Question

For the following exercises, use the fact that a falling body with friction equal to velocity squared obeys the equation \(d v / d t=g-v^{2}\).Derive the previous expression for \(v(t)\) by integrating \(\frac{d v}{g-v^{2}}=d t\).

Short answer

Integrate the equation using logarithm, then solve for \( v(t) \).

Step-by-step solution

Express the Integral

To start solving the given equation \( \frac{d v}{g-v^{2}}=d t \), we need to express the integral on one side as \( \int \frac{1}{g-v^2} \, dv = \int \, dt \). This sets us up to integrate both sides.

Tackle the Left-Hand Side

We recognize that the left side, \( \int \frac{1}{g-v^2} \, dv \), can be handled using a trigonometric substitution or by rewriting it in a more recognizable form. This integral resembles \( \int \frac{1}{a^2-x^2} \, dx \), which is \( \frac{1}{2a} \ln \left| \frac{a+x}{a-x} \right| + C \) if \( a^2=g \), with \( a=\sqrt{g} \).

Perform the Integration

Using the result from Step 2, we have \[ \int \frac{1}{g-v^2} \, dv = \frac{1}{2\sqrt{g}} \ln \left| \frac{\sqrt{g} + v}{\sqrt{g} - v} \right| + C \]. Integrating \( \int \, dt \) gives \( t + C \). Hence, \[ \frac{1}{2\sqrt{g}} \ln \left| \frac{\sqrt{g} + v}{\sqrt{g} - v} \right| = t + C_0 \].

Solve for v(t)

We solve for \( v \) in terms of \( t \) by isolating \( v \) in the equation from Step 3. First, multiply by \( 2\sqrt{g} \) to get \[ \ln \left| \frac{\sqrt{g} + v}{\sqrt{g} - v} \right| = 2\sqrt{g}t + C_1 \]. Exponentiate both sides to eliminate the natural logarithm: \[ \frac{\sqrt{g} + v}{\sqrt{g} - v} = e^{2\sqrt{g}t + C_1} \].

Clear the Fraction

Clear the fraction by cross-multiplying: \[ (\sqrt{g} + v) = e^{2\sqrt{g}t + C_1} (\sqrt{g} - v) \].Solve for \( v \) to get: \[ v = \frac{ \sqrt{g}(e^{2\sqrt{g}t + C_1} - 1) }{ e^{2\sqrt{g}t + C_1} + 1 }. \]Once \( C_1 \) is determined by initial conditions, you'll have \( v(t) \).

Key concepts

Integration

Integration is a fundamental concept in calculus, playing a critical role in solving differential equations. The process of integration involves finding the antiderivative or the integral of a function. In the given exercise, we aim to integrate the expression \( \frac{d v}{g-v^{2}} = d t \).

To solve this kind of equation, we set up the integral as \( \int \frac{1}{g-v^2} \, dv = \int \, dt \). On the left side, we're looking for the antiderivative of \( \frac{1}{g-v^2} \) with respect to \( v \), while the right side is simply the antiderivative of 1 with respect to \( t \).

Integration here is the key step to translating the problem from a differential equation to an expression we can solve directly for \( v(t) \). This transformation allows us to interpret or predict how the velocity of the falling object changes over time.

Trigonometric Substitution

Trigonometric substitution is a technique utilized to simplify integrals involving square roots. It is particularly handy when dealing with integrals resembling the form \( \int \frac{dx}{a^2-x^2} \), which can be tricky using standard methods.

For the exercise at hand, the integral \( \int \frac{1}{g-v^2} \, dv \) resembles this form. By recognizing that it can be rewritten and solved using the identity \( \int \frac{1}{a^2-x^2} \, dx = \frac{1}{2a} \ln \left| \frac{a+x}{a-x} \right| + C \), we can make the problem much more manageable.

Setting \( a = \sqrt{g} \), we use this formula to solve for the integral, which leads us to an expression involving the natural logarithm. Trigonometric substitution here is crucial for handling the intricacies of the math and getting us closer to the solution.

Natural Logarithm

The natural logarithm, denoted as \( \ln \), is the logarithm to the base of Euler's number \( e \). It is a vital function in calculus, especially when dealing with growth and decay problems.

In this exercise, after making a trigonometric substitution, the integral \( \int \frac{1}{g-v^2} \, dv \) transforms into \( \frac{1}{2\sqrt{g}} \ln \left| \frac{\sqrt{g} + v}{\sqrt{g} - v} \right| + C \).

The natural logarithm aids in simplifying the integration result and ultimately helps solve for \( v(t) \). After obtaining an equation involving \( \ln \), exponentiation is used to clear the logarithm, progressing towards isolating \( v \). This process makes use of the inverse relationship between exponential functions and logarithms.

Initial Conditions

In solving differential equations, initial conditions are crucial for determining specific solutions out of a family of potential solutions. They define the value of the dependent variable (in this case, \( v \)) at a specific point in time (\( t \)).

Our solution for \( v(t) \) involves a constant \( C_1 \), which can be specified using given initial conditions. For instance, if we know that the velocity \( v(0) \) is a certain value \( V_0 \), we can substitute \( t = 0 \) and \( v = V_0 \) into our expression for \( v(t) \) to solve for \( C_1 \).

Using initial conditions allows us to tailor the general solution to match the scenario described in the problem, ensuring accurate and relevant predictions of the behavior of the system over time.