/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 5 Let \(A=A(t)\) be the amount of ... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 5

Let \(A=A(t)\) be the amount of lead present at time \(t .\) From \(d A / d t=k A\) and \(A(0)=1\) we obtain \(A=e^{k t}\) \(\operatorname{Using} A(3.3)=1 / 2\) we find \(k=\frac{1}{3.3} \ln (1 / 2) .\) When \(90 \%\) of the lead has decayed, 0.1 grams will remain. Setting \(A(t)=0.1\) we have \(e^{t(1 / 3.3) \ln (1 / 2)}=0.1,\) so $$\frac{t}{3.3} \ln \frac{1}{2}=\ln 0.1 \quad \text { and } \quad t=\frac{3.3 \ln 0.1}{\ln (1 / 2)} \approx 10.96 \text { hours. }$$

Short Answer

Expert verified
When 90% of lead has decayed, it will take approximately 10.96 hours.

Step by step solution

01

Interpret Given Information

We have a differential equation \( \frac{dA}{dt} = kA \), which describes the rate of change of an amount \( A \) (such as a chemical quantity) over time \( t \). Initially at time \( t = 0 \), the amount \( A(0) = 1 \). Given the function \( A = e^{kt} \) satisfies these conditions.
02

Using Initial Condition

Use the additional information \( A(3.3) = \frac{1}{2} \) to determine \( k \). In the exponential decay model, at \( t = 3.3 \), half of the lead remains, thus \( A(3.3) = e^{3.3k} = \frac{1}{2} \). Solving for \( k \), we have \( e^{3.3k} = \frac{1}{2} \), so \( 3.3k = \ln(\frac{1}{2}) \) and \( k = \frac{\ln(1/2)}{3.3} \).
03

Setting Up When 90% Decays

Since 90% of the lead has decayed, 10% remains, making \( A(t) = 0.1 \). This setup needs to plug into the model equation; thus, \( e^{kt} = 0.1 \).
04

Solve for Time \( t \)

Substitute \( k = \frac{\ln(1/2)}{3.3} \) and solve \( e^{t(\ln(1/2)/3.3)} = 0.1 \). This can be rearranged to \( \frac{t}{3.3} \ln(1/2) = \ln(0.1) \). Solve for \( t \) by isolating it, giving \( t = 3.3 \cdot \frac{\ln(0.1)}{\ln(1/2)} \).
05

Compute the Value of \( t \)

Calculate the value: \( t = 3.3 \times \frac{-2.3026}{-0.6931} \approx 10.96 \) hours, thus determining the time when 90% has decayed.

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Key Concepts

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

Differential Equations
Differential equations form the cornerstone of modeling dynamic systems in mathematics and science. They describe how a quantity changes over time. In the case of exponential decay, the differential equation given is \( \frac{dA}{dt} = kA \). Here, \( A \) denotes the quantity that decays over time, and \( k \) is a constant that determines the rate of decay.

When solving these equations, the goal is to find a function \( A(t) \) that satisfies the equation, representing the amount at any time \( t \). Using the technique of separation of variables or recognizing the form, we can verify that the solution is \( A = e^{kt} \). This expresses how the amount \( A \) changes exponentially with respect to time.

Understanding this basic framework is crucial for effectively applying differential equations to real-world situations, like calculating half-life or predicting decay in various scientific fields.
Half-Life Calculation
The concept of half-life is a way to quantify how long it takes for a substance to reduce to half its initial amount. In exponential decay scenarios, it is a critical aspect that helps us understand the speed of the decay process. For the example given, at time \( t = 3.3 \), half of the lead has decayed, which is why \( A(3.3) = \frac{1}{2} \).

To find the decay constant \( k \), we use this half-life information in our exponential function. Solving for \( k \) from the equation \( e^{3.3k} = \frac{1}{2} \), gives an expression \( k = \frac{\ln(1/2)}{3.3} \). This constant \( k \) is then used to calculate other time-dependent quantities, such as how long it takes for 90% of the lead to decay.

Half-life calculations are not only used in chemistry and physics but also in many fields like archaeology for radiocarbon dating and medicine for the breakdown of drugs in the body.
Exponential Functions
Exponential functions are mathematical functions of the form \( f(x) = a \times e^{bx} \), where \( e \) is the base of the natural logarithm, and \( a \) and \( b \) are constants. These functions are fundamental in modeling processes that grow or decay at a rate proportional to their current value.

In the context of the exercise, \( A = e^{kt} \) is an example of an exponential decay function with a constant \( k \). The function describes how the amount of lead decreases exponentially over time. The decay occurs at a continuous rate described by \( k \).

Exponential functions are key in a wide variety of applications, including physics for radioactive decay, finance for compound interest calculations, and biology for population growth models. Understanding how they work enables us to apply them effectively to solve practical problems.

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Most popular questions from this chapter

For \(\frac{d T}{d t}-k T=-T_{m} k\) an integrating factor is \(e^{\int(-k) d t}=e^{-k t}\) so that \(\frac{d}{d t}\left[e^{-k t} T\right]=-T_{m} k e^{-k t}\) and \(T=T_{m}+c e^{k t}\) for \(-\infty< t<\infty .\) If \(T(0)=T_{0}\) then \(c=T_{0}-T_{m}\) and \(T=T_{m}+\left(T_{0}-T_{m}\right) e^{k t}\).

From \(\frac{1}{x^{2}+1} d x=4 d t\) we obtain \(\tan ^{-1} x=4 t+c .\) Using \(x(\pi / 4)=1\) we find \(c=-3 \pi / 4 .\) The solution of the initial-value problem is \(\tan ^{-1} x=4 t-\frac{3 \pi}{4}\) or \(x=\tan \left(4 t-\frac{3 \pi}{4}\right)\).

(a) The second derivative of \(y\) is \\[\frac{d^{2} y}{d x^{2}}=-\frac{d y / d x}{(y-1)^{2}}=-\frac{1 /(y-3)}{(y-3)^{2}}=-\frac{1}{(y-3)^{3}}\\] The solution curve is concave down when \(d^{2} y / d x^{2}<0\) or \(y>3\), and concave up when \(d^{2} y / d x^{2}>0\) or \(y<3 .\) From the phase portrait we see that the solution curve is decreasing when \(y<3\) and increasing when \(y>3\). (b) Separating variables and integrating we obtain \\[ \begin{aligned} (y-3) d y &=d x \\ \frac{1}{2} y^{2}-3 y &=x+c \\ y^{2}-6 y+9 &=2 x+c_{1} \\ (y-3)^{2} &=2 x+c_{1} \\ y &=3 \pm \sqrt{2 x+c_{1}}.\end{aligned}\\] The initial condition dictates whether to use the plus or minus sign. When \(y_{1}(0)=4\) we have \(c_{1}=1\) and \(y_{1}(x)=3+\sqrt{2 x+1}\). When \(y_{2}(0)=2\) we have \(c_{1}=1\) and \(y_{2}(x)=3-\sqrt{2 x+1}.\) When \(y_{3}(1)=2\) we have \(c_{1}=-1\) and \(y_{3}(x)=3-\sqrt{2 x-1}.\) When \(y_{4}(-1)=4\) we have \(c_{1}=3\) and \(y_{4}(x)=3+\sqrt{2 x+3}\).

Let \(u=x+y\) so that \(d u / d x=1+d y / d x .\) Then \(\frac{d u}{d x}-1=\cos u\) and \(\frac{1}{1+\cos u} d u=d x .\) Now $$\frac{1}{1+\cos u}=\frac{1-\cos u}{1-\cos ^{2} u}=\frac{1-\cos u}{\sin ^{2} u}=\csc ^{2} u-\csc u \cot u$$ so we have \(\int\left(\csc ^{2} u-\csc u \cot u\right) d u=\int d x\) and \(-\cot u+\csc u=x+c .\) Thus \(-\cot (x+y)+\csc (x+y)=x+c\) Setting \(x=0\) and \(y=\pi / 4\) we obtain \(c=\sqrt{2}-1 .\) The solution is $$\csc (x+y)-\cot (x+y)=x+\sqrt{2}-1.$$

(a) Let \(\rho\) be the weight density of the water and \(V\) the volume of the object. Archimedes' principle states that the upward buoyant force has magnitude equal to the weight of the water displaced. Taking the positive direction to be down, the differential equation is \\[ m \frac{d v}{d t}=m g-k v^{2}-\rho V \\] (b) Using separation of variables we have \\[ \begin{aligned} \frac{m d v}{(m g-\rho V)-k v^{2}} &=d t \\ \frac{m}{\sqrt{k}} \frac{\sqrt{k} d v}{(\sqrt{m g-\rho V})^{2}-(\sqrt{k} v)^{2}} &=d t \\ \frac{m}{\sqrt{k}} \frac{1}{\sqrt{m g-\rho V}} \tanh ^{-1} \frac{\sqrt{k} v}{\sqrt{m g-\rho V}} &=t+c \end{aligned} \\] Thus \\[ v(t)=\sqrt{\frac{m g-\rho V}{k}} \tanh \left(\frac{\sqrt{k m g-k \rho V}}{m} t+c_{1}\right) \\] (c) since \(\tanh t \rightarrow 1\) as \(t \rightarrow \infty,\) the terminal velocity is \(\sqrt{(m g-\rho V) / k}.\)
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