Question

The following models were discussed in Section 1 and reappear in later sections of this chapter. In each case carry out the indicated analysis using direction fields. A model that describes the free fall of an object in a gravitational field subject to air resistance uses the equation \(v^{\prime}(t)=g-b v,\) where \(v(t)\) is the velocity of the object, for \(t \geq 0, g=9.8 \mathrm{m} / \mathrm{s}^{2}\) is the acceleration due to gravity, and \(b>0\) is a constant that involves the mass of the object and the air resistance. Let \(b=0.1 \mathrm{s}^{-1}\). a. Draw the direction field for \(0 \leq t \leq 60,0 \leq y \leq 150\). b. For what initial values \(v(0)=A\) are solutions increasing? Decreasing? c. What is the equilibrium solution?

Short answer

Also, identify the equilibrium solution. Answer: For initial values \(v(0) = A < 98\), the solutions are increasing. For initial values \(v(0) = A > 98\), the solutions are decreasing. The equilibrium solution is \(v(t) = 98\).

Step-by-step solution

Setting up the given equation

The given equation is: \(v'(t) = g - bv\) Where \(v(t)\) = velocity of the object at time t, \(g = 9.8 m/s^2\) = acceleration due to gravity, \(b = 0.1 s^{-1}\) = air resistance constant.

Draw the direction field for specified range of values

To draw the direction field for \(0 \leq t \leq 60, 0 \leq y \leq 150\), we can use a software or draw it by hand using the slope values at different points. Here, we will not provide actual direction field drawing, but you can easily do that using software like MATLAB or Mathematica.

Determine initial values for increasing and decreasing solutions

To determine whether the solutions are increasing or decreasing, we can analyze the sign of \(v'(t)\). If \(v'(t) > 0\), the solution is increasing. If \(v'(t) < 0\), the solution is decreasing. Using the given equation, \(v'(t) = g - bv\) Let's set \(v'(t) > 0\) and solve for \(v(t)\): \(g - bv > 0\) \(v(t) < \frac{g}{b}\) With g = 9.8 and b = 0.1, \(v(t) < \frac{9.8}{0.1}\) \(v(t) < 98\) So, for initial values \(v(0) = A < 98\), the solutions are increasing. Let's set \(v'(t) < 0\) and solve for \(v(t)\): \(g - bv < 0\) \(v(t) > \frac{g}{b}\) With g = 9.8 and b = 0.1, \(v(t) > \frac{9.8}{0.1}\) \(v(t) > 98\) So, for initial values \(v(0) = A > 98\), the solutions are decreasing.

Find the equilibrium solution

The equilibrium solution occurs when the rate of change is constant, meaning \(v'(t) = 0\). Using the given equation, \(v'(t) = g - bv\) Setting \(v'(t) = 0\) and solve for \(v(t)\): \(0 = g - bv\) \(v(t) = \frac{g}{b}\) With g = 9.8 and b = 0.1, \(v(t) = \frac{9.8}{0.1}\) \(v(t) = 98\) The equilibrium solution is \(v(t) = 98\).

Key concepts

Direction Fields

Imagine you have a map showing tiny arrows, guiding you how to draw curves along a plane. This is what direction fields do for differential equations. They're like the GPS for equations! By looking at a direction field, you can visualize how a function behaves over time. It's a sketch of slope directions showing the future possibilities of a graph.

For the equation given in the exercise, the direction field helps us understand how velocity changes with time when an object falls with air resistance. By plotting the slope of the velocity at various points, you can see if the velocity is increasing or decreasing at those points. As a student, this visual tool can make solving differential equations much more intuitive and less abstract. You can use software like MATLAB or Mathematica to easily generate these fields. Experimenting with this can build your understanding of solutions to differential equations beyond just calculations.

Equilibrium Solutions

In the world of mathematics, an equilibrium solution represents a state where things remain constant over time. It's like finding a perfect balance where no change occurs. For the differential equation model in this problem, the equilibrium solution is where the change in velocity, or the rate of speeding up or slowing down, becomes zero.

Given the equation \(v'(t) = g - bv\), setting \(v'(t) = 0\) results in an equilibrium. Here, it leads us to the expression: \(v(t) = \frac{g}{b}\). By substituting the values we have \(g = 9.8\) and \(b = 0.1\), resulting in an equilibrium velocity \(v(t) = 98\). This means if an object is moving at this velocity, it will continue to move at this velocity unless there's a change in conditions or parameters.

Understanding equilibrium helps describe physical situations accurately and serves as a benchmark to compare changes or deviations.

Air Resistance Modeling

When you drop an object from a height, it does not keep accelerating infinitely because of forces like air resistance. Air resistance modeling captures the tug-of-war between gravity pulling objects down and air drag slowing them down.

The exercise models this situation with the differential equation \(v'(t) = g - bv\), indicating the dynamic relationship between gravitational pull and resistive forces. Here, \(g\) represents constant gravity and \(b\) incorporates the air resistance depending on an object's properties.

This type of modeling is crucial in physics because it describes the realistic behavior of falling objects. Real-world applications include designing parachutes, predicting free-fall speeds, and even understanding how raindrops fall. Using such a model is important for accurate predictions and safe engineering.

Initial Value Problems

Initial Value Problems (IVPs) are like solving a mystery with a good starting clue. They focus on finding a function given some starting conditions, or initial values. For the problem in the exercise, it revolves around determining the velocity of a falling object with a known starting velocity.

When you know your starting point, solving the differential equation becomes more straightforward. You'll check if the solution increases or decreases based on these values. In our exercise, the solutions are increasing or decreasing depending on whether initial velocity \(v(0)\) is greater or less than \(98\).

IVPs are everywhere in real-life scenarios, from engineering problems like vehicle acceleration to predicting population growth or cooling rates in physics. Once you master these, you'll find it easier to address a wide range of practical problems by setting initial conditions and solving through them.