This is a module for beginning calculus students to study differential equations that model falling objects subject to air resistance -- for example, raindrops -- and to develop Euler's Method, a tool for approximating solutions of initial value problems.
About this Module and its Authors
Click on the button corresponding to your preferred computer algebra system (CAS). This will download a file which you may open with your CAS.
Ver. 7 or higher |
Ver. 5.0 or higher |
Ver. 5.1 or higher |
Published June, 2004
This is a module for beginning calculus students to study differential equations that model falling objects subject to air resistance -- for example, raindrops -- and to develop Euler's Method, a tool for approximating solutions of initial value problems.
About this Module and its Authors
Click on the button corresponding to your preferred computer algebra system (CAS). This will download a file which you may open with your CAS.
Ver. 7 or higher |
Ver. 5.0 or higher |
Ver. 5.1 or higher |
Published June, 2004
Our models for velocity of a falling object will all be based on Newton's Second Law of Motion, which states that force equals mass times acceleration:
Here F is the force exerted on an object of mass m, causing the object to have an acceleration a.
Acceleration is defined to be the derivative of velocity, i.e.,
where v = v(t) is the velocity at time t. Thus Newton's Law can be rewritten as
The primary force on a falling body is gravity, the pull of the Earth's mass on the object. Our first model for a falling body will consider gravity to be the only force on the object.
It is known through experimental observation that (near the surface of the Earth) the force of gravity on an object is proportional to the mass of the object, i.e., there is a constant g such that
The value of the constant g is known by experimentation to be approximately 32.2 sec^{2}.
Equating our two formulas for the force F and dividing by m, we find a differential equation:
If we assume that our object was initially at rest at time t = 0, then our initial condition is
v(0) = 0. Together with the differential equation, we have an initial value problem for the velocity function v = v(t):
This solution for a velocity function leads to a second differential equation: The velocity v is itself the derivative of the distance function s = s(t), i.e.,
We now investigate a more sophisticated model for a falling body, one that takes into account the resisting force of the air through which the object falls. The usual physical assumption is that the force of air resistance is proportional to some power of the velocity, but the particular power (first, second, or other) depends on the particular object.
We consider raindrops falling from a cloud 3000 feet above the ground. If the raindrop is small, say a drop of diameter 0.003 inches (or 0.00025 feet), a size found in a drizzle, the force of air resistance is modeled well by a multiple of the first power of the velocity. In other words, the resisting force can be described by
for some positive constant k. (The minus sign indicates that the force is in the direction opposite to the velocity, i.e., upward.) When combined with the force of gravity,
this yields the total force on the raindrop:
We recall Newton's Second Law of Motion:
Equating our two formulas for the force F and dividing by m, we find a new differential equation for velocity:
We'll let c represent the quotient k/m. When we attach our initial condition, v(0) = 0, we obtain our new initial value problem:
Experimental evidence gives an approximate value of 52.6 sec^{-1 } for c, when distances are in feet and the drops are drizzle size.
Guessing a formula for v as a function of time may be more difficult for this problem than actual calculation was on Page 2. You may or may not know a systematic way to solve this type of problem at this point. On the next page we will take a different approach and use a numerical technique called Euler's Method. This technique will only approximate the desired solution, but it has the distinct advantage of applicability to any initial value problem.
We ended Page 3 with an initial value problem to be solved: Find v = v(t) so that
More generally, our problem is to solve any initial value problem of the form
We will calculate approximate values for the velocity v at n equally spaced points in some fixed time interval. Our procedure is simple: We repeatedly calculate a rise in v as slope x run. Then we add the rise to the current value of v to get the next value of v.
Our goal is to estimate the velocity v(t) at times
Our estimated velocity values at these times will be denoted by
Our method for estimating the velocity values will be recursive, i.e., v_{k} will be calculated from the preceding v_{k--1} for each k = 1, 2, 3, ... .
How do we obtain v_{1} from v_{0}, the initial velocity? We will answer this in a geometric fashion. We will look at the graph of velocity versus time on the (t,v)-plane. The following figure shows a graph of the starting situation: the initial velocity v_{0 } is shown as a vertical line segment of length v_{0 } at the starting time t_{0 } = 0.
We now include the graph of v versus t. Our next velocity value, v_{1}, is shown as the length of a vertical line segment at time t_{1}.
However, the value of v_{1 } is not known to us, and hence we will estimate its value. We do this by drawing the tangent line to the graph at t_{ } = t_{0}. Follow this tangent line to the point P, the top of a vertical line segment that approximates v_{1}.
We can compute the length of this new line segment: We separate the line segment into two pieces -- the bottom piece having length v_{0}, and the top piece being the rise of a right triangle with Using
we see that rise equals slope times . Hence,
(We use the symbol to mean "almost equal to".)
This is the key to Euler's Method for approximating the solution of an initial value problem. It's valuable because the slope (of the tangent line) equals the derivative dv/dt, which is given by our original differential equation when t_{ } = t_{0 } and v_{ } = v_{0}:
Substituting this value of the slope into the preceding equation, we find
Great! This gives us a method for going from v_{0 } to v_{1}. But how do we go from v_{1 } to v_{2}? Easy -- we use the same equation, only with v_{0 } and v_{1 } replaced by v_{1 } and v_{2}:
In general, to go from v_{k - 1 } to v_{k } we have
This equation, along with the initial value v_{0 } = 0 and the assignment of a value to the step size , plays the central role in our computations.
Here again is our initial value problem: Find v = v(t) so that
We have specific values for g and c, both obtained experimentally: and .
We will use Euler's Method to calculate approximate values for the velocity v at n equally spaced points in a fixed time interval. The Euler procedure gives a better approximation to the exact solution if n is large rather than small. Thus, for convenience, we set n = 100. Our time interval will be 0 < t < 0.2 seconds -- the reason for this choice is suggested by the slope field on Page 3. Thus the distance between consecutive t values will be
For large raindrops, say with diameter 0.05 inches (or 0.004 feet), a size typical of drops in a thunderstorm), the force of air resistance is better modeled as a multiple of the square of the velocity. The differential equation now has the form
where a is another constant. In this case, the experimental evidence yields a value for a of 0.115. With the same initial condition, v(0) = 0, we have a new initial value problem. We will use Euler's Method to approximate the solution of this new problem, this time over the time interval from 0 to 2 seconds.
is an exact solution of the drizzle drop problem. How does this formula reveal the terminal velocity you know already? If you guessed the form of the solution on Page 3, compare this symbolic form with your guess. Are the two proposed solutions the same? If not, describe how they differ.