Problem 24 - Advanced Engineering Mathematics

Question

(a) Solving \(v_{t}=\sqrt{m g / k}\) for \(k\) we obtain \(k=m g / v_{t}^{2} .\) The differential equation then becomes \\[ m \frac{d v}{d t}=m g-\frac{m g}{v_{t}^{2}} v^{2} \\] or \(\frac{d v}{d t}=g\left(1-\frac{1}{v_{t}^{2}} v^{2}\right)\) Separating variables and integrating gives \\[ v_{t} \tanh ^{-1} \frac{v}{v_{t}}=g t+c_{1} \\] The initial condition \(v(0)=0\) implies \(c_{1}=0,\) so \\[ v(t)=v_{t} \tanh \frac{g t}{v_{t}} \\] We find the distance by integrating: \\[ s(t)=\int v_{t} \tanh \frac{g t}{v_{t}} d t=\frac{v_{t}^{2}}{g} \ln \left(\cosh \frac{g t}{v_{t}}\right)+c_{2} \\] The initial condition \(s(0)=0\) implies \(c_{2}=0,\) so \\[ s(t)=\frac{v_{t}^{2}}{g} \ln \left(\cosh \frac{g t}{v_{t}}\right) \\] In 25 seconds she has fallen \(20,000-14,800=5,200\) feet. Using a CAS to solve \\[ 5200=\left(v_{t}^{2} / 32\right) \ln \left(\cosh \frac{32(25)}{v_{t}}\right) \\] for \(v_{t}\) gives \(v_{t} \approx 271.711 \mathrm{ft} / \mathrm{s} .\) Then \\[ s(t)=\frac{v_{t}^{2}}{g} \ln \left(\cosh \frac{g t}{v_{t}}\right)=2307.08 \ln (\cosh 0.117772 t) \\] (b) \(\operatorname{At} t=15, s(15)=2,542.94 \mathrm{ft}\) and \(v(15)=s^{\prime}(15)=256.287 \mathrm{ft} / \mathrm{sec}.\)

Step-by-Step Solution

Answer

v_t ≈ 271.711 ft/s; s(15) = 2,542.94 ft; v(15) = 256.287 ft/s.

Step 1: Simplify the Given Expression for v_t

The problem starts with the equation \( v_{t} = \sqrt{\frac{m g}{k}} \). To solve for \( k \), we square both sides to get rid of the square root, resulting in \( v_{t}^2 = \frac{m g}{k} \). Rearranging gives \( k = \frac{m g}{v_{t}^2} \).

Step 2: Derive the Differential Equation

Using the expression for \( k \) found in Step 1, substitute \( k = \frac{m g}{v_{t}^2} \) into the differential equation \( m \frac{dv}{dt} = mg - \frac{m g v^2}{v_{t}^2} \), which simplifies to \( \frac{dv}{dt} = g \left( 1 - \frac{v^2}{v_{t}^2} \right) \).

Step 3: Separate Variables and Integrate

Separate variables in the equation \( \frac{dv}{dt} = g \left(1 - \frac{v^2}{v_{t}^2}\right) \) to get \( \int \frac{dv}{1 - \frac{v^2}{v_{t}^2}} = \int g \, dt \). Integrating both sides results in \( v_{t} \tanh^{-1} \frac{v}{v_{t}} = g t + c_1 \).

Step 4: Apply Initial Condition for Velocity

Given \( v(0) = 0 \), set \( t = 0 \) in the integrated equation \( v_{t} \tanh^{-1} \frac{v}{v_{t}} = g t + c_1 \). Solving yields \( c_1 = 0 \). Therefore, \( v(t) = v_{t} \tanh \frac{g t}{v_{t}} \).

Step 5: Integrate Velocity to Find the Distance

Integrate velocity function \( v(t) = v_{t} \tanh \frac{g t}{v_{t}} \) to find distance. This gives \( s(t) = \int v_{t} \tanh \frac{g t}{v_{t}} \, dt = \frac{v_{t}^2}{g} \ln \left( \cosh \frac{g t}{v_{t}} \right) + c_2 \).

Step 6: Apply Initial Condition for Distance

The initial condition \( s(0) = 0 \) implies \( c_2 = 0 \). Hence, distance as a function of time is \( s(t) = \frac{v_{t}^2}{g} \ln \left( \cosh \frac{g t}{v_{t}} \right) \).

Step 7: Solve for Terminal Velocity v_t

Given the drop of 5,200 feet in 25 seconds, use \( 5200 = \frac{v_{t}^2}{32} \ln \left( \cosh \frac{32 \times 25}{v_{t}} \right) \) to solve for \( v_{t} \) using a computational tool, finding \( v_{t} \approx 271.711 \) ft/s.

Step 8: Calculate Distance and Velocity at t = 15

Substitute \( t = 15 \) into distance formula \( s(t) = \frac{v_{t}^2}{32} \ln \left( \cosh \frac{32 t}{v_{t}} \right) \) to get \( s(15) = 2,542.94 \) ft, and use the derivative \( v(t) = v_{t} \tanh \frac{32 t}{v_{t}} \) to find \( v(15) = 256.287 \) ft/s.

Key Concepts

Separation of Variables

In differential equations, the method of separation of variables is a powerful tool. This technique allows us to simplify the process of solving differential equations by separating the variables. In this exercise, our goal is to solve the differential equation \( \frac{d v}{d t} = g\left(1-\frac{1}{v_{t}^{2}} v^{2}\right) \).
To apply separation of variables, we rearrange the equation such that each side depends on only one variable. This becomes \( \int \frac{dv}{1 - \frac{v^2}{v_{t}^2}} = \int g \, dt \). The idea is to integrate both sides of the equation separately.
By handling the left side, which involves the function of \( v \), we end up expressing the integrated form involving an inverse hyperbolic tangent function, specifically \( v_{t} \tanh^{-1} \frac{v}{v_{t}} \). On the right side, integrating simply results in \( gt \), plus a constant \( c_1 \). This constant can later be determined using initial conditions.

Tanh and Hyperbolic Functions

In mathematics, hyperbolic functions are analogs of the trigonometric functions but for the hyperbola, rather than the circle. In solving the equation, we come across the function \( \tanh^{-1} \), known as inverse hyperbolic tangent. This function appears after separating and integrating variables.
The inverse hyperbolic tangent, \( \tanh^{-1}(x) \), takes an input, and effectively tells us the hyperbolic angle, whose hyperbolic tangent is \( x \). This function is important in our equation since it connects the velocity \( v \) and time \( t \) through the relationship \( v_{t} \tanh^{-1} \frac{v}{v_{t}} = gt + c_1 \).
Using the properties of hyperbolic functions, when applying initial conditions, it allows us to solve for constants and eventually express the velocity \( v(t) \) straightforwardly in terms of \( \tanh \), which is much more convenient to work with.

Initial Conditions

Initial conditions are vital in the context of differential equations as they help us determine the unique function that satisfies the differential equation for a particular problem. They are like a starting point or boundary condition that helps find specific solutions.
In this scenario, the initial condition \( v(0) = 0 \) tells us that the velocity is zero at time \( t = 0 \). This information is essential to solve for the constant \( c_1 \) in our solution after integration. Plugging \( t = 0 \) into the equation \( v_{t} \tanh^{-1} \frac{v}{v_{t}} = gt + c_1 \) simplifies to \( c_1 = 0 \), because \( \tanh^{-1}(0) = 0 \) when \( v = 0 \).
Similarly, when considering distance \( s(t) \), the initial condition \( s(0) = 0 \) allows us to determine another constant, \( c_2 \), ensuring the equation \( s(t) = \frac{v_{t}^{2}}{g} \ln \left( \cosh \frac{gt}{v_{t}} \right) \) fits the specific situation described.