Problem 15 - Advanced Engineering Mathematics

Question

(a) After separating variables we obtain \\[ \begin{aligned} \frac{m d v}{m g-k v^{2}} &=d t \\ \frac{1}{g} \frac{d v}{1-(\sqrt{k} v / \sqrt{m g})^{2}} &=d t \\ \frac{\sqrt{m g}}{\sqrt{k} g} \frac{\sqrt{k / m g} d v}{1-(\sqrt{k} v / \sqrt{m g})^{2}} &=d t \\ \sqrt{\frac{m}{k g}} \tanh ^{-1} \frac{\sqrt{k} v}{\sqrt{m g}} &=t+c \\ \tanh ^{-1} \frac{\sqrt{k} v}{\sqrt{m g}} &=\sqrt{\frac{k g}{m}} t+c_{1} \end{aligned} \\] Thus the velocity at time \(t\) is \\[ v(t)=\sqrt{\frac{m g}{k}} \tanh \left(\sqrt{\frac{k g}{m}} t+c_{1}\right) \\] Setting \(t=0\) and \(v=v_{0}\) we find \(c_{1}=\tanh ^{-1}\left(\sqrt{k} v_{0} / \sqrt{m g}\right).\) (b) since \(\tanh t \rightarrow 1\) as \(t \rightarrow \infty,\) we have \(v \rightarrow \sqrt{m g / k}\) as \(t \rightarrow \infty\) (c) Integrating the expression for \(v(t)\) in part (a) we obtain an integral of the form \(\int d u / u:\) \\[ s(t)=\sqrt{\frac{m g}{k}} \int \tanh \left(\sqrt{\frac{k g}{m}} t+c_{1}\right) d t=\frac{m}{k} \ln \left[\cosh \left(\sqrt{\frac{k g}{m}} t+c_{1}\right)\right]+c_{2} \\] Setting \(t=0\) and \(s=0\) we find \(c_{2}=-(m / k) \ln \left(\cosh c_{1}\right),\) where \(c_{1}\) is given in part \((\mathrm{a}).\)

Step-by-Step Solution

Answer

The velocity function is \( v(t)=\sqrt{\frac{m g}{k}} \tanh \left(\sqrt{\frac{k g}{m}} t+c_{1}\right) \) and the position function is \( s(t)=\frac{m}{k} \ln \left[\cosh \left(\sqrt{\frac{k g}{m}} t+c_{1}\right)\right] - \frac{m}{k} \ln(\cosh c_{1}) \).

Step 1: Separating Variables and Rearranging

We start with the given expression: \[ \frac{m d v}{m g-k v^{2}} =d t \] and manipulate it to separate the variables so we can integrate both sides independently.

Step 2: Express in Terms of Substitution

Rewrite the equation in a form that simplifies integration: \[ \frac{1}{g} \frac{d v}{1-(\sqrt{k} v / \sqrt{m g})^{2}} =d t \].

Step 3: Recognize the Integral Form

This manipulation turns the left side into a form that resembles the derivative of an inverse hyperbolic function, allowing us to perform a straightforward integration.

Step 4: Integrate Both Sides

Integrate the left side: \[ \int \frac{d v}{1-(\sqrt{k} v / \sqrt{m g})^{2}} = \tanh^{-1} \frac{\sqrt{k} v}{\sqrt{m g}} \]. On the right side, integrate to get \( t + c \).

Step 5: Solve for the Velocity Function

Combine both integrated expressions to solve for \( \tanh^{-1} \frac{\sqrt{k} v}{\sqrt{m g}} = \sqrt{\frac{k g}{m}} t + c_1 \) and solve for \( v(t) \), yielding \[ v(t)=\sqrt{\frac{m g}{k}} \tanh \left(\sqrt{\frac{k g}{m}} t+c_{1}\right) \].

Step 6: Calculating Constant \(c_1\)

Setting \( t=0 \) and \( v=v_0 \), solve for \( c_1 \): \[ c_1 = \tanh^{-1} \left(\sqrt{k} v_0 / \sqrt{m g}\right) \].

Step 7: Behavior of Velocity at Infinity

By considering the limits, as \( t \rightarrow \infty \), and using the property \( \tanh t \rightarrow 1 \), conclude that \( v \rightarrow \sqrt{m g / k} \).

Step 8: Integrating to Find Position

Integrate the velocity expression from part (a): \[ s(t)=\sqrt{\frac{m g}{k}} \int \tanh \left(\sqrt{\frac{k g}{m}} t+c_{1}\right) d t \]. This simplifies using the integral \( \int \tanh u \, du = \ln(\cosh u) + C \).

Step 9: Solve for Position Function

Compute the integration: \[ s(t) = \frac{m}{k} \ln \left[\cosh \left(\sqrt{\frac{k g}{m}} t+c_{1}\right)\right] + c_2 \].

Step 10: Finding Constant \(c_2\)

Set \( t=0 \) and \( s=0 \) to find \( c_2 \): \[ c_2 = -\frac{m}{k} \ln \left(\cosh c_{1}\right) \]. This uses the \( c_1 \) from Step 6.

Key Concepts

Separation of Variables

Separation of Variables is a fundamental technique used to solve differential equations. It is particularly handy when you can write a differential equation so that all terms involving one variable, say \(v\), are on one side of the equation, and all terms involving the other variable, often \(t\) for time, are on the other side. This method allows for the individual integration of each variable, thus simplifying the solution process.

• First, identify which terms in your equation belong to each variable. Group these terms on separate sides of the equation.
• For example, from the given expression \( \frac{m d v}{m g-k v^{2}} =d t \), we can manipulate and rearrange to isolate the terms associated with \(v\) and \(t\) on opposite sides.
• Once separated, temporarily treat each side of the equation as distinct integrals. This means integrating the side with \(v\) with respect to \(v\) and the other side with respect to \(t\).

The main advantage of separation of variables is that it often turns a complex differential problem into simpler single-variable integrals, making it easier to manage and solve. It is a popular and powerful tool in mathematical analysis and engineering.

Hyperbolic Functions

Hyperbolic functions, such as \(\tanh\) and its inverse \(\tanh^{-1}\), appear frequently in various branches of mathematics, including when solving differential equations. These functions are analogous to trigonometric functions but relate to hyperbolas instead of circles.

• Functions like \(\tanh(x) = \frac{\sinh(x)}{\cosh(x)}\) provide valuable relationships, particularly in physics and calculus, due to their growth and limit properties.
• When you see a term like \(\tanh^{-1} y \), you can interpret it as finding the \(x\) such that \(\tanh(x)=y\).
• Hyperbolic functions can transform the form and understandability of integrals and algebraic expressions due to their inherent properties.

A useful property of \(\tanh(x)\) used in this exercise is that it approaches 1 as \(x\) goes to infinity. This property helps in analyzing the limiting behavior of functions involving hyperbolic functions, such as velocity and displacement over time in the given problem.

Initial Value Problems

Initial Value Problems (IVPs) are differential equations along with specified values at the start (or at a specific value) of the independent variable. This provides a unique solution to a differential equation by defining the exact starting conditions.

• In many real-world applications, knowing the initial value allows us to predict future behavior of the system being modeled, like predicting the velocity of an object over time given an initial velocity.
• For our problem, when set to \(t=0\) with \(v=v_0\), it helps to calculate the constant \(c_1\) in the velocity equation: \[c_1 = \tanh^{-1} \left(\frac{\sqrt{k} v_0}{\sqrt{m g}}\right)\].
• It's crucial because this constant captures the initial state of the system, which influences the entire solution trajectory.

By providing initial conditions, we can seamlessly solve the differential and its accompanying equation, ensuring a specific path, thus making predictions accurate and reliable for future values.