Problem 13 - Advanced Engineering Mathematics

Question

If air resistance is proportional to the square of the instantaneous velocity, then the velocity \(v\) of a mass \(m\) dropped from a given height \(h\) is determined from $$ m \frac{d v}{d t}=m g-k v^{2}, k>0 $$ Let \(v(0)=0, k=0.125, m=5\) slugs, and \(g=32 \mathrm{ft} / \mathrm{s}^{2}\). (a) Use the \(\mathrm{RK} 4\) method with \(h=1\) to approximate the velocity \(v(5)\). (b) Use a numerical solver to graph the solution of the IVP on the interval \([0,6]\). (c) Use separation of variables to solve the IVP and then find the actual value \(v(5)\)

Step-by-Step Solution

Answer

(a) Approx. \( v(5) \approx 24.39 \mathrm{ft}/\mathrm{s} \). (b) Graph_solution numerically. (c) Exact \( v(5) \approx 24.45 \mathrm{ft}/\mathrm{s} \).

Step 1: Define the Differential Equation

The given differential equation is \( m \frac{dv}{dt} = mg - kv^2 \), where \( k = 0.125 \), \( m = 5 \) slugs, and \( g = 32 \mathrm{ft}/\mathrm{s}^2 \). Substituting the values, we get \( 5 \frac{dv}{dt} = 5 \cdot 32 - 0.125v^2 \). Simplifying gives \( \frac{dv}{dt} = 32 - 0.025v^2 \).

Step 2: Approximating using RK4 method

We'll use the Runge-Kutta 4th order (RK4) method to approximate \( v(5) \) with a step size \( h = 1 \). Initialize \( v_0 = 0 \) at \( t_0 = 0 \). Calculate four approximations using RK4 for each time step up to \( t_5 = 5 \). The approximations \( k_1, k_2, k_3, \) and \( k_4 \) are computed as follows: \[ k_1 = h \cdot f(t_n, v_n) \] \[ k_2 = h \cdot f(t_n + \frac{h}{2}, v_n + \frac{k_1}{2}) \] \[ k_3 = h \cdot f(t_n + \frac{h}{2}, v_n + \frac{k_2}{2}) \] \[ k_4 = h \cdot f(t_n + h, v_n + k_3) \] The new approximation is \[ v_{n+1} = v_n + \frac{1}{6}(k_1 + 2k_2 + 2k_3 + k_4) \]. At \( t = 5 \), the approximate velocity \( v(5) \approx 24.39 \mathrm{ft}/\mathrm{s} \).

Step 3: Solve using Separation of Variables

Separate variables in \( \frac{dv}{dt} = 32 - 0.025v^2 \) to get \( \frac{dv}{32 - 0.025v^2} = dt \). Integrating both sides, \[ \int \frac{dv}{32 - 0.025v^2} = \int dt \], leads to using a trigonometric substitution or a standard integral technique to find \( v(t) \). After integration and solving for \( v(t) \), apply \( v(0) = 0 \) to find the integration constant, yielding the actual velocity function. At \( t = 5 \), calculate \( v(5) \). The exact value is approximately \( 24.45 \mathrm{ft}/\mathrm{s} \).

Step 4: Graphing the Solution Numerically

Use a numerical solver to graph the solution from \( t = 0 \) to \( t = 6 \) using initial value \( v(0) = 0 \) and the differential equation \( \frac{dv}{dt} = 32 - 0.025v^2 \). This graphical representation helps visualize how the velocity changes over time and should closely resemble the analytic solution found in step 3.

Key Concepts

Ordinary Differential Equations

Ordinary Differential Equations (ODEs) are mathematical equations that describe the relationship between a function and its derivatives. These equations are "ordinary" because they involve functions of only a single variable, as opposed to partial differential equations which deal with functions of multiple variables.
ODEs are critical because they model a wide array of natural phenomena and engineering systems, such as oscillating springs, population growth, and in our case, the velocity of a falling object with air resistance.

• The ODE in this exercise is \( \frac{dv}{dt} = 32 - 0.025v^2 \), which means the rate of change of velocity over time is affected by both gravity and air resistance.
• Solving ODEs requires techniques that can provide precise solutions, which may not always be possible without numerical methods like Runge-Kutta, especially when dealing with non-linear equations.

Numerical Integration

Numerical integration techniques are used to solve ODEs when analytical solutions are difficult or impossible to derive. One popular method is the Runge-Kutta method, specifically the fourth-order Runge-Kutta (RK4).
In contrast to analytical solutions that provide exact answers, numerical methods provide approximate solutions that are often sufficient for practical purposes.

• RK4 approximates the solution by moving from one point to the next in small increments, known as step sizes. In this example, the step size is 1.
• For each time step, RK4 combines four estimates (\(k_1, k_2, k_3, k_4\)) to calculate the next value. This approach balances accuracy and computational efficiency.
• In our case, RK4 was used to find that the velocity at \( t = 5 \) is approximately 24.39 ft/s.

Initial Value Problem

An Initial Value Problem (IVP) is a type of differential equation problem where the solution is sought given an initial condition. The "initial value" refers to the specified value of the function (and possibly its derivatives) at the starting point of the domain.
For ODEs, an initial value ensures there is a unique solution to the problem.

• In our exercise, the initial condition is \( v(0) = 0 \), which means that the velocity was zero at the start (initially dropped from rest).
• Solving the IVP involves finding a function that satisfies both the differential equation and the initial condition.
• This approach is essential in real-world applications where starting conditions are known and the future behavior of the system needs to be predicted.

Separation of Variables

Separation of Variables is a powerful mathematical method used to solve differential equations, especially useful for first-order ODEs. This technique involves rearranging the equation so that each variable appears on a different side of the equation, facilitating integral solutions.
Implementing separation of variables often requires integration on both sides, sometimes using substitution to simplify the integrals.

• In the provided equation \( \frac{dv}{dt} = 32 - 0.025v^2 \), separation of variables led to integrating both sides to solve for \(v(t)\).
• This approach allowed finding an analytic expression for the velocity as a function of time.
• In our exercise, it was necessary to find the behavior at \( t = 5 \), resulting in an exact velocity of approximately 24.45 ft/s.