Question

A quantity grows with time such that its rate of growth \(d y / d t\) is proportional to the present amount \(y .\) Use this statement to derive the equation for exponential growth, \(y=a e^{n t}\)

Short answer

The exponential growth equation is derived by setting up the differential equation \(\frac{dy}{dt} = ky\), separating variables, integrating, and applying the constant of integration to get \(y = ae^{nt}\).

Step-by-step solution

Express the proportional relationship

Given that the rate of growth \(\frac{dy}{dt}\) is proportional to the present amount \(y\), we can express this relationship as a differential equation: \[\frac{dy}{dt} = ky\], where \(k\) is the proportionality constant.

Separate variables

To solve the differential equation, we need to separate the variables, \(y\) and \(t\). We rearrange the equation to get all the \(y\)-terms on one side and all the \(t\)-terms on the other side: \[\frac{1}{y} dy = k dt\].

Integrate both sides

Integrate both sides of the equation to get the antiderivatives. The integral of \(\frac{1}{y}dy\) is \(\ln|y|\) and the integral of \(kdt\) is \(kt\). So, we obtain: \[\int \frac{1}{y} dy = \int k dt\] which gives us \[\ln|y| = kt + C\], where \(C\) is the constant of integration.

Solve for \(y\)

Exponentiate both sides to solve for \(y\). Note that \(e^{\ln|y|} = |y|\) and the absolute value can be dropped because the amount \(y\) is always positive. Thus we get: \[y = e^{kt + C}\].

Express the constant \(C\) as \(\ln(a)\)

We can express the constant \(C\) as \(\ln(a)\) for some constant \(a\), since the constant of integration can be any real number: \[y = e^{kt + \ln(a)}\].

Combine constants

Combine the exponentiated constants using properties of exponents: \[y = e^{\ln(a)}e^{kt}\]. Since \(e^{\ln(a)} = a\), we get the exponential growth equation: \[y = ae^{kt}\].

Introduce the notational change

Since the proportionality constant \(k\) can be denoted by any symbol, we replace \(k\) with \(n\) to get the final form of the equation: \[y = ae^{nt}\].

Key concepts

Differential Equations

Differential equations are a type of equation that involve derivatives, expressing how a quantity changes in relation to another quantity. For instance, the differential equation \( \frac{dy}{dt} = ky \) from our exercise shows that the rate of change of the quantity \( y \) over time \( t \) is proportional to the quantity itself. To solve such equations, one typically identifies the type (such as linear, separable, exact, etc.) and applies the appropriate method to find the solution.

Separation of Variables

The technique of separation of variables involves rearranging a differential equation so that each variable and its differential appear on opposite sides of the equation. In our example, we have \( \frac{1}{y} dy = k dt \), where all instances of \( y \) and its differential \( dy \) are on one side, while those of \( t \) and \( dt \) are on the other. This method sets the stage for integration, which will bring us closer to the solution of the differential equation.

Integration

Integration is the process of finding the integral or antiderivative of a function. This is the reverse operation of differentiation. When we integrate both sides of our separated equation, we get \( \ln|y| = kt + C \), where \( C \) is the constant of integration. Integration helps us find a general solution to differential equations, though additional steps may be needed to determine a specific solution depending on initial conditions or additional information provided in the problem.

Natural Logarithm

The natural logarithm, denoted as \( \ln(x) \), is the inverse operation of exponentiating with the base \( e \) (Euler's number, approximately 2.718). When we have \( \ln|y| = kt + C \) after integrating, we need to exponentiate to solve for \( y \) since \( e^{\ln|y|} = |y| \). This step harnesses the relationship between logarithms and exponents, enabling us to solve for the unknown quantity \( y \) in terms of time \( t \) and thereby reveal the law governing its exponential growth.