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In Problems 1-22, solve the given differential equation by separation of variables. \(\frac{d S}{d r}=k S\) 16\. \(\frac{d Q}{d t}=k(Q-70)\)

Short Answer

Expert verified
The solution is \(Q = Ce^{kt} + 70\).

Step by step solution

01

Identify the variables

The given differential equation is \(\frac{d Q}{d t} = k(Q - 70)\). Here, \(Q\) is the dependent variable and \(t\) is the independent variable.
02

Separate the variables

To separate the variables, we need all terms involving \(Q\) on one side of the equation and all terms involving \(t\) on the other side. Start by rewriting the equation: \(\frac{dQ}{(Q - 70)} = k dt\).
03

Integrate both sides

Integrate both sides of the separated equation. For the left side, integrate with respect to \(Q\), and for the right side, integrate with respect to \(t\):\[\int \frac{1}{Q-70} \, dQ = \int k \, dt\].This gives:\[\ln |Q-70| = kt + C\], where \(C\) is the constant of integration.
04

Solve for Q

Exponentiate both sides to remove the natural logarithm:\[Q - 70 = e^{kt+C}\].This can be rewritten as:\[Q - 70 = Ce^{kt}\].Finally, solve for \(Q\):\[Q = Ce^{kt} + 70\].

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Key Concepts

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

Differential Equations
A differential equation is a mathematical equation that involves functions and their derivatives. It tells us how one variable changes with respect to another. In simpler terms, it's an equation that involves rates of change. Think about how the speed of your car changes over time; that's a real-world example of a differential equation at work. In our exercise, the equation \(\frac{dQ}{dt} = k(Q-70)\) represents how the quantity \(Q\) changes over time \(t\). Since it involves both the function \(Q\) and its derivative with respect to time, it is classified as a differential equation.Differential equations are extremely powerful as they can describe various physical phenomena, such as motion, growth, decay, and more. By solving these equations, we can predict future behavior or understand underlying processes.
Separation of Variables
Separation of variables is a technique used to simplify and solve differential equations. It involves rearranging an equation so that each variable appears on its own side of the equation. The goal is to isolate the dependent variable (for example, \(Q\) in our exercise) and its derivatives on one side of the equation. Meanwhile, move the independent variable (like \(t\), in this case) to the other side. Once the variables are separated, you can integrate each side independently. In our example, we separate the equation \(\frac{dQ}{dt} = k(Q-70)\) into \(\frac{dQ}{Q-70} = k \, dt\). This makes it easier to solve because we can now focus on integrating each part separately.
Integration
Integration is a fundamental concept in calculus. It is the process of finding the integral of a function, which can be understood as the reverse of differentiation. In the context of differential equations, integration helps us find a function from its derivative. After separating the variables in a differential equation, the next step is to integrate both sides. In our example, we integrate \(\int \frac{1}{Q-70} \, dQ\) and \(\int k \, dt\). The integration involves finding antiderivatives, resulting in expressions like \[\ln |Q-70| = kt + C\]. Here, \(C\) is called the constant of integration. Integration can solve problems related to area under curves, total change, and more, making it versatile and essential in mathematical analysis.
Exponential Growth and Decay
Exponential growth and decay are processes described by specific types of differential equations, usually involving continuous change. In exponential growth, the rate of change of a quantity is directly proportional to the quantity itself. This leads to rapid increase over time. Conversely, exponential decay refers to processes where the quantity decreases over time, such as radioactive decay.In our exercise solution, we reached \(Q = Ce^{kt} + 70\), an equation that describes how \(Q\) behaves over time. If \(k > 0\), the function exhibits exponential growth. If \(k < 0\), it demonstrates exponential decay. This concept is widely applied in science and finance, such as predicting population growth or calculating compound interest. Understanding exponential behavior helps in predicting how systems evolve over time.

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Most popular questions from this chapter

The number \(N(t)\) of people in a community who are exposed to a particularadvertisement is governed by the logisticequation. Initially \(N(0)=500\), and it is observed that \(N(1)=1000\). Solve for \(N(t)\) if it is predicted that the limiting number of people in the community who will see the advertisement is 50,000 .

(a) Use a CAS and the concept of level curves to plot representative graphs of members of the family of solutions of the differential equation \(\frac{d y}{d x}=-\frac{8 x+5}{3 y^{2}+1} .\) Experiment with different numbers of level curves as well as various rectangular regions defined by \(a \leq x \leq b, c \leq y \leq d\). (b) On separate coordinate axes plot the graphs of the particular solutions corresponding to the initial conditions: \(y(0)=-1 ; y(0)=2 ; y(-1)=4 ; y(-1)=-3\)

A model for the population \(P(t)\) in a suburb of a large city is given by the initial-value problem $$ \frac{d P}{d t}=P\left(10^{-1}-10^{-7} P\right), \quad P(0)=5000, $$ where \(t\) is measured in months. What is the limiting value of the population? At what time will the population be equal to one-half of this limiting value?

A tank in the form of a right circular cylinder standing on end is leaking water through a circular hole in its bottom. As we saw in (10) of Section 1.3, when friction and contraction of water at the hole are ignored, the height \(h\) of water in the tank is described by $$ \frac{d h}{d t}=-\frac{A_{h}}{A_{w}} \sqrt{2 g h}, $$ where \(A_{w}\) and \(A_{h}\) are the cross-sectional areas of the water and the hole, respectively. (a) Solve for \(h(t)\) if the initial height of the water is \(H\). By hand, sketch the graph of \(h(t)\) and give its interval \(I\) of definition in terms of the symbols \(A_{w}, A_{h}\), and \(H\). Use \(g=32 \mathrm{ft} / \mathrm{s}^{2}\) (b) Suppose the tank is \(10 \mathrm{ft}\) high and has radius \(2 \mathrm{ft}\) and the circular hole has radius \(\frac{1}{2}\) in. If the tank is initially full, how long will it take to empty?

Solve the given initial-value problem. Give the largest interval \(I\) over which the solution is defined. $$ y \frac{d x}{d y}-x=2 y^{2}, \quad y(1)=5 $$

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