrm{gal})-\left(\frac{A}{400} \mathrm{lb} / \mathrm{gal}\right)(3 \mathrm{gal} / \mathrm{min}) \\ &=6-\frac{3 A}{400}, \quad A(100)=490.625 \end{aligned}$$ Solving the equation, we obtain \(A(t)=800+c e^{-3 t / 400} .\) The initial condition yields \(c=-654.947,\) so that $$A(t)=800-654.947 e^{-3 t / 400}$$. When \(t=150, A(150)=587.37\) lbs. (d) As \(t \rightarrow \infty,\) the amount of salt is 800 lbs, which is to be expected since \((400 \text { gal })(2 \mathrm{lb} / \mathrm{gal})=800 \mathrm{lbs}\) (e)

The differential equation for the first container is \(d T_{1} / d t=k_{1}\left(T_{1}-0\right)=k_{1} T_{1},\) whose solution is \(T_{1}(t)=c_{1} e^{k_{1} t}\) since \(T_{1}(0)=100\) (the initial temperature of the metal bar), we have \(100=c_{1}\) and \(T_{1}(t)=100 e^{k_{1} t}\). After 1 minute, \(T_{1}(1)=100 e^{k_{1}}=90^{\circ} \mathrm{C},\) so \(k_{1}=\ln 0.9\) and \(T_{1}(t)=100 e^{t \ln 0.9} .\) After 2 minutes, \(T_{1}(2)=100 e^{2 \ln 0.9}=\) \(100(0.9)^{2}=81^{\circ} \mathrm{C}\). The differential equation for the second container is \(d T_{2} / d t=k_{2}\left(T_{2}-100\right),\) whose solution is \(T_{2}(t)=\) \(100+c_{2} e^{k_{2} t} .\) When the metal bar is immersed in the second container, its initial temperature is \(T_{2}(0)=81,\) so $$T_{2}(0)=100+c_{2} e^{k_{2}(0)}=100+c_{2}=81$$ and \(c_{2}=-19 .\) Thus, \(T_{2}(t)=100-19 e^{k_{2} t} .\) After 1 minute in the second tank, the temperature of the metal bar is \(91^{\circ} \mathrm{C},\) so $$\begin{aligned} T_{2}(1) &=100-19 e^{k_{2}}=91 \\ e^{k_{2}} &=\frac{9}{19} \\ k_{2} &=\ln \frac{9}{19} \end{aligned}$$ and \(T_{2}(t)=100-19 e^{t \ln (9 / 19)} .\) Setting \(T_{2}(t)=99.9\) we have $$\begin{aligned} 100-19 e^{t \ln (9 / 19)} &=99.9 \\ e^{t \ln (9 / 19)} &=\frac{0.1}{19} \\ t &=\frac{\ln (0.1 / 19)}{\ln (9 / 19)} \approx 7.02 \end{aligned}$$ Thus, from the start of the "double dipping" process, the total time until the bar reaches \(99.9^{\circ} \mathrm{C}\) in the second container is approximately 9.02 minutes.
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