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Chapter 2: Problem 29

Assume \(R d q / d t+(1 / C) q=E(t), R=200, C=10^{-4},\) and \(E(t)=100\) so that \(q=1 / 100+c e^{-50 t} .\) If \(q(0)=0\) then \(c=-1 / 100\) and \(i=\frac{1}{2} e^{-50 t}\).

Short Answer

Expert verified
The solution is consistent with \( c = -\frac{1}{100} \) and \( i(t) = \frac{1}{2} e^{-50t} \).

Step by step solution

01

Identify the Given Equation and Parameters

We are given the differential equation \( R \frac{dq}{dt} + \frac{1}{C} q = E(t) \) with parameters \( R = 200 \), \( C = 10^{-4} \), and \( E(t) = 100 \). We also have the solution form \( q = \frac{1}{100} + c e^{-50t} \).
02

Determine the Value of c Using Initial Condition

Since \( q(0) = 0 \), we substitute this into the solution form: \( 0 = \frac{1}{100} + c e^{0} \). This simplifies to \( 0 = \frac{1}{100} + c \), leading to \( c = -\frac{1}{100} \).
03

Confirm Charge Function q(t)

Substitute \( c = -\frac{1}{100} \) back into the general solution to get \( q(t) = \frac{1}{100} - \frac{1}{100} e^{-50t} \). This satisfies the initial condition \( q(0) = 0 \).
04

Define and Confirm Current Function i(t)

Based on the solution, the current \( i(t) = \frac{dq}{dt} \) is given as \( i = \frac{1}{2} e^{-50t} \). We differentiate \( q(t) \): \( \frac{d}{dt} \left( \frac{1}{100} - \frac{1}{100} e^{-50t} \right) = 50 \times \frac{1}{100} e^{-50t} = \frac{1}{2} e^{-50t} \). This confirms \( i(t) \) as computed.

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Key Concepts

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

First-order differential equations
A first-order differential equation involves derivatives of the first degree. In these equations, the solution depends on the value of one derivative.
These types of equations model a variety of real-world phenomena such as electrical circuits, population growth, and fluid movement. They are typically written in the form:
  • \( \frac{dy}{dt} = f(t, y) \)
In our specific exercise, the given first-order differential equation is:
  • \( R \frac{dq}{dt} + \frac{1}{C} q = E(t) \)
Here, \( R \) and \( C \) are constants, \( E(t) \) represents an external input, often referred to as a driving function, and \( q \) represents the charge.
Solving such equations often involves identifying the dependent and independent variables, and using initial conditions to determine the particular solution.
Initial value problem
An initial value problem (IVP) in differential equations emphasizes finding a solution when some initial conditions are given.
It is crucial because a differential equation can have many possible solutions; the initial conditions help pinpoint a unique one. In our context, the initial condition states that the charge \( q(0) = 0 \) when \( t = 0 \).
This condition is essential to determining the constant \( c \) in the equation \( q = \frac{1}{100} + c e^{-50t} \). By substituting \( q(0) = 0 \) into the expression, we find that:
  • \( 0 = \frac{1}{100} + c \times e^{0} \)
  • This simplifies to \( c = -\frac{1}{100} \)
Thus, the initial condition helps us adjust our equation to accurately reflect the starting scenario of our model.
Exponential solutions
In many differential equations, exponential functions emerge as part of the solution due to their unique derivatives.
  • The exponential function is often in the form \( e^{kt} \), where \( k \) is a constant.
In our case, the solution possesses an exponential component \( c e^{-50t} \).
The properties of the exponential function are useful because they model decay or growth processes efficiently. Here, the exponential term represents how the charge \( q(t) \) changes over time, decaying as time progresses.
Exponential solutions are significant in real life for processes that involve rapid change or decay, such as radioactive decay, cooling, and electrical discharge in circuits. By understanding the rate of change determined by the exponent, one can predict future behavior more accurately.

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Most popular questions from this chapter

Assume that \(A=A_{0} e^{k t}\) and \(k=-0.00012378\). If \(A(t)=0.145 A_{0}\) then \(t \approx 15,600\) years.

Separating variables and integrating we obtain \\[\frac{d x}{\sqrt{1-x^{2}}}-\frac{d y}{\sqrt{1-y^{2}}}=0 \quad \text { and } \quad \sin ^{-1} x-\sin ^{-1} y=c\\]. Setting \(x=0\) and \(y=\sqrt{3} / 2\) we obtain \(c=-\pi / 3 .\) Thus, an implicit solution of the initial-value problem is \(\sin ^{-1} x-\sin ^{-1} y=\pi / 3 .\) Solving for \(y\) and using an addition formula from trigonometry, we get \\[y=\sin \left(\sin ^{-1} x+\frac{\pi}{3}\right)=x \cos \frac{\pi}{3}+\sqrt{1-x^{2}} \sin \frac{\pi}{3}=\frac{x}{2}+\frac{\sqrt{3} \sqrt{1-x^{2}}}{2}\\].

(a) Separating variables we have \(2 y d y=(2 x+1) d x .\) Integrating gives \(y^{2}=x^{2}+x+c .\) When \(y(-2)=-1\) we find \(c=-1,\) so \(y^{2}=x^{2}+x-1\) and \(y=-\sqrt{x^{2}+x-1} .\) The negative square root is chosen because of the initial condition. (b) From the figure, the largest interval of definition appears to be approximately \((-\infty,-1.65)\). (c) Solving \(x^{2}+x-1=0\) we get \(x=-\frac{1}{2} \pm \frac{1}{2} \sqrt{5},\) so the largest interval of definition is \(\left(-\infty,-\frac{1}{2}-\frac{1}{2} \sqrt{5}\right)\) The right-hand endpoint of the interval is excluded because \(y=-\sqrt{x^{2}+x-1}\) is not differentiable at this point.

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.

While the object is in the air its velocity is modelled by the linear differential equation \(m d v / d t=m g-k v .\) Using \(m=160, k=\frac{1}{4},\) and \(g=32,\) the differential equation becomes \(d v / d t+(1 / 640) v=32 .\) The integrating factor is \(e^{\int d t / 640}=e^{t / 640}\) and the solution of the differential equation is \(e^{t / 640} v=\int 32 e^{t / 640} d t=20,480 e^{t / 640}+c\) Using \(v(0)=0\) we see that \(c=-20,480\) and \(v(t)=20,480-20,480 e^{-t / 640} .\) Integrating we get \(s(t)=20,480 t+\) \(13,107,200 e^{-t / 640}+c .\) since \(s(0)=0, c=-13,107,200\) and \(s(t)=-13,107,200+20,480 t+13,107,200 e^{-t / 640}\) To find when the object hits the liquid we solve \(s(t)=500-75=425,\) obtaining \(t_{a}=5.16018 .\) The velocity at the time of impact with the liquid is \(v_{a}=v\left(t_{a}\right)=164.482 .\) When the object is in the liquid its velocity is modeled by the nonlinear differential equation \(m d v / d t=m g-k v^{2} .\) Using \(m=160, g=32,\) and \(k=0.1\) this becomes \(d v / d t=\left(51,200-v^{2}\right) / 1600 .\) Separating variables and integrating we have \\[ \frac{d v}{51,200-v^{2}}=\frac{d t}{1600} \quad \text { and } \quad \frac{\sqrt{2}}{640} \ln \left|\frac{v-160 \sqrt{2}}{v+160 \sqrt{2}}\right|=\frac{1}{1600} t+c \\] Solving \(v(0)=v_{a}=164.482\) we obtain \(c=-0.00407537 .\) Then, for \(v<160 \sqrt{2}=226.274\) \\[ \left|\frac{v-160 \sqrt{2}}{v+160 \sqrt{2}}\right|=e^{\sqrt{2} t / 5-1.8443} \quad \text { or } \quad-\frac{v-160 \sqrt{2}}{v+160 \sqrt{2}}=e^{\sqrt{2} t / 5-1.8443} \\] Solving for \(v\) we get \\[ v(t)=\frac{13964.6-2208.29 e^{\sqrt{2} t / 5}}{61.7153+9.75937 e^{\sqrt{2} t / 5}} \\] Integrating we find \\[ s(t)=226.275 t-1600 \ln \left(6.3237+e^{\sqrt{2} t / 5}\right)+c \\] Solving \(s(0)=0\) we see that \(c=3185.78,\) so \\[ s(t)=3185.78+226.275 t-1600 \ln \left(6.3237+e^{\sqrt{2} t / 5}\right) \\] To find when the object hits the bottom of the tank we solve \(s(t)=75,\) obtaining \(t_{b}=0.466273 .\) The time from when the object is dropped from the helicopter to when it hits the bottom of the tank is \(t_{a}+t_{b}=\) 5.62708 seconds.
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