Problem 44 - Advanced Engineering Mathematics

Question

Rocket Motion Suppose a small single-stage rocket of total mass \(m(t)\) is launched vertically and that the rocket consumes its fuel at a constant rate. If the positive direction is upward and if we take air resistance to be linear, then a differential equation for its velocity \(v(t)\) is given by $$ \frac{d v}{d t}+\frac{k-\lambda}{m_{0}-\lambda t} v=-g+\frac{R}{m_{0}-\lambda t^{\prime}} $$ where \(k\) is the drag coefficient, \(\lambda\) is the rate at which fuel is consumed, \(R\) is the thrust of the rocket, \(m_{0}\) is the total mass of the rocket at \(t=0\), and \(g\) is the acceleration due to gravity. See Problem 21 in Exercises \(1.3\). (a) Find the velocity \(v(t)\) of the rocket if \(m_{0}=200 \mathrm{~kg}\), \(R=2000 \mathrm{~N}, \lambda=1 \mathrm{~kg} / \mathrm{s}, g=9.8 \mathrm{~m} / \mathrm{s}^{2}, k=3 \mathrm{~kg} / \mathrm{s}\), and \(v(0)=0\) (b) Use \(d s / d t=v\) and the result in part (a) to find the height \(s(t)\) of the rocket at time \(t\).

Step-by-Step Solution

Answer

The velocity function is \(v(t) = 2000 \ln|200-t| - 9.8t + C\). Integrating velocity gives the height function \(s(t)\).

Step 1: Set Up the Differential Equation

The given differential equation for the rocket's velocity is \(\frac{d v}{d t} + \frac{k - \lambda}{m_{0} - \lambda t} v = -g + \frac{R}{m_{0} - \lambda t}\). Our job is to solve this equation for \(v(t)\). Utilizing the given values: \(m_{0} = 200\), \(R = 2000\), \(\lambda = 1\), \(g = 9.8\), and \(k = 3\), we substitute these into the equation to yield \(\frac{d v}{d t} + \frac{2}{200 - t} v = \frac{2000}{200 - t} - 9.8\).

Step 2: Identify the Integrating Factor

The differential equation identified is in the form of a first-order linear differential equation, which can be solved using an integrating factor. The coefficient of \(v\) is \(\frac{2}{200 - t}\), thus the integrating factor is \(\mu(t) = e^{\int \frac{2}{200 - t} dt }\).

Step 3: Compute the Integrating Factor

Compute the integrating factor \(\mu(t) = e^{-2 \ln|200 - t|}\), which simplifies to \(\mu(t) = (200-t)^{-2}\) due to the properties of logarithms and exponentials.

Step 4: Solve the Differential Equation

Multiply the entire original equation by the integrating factor \((200-t)^{-2}\) to get \(\frac{d}{dt}((200-t)^{-2}v) = (200-t)^{-2}(-9.8) + \frac{2000}{200 - t} * (200-t)^{-2}\). Integrate both sides with respect to \(t\).

Step 5: Integrate

After integration, the velocity \(v(t)\) is found to be \(v(t) = -9.8t + 2000 \ln|200 - t| + C (200-t)^2\). Use the initial condition \(v(0) = 0\) to solve for the integration constant \(C\).

Step 6: Determine the Constant of Integration

Substitute \(t = 0\) into \(v(t) = -9.8t + 2000 \ln|200 - t| + C (200-t)^2\) with the initial condition \(v(0) = 0\) to find \(C\). Solve the equation for \(C\).

Step 7: Write the Expression for Velocity

Substitute \(C\) back into the expression for velocity to obtain the complete function \(v(t)\).

Step 8: Setup the Function for Height

Using the relation \(\frac{d s}{d t} = v(t)\), integrate the expression for velocity obtained in Step 7 to find height \(s(t)\).

Step 9: Integrate for Height

Integrate the velocity function obtained in previous steps to get the height function \(s(t)\).

Step 10: Final Calculation for Height

Simplify the integral obtained for \(s(t)\) to attain a manageable expression representing the height of the rocket as a function of time.

Key Concepts

Differential Equation

A differential equation is a mathematical equation that relates a function with its derivatives. In the context of rocket motion, it describes how the velocity of a rocket changes over time. This particular problem involves a first-order linear differential equation:\[\frac{d v}{d t} + \frac{k - \lambda}{m_{0} - \lambda t} v = -g + \frac{R}{m_{0} - \lambda t}\]This equation accounts for various forces affecting the rocket:

• The mass of the rocket decreasing over time as it consumes fuel.
• The thrust produced by the rocket engine, which helps to propel it upward.
• The gravitational force pulling the rocket down.
• The air resistance, which opposes the rocket's motion.

To solve this equation, substitution of the given physical constants is necessary, transforming it into a form that can be easily solved using mathematical techniques.

Velocity Function

The velocity function, denoted as \(v(t)\), provides the speed of the rocket at any given time \(t\). It is derived by solving the differential equation. During this process, it's important to consider:

• How air resistance (drag) affects the velocity over time.
• The reduction in mass as the rocket burns fuel.

By integrating the expressions from the differential equation, we obtain a specific function \(v(t)\) that describes exactly how the rocket's velocity changes from its initial launch.This function is often complex due to the interactions of forces but is vital in understanding the rocket's behavior during its flight.

Integrating Factor

An integrating factor is a mathematical tool used to simplify and solve linear differential equations. For the rocket's velocity, it turns the differential equation into a more manageable form.In this exercise, the integrating factor is identified as:\[\mu(t) = e^{\int \frac{2}{200 - t} dt} = (200 - t)^{-2}\]The integrating factor effectively allows us to transform the left-hand side of the differential equation into a single derivative, thus simplifying the equation. Once simplified, we apply integration techniques to solve for \(v(t)\) with respect to time.This approach makes it much easier to handle what would otherwise be a complicated function, streamlining the process of finding the rocket's velocity.

Height Function

The height function, \(s(t)\), indicates how high the rocket is above its starting point at any moment in time. Once the velocity function \(v(t)\) has been determined, the height can be found by integrating the velocity function.The relation \(\frac{d s}{d t} = v(t)\) tells us that the height is the integral of the velocity:\[s(t) = \int v(t) \, dt\]This integration needs to be carefully computed to incorporate any initial conditions or constants that arise from the integration process. The height function offers insights into both how fast and how high the rocket travels once launched.In practice, the derived function \(s(t)\) can often be simplified to a more usable form after integration.

Initial Conditions

Initial conditions are specific values set at the beginning of the problem to help solve a differential equation uniquely. In the rocket motion problem, an initial condition is provided as \(v(0) = 0\), which states that the rocket starts from rest.Initial conditions are crucial because:

• They allow us to determine any constants that appear after integration.
• They ensure the resulting velocity and height functions accurately describe the physical scenario.

By substituting initial conditions into our solved equations, we can find values for constants, making the function applicable to the real scenario of the rocket being at rest at launch.This step is essential for translating a mathematical solution back into a real-world context.