This document is supplemented by Chapter 1 of the text.
The Class Notes document is followed by a set of q_a_ questions.
The questions in this first q_a_ assume familiarity with first-semester physics, first-year calculus, and precalculus. Most students will begin a differential equations course needing review of some topics which are usually covered in prerequisite courses, and this will be considered during the first couple of weeks of the course. Your instructor will note apparent these areas and make suggestions for review. By the third or fourth week of the course (or in the case of a summer course by the end of the second or beginning of the third week) you will be responsible for having reviewed these topics.
Class Notes 110110
The equation
F_net = m a
expresses Newton's Second Law, which relates the acceleration of a mass m to the net force acting on that mass.
Since the acceleration function a(t) is equal to the second derivative x '' (t) of the position function x(t), Newton's Second Law can be written in terms of derivatives as
F_net = m x '' ,
This is always so.
So the equation
m x '' = F_net
is very important in applications.
We will use this equation to understand some basic things about differential equations.
SHM (simple harmonic motion) occurs then F_net = - k x. In this case our equation becomes
m x '' = - k x.
We rearrange this to the form
x '' = - k / m * x.
To clarify some terminology:
This equation relates a function x(t) to at least on of its derivatives, so we call it a differential equation.
The highest derivative in this equation is the second derivative. So we call this a second order differential equation.
An example of a first-order differential equation might be
x ' = t + x.
Note that this equation contains the variable t as well as the function x and its derivative x '.
We could write this equation as
x ' = f(t, x),
where f(t, x) = t + x.
This equation looks very simple, and it isn't all that complicated, but it's harder than it looks. You are unlikely to be able to find a solution to this equation until we have developed some more terminology to classify equations, and the the machinery to solve this particular type of equation.
An equation we can easily solve is
x ' = t.
Remember that x ' is the derivative of the x(t) function with respect to t. Thus x(t) is an antiderivative of x ' (t), and if we integrate x ' (t) with respect to t we get x(t). There is also a constant involved in the integration, but we will take care of that when we integrate the equation.
Integrating both sides of the equation, we therefore get
x (t) = 1/2 t^2 + c,
where c is a constant. Both sides of the equation would have integration constants, so c represents the integration constant from the right-hand side, minus the constant from the left-hand side.
So x(t) = 1/2 t^2 + c is the general solution to our equation x ' = t.
The integration constant allows us to impose one condition on our solution. For example if we want to impose an initial condition that x(0) = 12 (we call this an initial condition because it applies at t = 0), we can accomodate that:
x(t) = 1/2 t^2 + c so
x(0) = 1/2 * 0^2 + c = c.
Thus, if x(0) = 12, we have
12 = c.
Our specific solution (called a particular solution) to the equation is therefore
x(t) = 1/2 t^2 + 12.
If x(t) represents the position function for a particle with constant acceleration a, then the fact that acceleration is the second derivative of the position leads us to the equation
x '' (t) = a = constant.
An antiderivative of x '' (t) is x ' (t), and an antiderivative of the constant quantity a is a t. So a single integration of our equation yields
x ' (t) = a t + c_1,
where c_1 is a constant number.
Integrating once more we get
x(t) = a * t^2 / 2 + c_1 * t + c_2,
where c_1 is the integration constant from our previous integration and c_2 is the constant from our present integration.
The two constants allow us to impose two conditions on our solution. For example we could impose the conditions x ' (0) = 5 and x(0) = 7. In terms of motion, x ' represents velocity and x represents position, so our conditions specify an initial velocity and an initial position on our motion.
The first of our integrals is x ' (t) = a t + c_1. You can verify that the condition x ' (0) = 5 gives us c_1 = 5.
Our position function is therefore x(t) = a t^2 / 2 + 5 t + c_2. You can verify that our condition x(0) = 7 gives us c_2 = 7 so that our position function becomes
x(t) = 1/2 a t^2 + 5 t + 7.
In general if x(0) = x_0 and v(0) = x ' (0) = v_0, our general solution x(t) = a * t^2 / 2 + c_1 * t + c_2 becomes
x(t) = 1/2 a t^2 + v_0 t + x_0.
You should recognize this as a standard equation of uniformly accelerated motion in introductory physics.
In general a first-order differential equation will yield one arbitrary constant, allowing us to impose one condition on the solution, and a second-order equation will yield two arbitrary constants, allowing us to impose two conditions.
If an equation is of order n, then a general solution will include n arbitrary constants, allowing us to impose n conditions on our solution.
Returning to the equation for SHM:
x '' = - k / m * x,
we don't yet have a method for solving this equation. There is a simple method (let x = A e^(r t) and see what this tells us about the two arbitrary constants A and r). You might have seen this method in your first-year calculus course, which often includes a brief introduction to differential equations. The method is not difficult, and we will see it in a later chapter.
Here we simply assert that the general solution to the equation can be expressed as
x(t) = B cos( sqrt(k/m) t ) + C sin ( sqrt(k/m) t ).
If you plug this function into the equation x '' = -k/m x, you can easily verify that x '' (t) is in fact equal to -k /m * x(t).
As expected, this solution includes two arbitrary constants B and C.
Basic trigonometric identities allow us to rearrange the expression B cos( sqrt(k/m) t ) + C sin ( sqrt(k/m) t ) into the expression A cos( sqrt(k/m) t + phi ), where A and phi are constants that can be expressed in terms of B and C in the original form. So another way of writing the solution is
x(t) = A cos( sqrt(k/m) t + phi).
You should verify that this is also a solution of the equation.
Both solutions can be written in terms of omega = sqrt(k /m):
x(t) = B cos( omega * t) + C sin( omega * t)
x(t) = A cos(omega * t + phi) ).
The second solution is easily interpreted as the x component of a position vector r ( t ) whose terminal point moves with angular velocity omega around a circle of radius A, with initial angular position phi.
x(t) is therefore the general equation of motion for a simple harmonic oscillator with amplitude A and angular frequency omega = sqrt(k / m).
You should verify that x(t) = A cos(omega * t + phi), with omega = sqrt(k/m), is a solution of the equation x '' = -k / m * x.
An object moving against a frictional force F_frict and an air resistance - k v which is proportional to its speed has
F_net = -F_frict - c v, so that
m x '' = -F_frict - c v and
x '' = -F_frict / m - c / m * v.
Since v = x ' this equation becomes
x '' = -F_frict / m - c / m * x '.
We won't at this point discuss how to classify or solve this equation. However note the following:
This is a second-order equation, with x '' expressed in terms of t, x, and x '.
In fact x '' is expressed only in terms of x ', with no specified dependence on t or x.
The equation is certainly of the form
x '' = f(t, x, x '),
though in this case t and x aren't actually part of the definition of the function f.
If the object in the preceding is also subject to a linear restoring force F = - k x (e.g., consider a mass oscillating on a smooth tabletop while attached to a spring; friction and air resistance are still present but now we have the restoring force of the spring to consider), the equation would be
x '' = -F_frict / m - c / m * x ' - k x.
This is still of the form
x '' = f(t, x, x').
In this case the function f(t, x, x') includes the expressions x and x ', but there is no t dependence.
Now suppose the table is gradually tilted. The gravitational force on the mass will then have a component parallel to the object's motion, which will increase in magnitude as the table gets steeper. The equation of motion could then be of the form
x '' = -F_frict / m - c / m * x ' - k x + m g cos ( b t ),
where g is the acceleration of gravity and b is a constant which determines how quickly the angle of the incline changes.
This equation is also of the form
x '' = f(t, x, x ' ),
where now all three of the quantities t, x and x ' are included in the expression for t.
[Physics students may note that the frictional force will no longer be constant when as table is tilted through different angles, so F_frict also becomes a function of t. This was not mentioned before for fear of overly complicating the equation, but the equation would be x '' = - mu g sin( b t) - c / m * x ' - k x + m g cos(b t). This is still of the form x '' = f(t, x, x').]
[Instructor note: Off the top of my head it doesn't appear that the two equations given here have a closed-form solution. Most differential equations don't. However we can always generate approximate solutions, and we can often solve a similar equation in closed form and then using reasonable approximations consider how solutions to our actual equation will differ.]
Let's return to the equation x ' = x + t. This equation can be solved, and we'll soon learn how to solve it. But unless you've studied differential equations to some extent, you probably can't solve it right now. So we can pretend that it's just not solvable.
However we can develop an approximate solution.
First note that the equation is of first order, so we can impose one condition on the equation. Let's say that our condition is that x = .37 when t = .30.
We can reason as follows:
If x = .37 and t = .30, then x ' = x + t = .37 + .30 = .67.
x ' tells us how quickly x changes with respect to t, so that as long as x ' doesn't change too significantly,
`dx = x ' * `dt
is a good approximation to how much x will change as a result of a change in t.
Without a lot of further analysis, let's just see what happens if we assume that a change `dt = .1 doesn't significantly affect the value of x '. This might or might not be the case, depending on your interpretation of the word 'significant' for the specific situation.
Now, if x ' = .67, then `dt = .1 gives us a slope of .67 and a 'run' of .1, which will result in a 'rise' equal to slope * run = .67 * .1 = .067.
Remember that we started with x = .37. Adding a 'run' of .067 we end up with x = .37 + .067 = .437.
An interval `dt = .1 increases the value of t from .30 to .40.
The approximate value of x, expected at t = .40, is therefore .437.
We can summarize the same series of calculations more formally as follows:
Our equation is x ' = f(t, x) = x + t.
At (t, x) = (.30, .37) we have x ' = f( .30, .37) = .67.
Using `dt = .1, our approximation for the change in x is
`dx = x ' * `dt = .67 * .1 = .067.
Our approximation is therefore
(t_new, x_new) = (30 + .1, .37 + .067) = (.40, .437).
There is no need to stop at this point, except to note that between our original point and our new point, x ' will indeed change. Remember that we did assume that x ' didn't change significantly. When x = .437 and t = .40, our value of x ' will be x ' = x + t = .837. Comparing this with .67, we see that there is about a 20% difference, and our estimate of `dx is probably off by about 10%. Depending on what we're trying to accomplish with our solution, this might or might not be significant. Taking into account other uncertainties in our situation we might decide that we need to do better. It would be easy to do so by reducing our `dt. For example, reducing `dt to .01 would result in the approximate point (.3767, .31), at which x ' would be .6867. We would expect our approximation to be off by only about 1%. The disadvantage is that we would not yet have a value of x corresponding to t = .40. Another option, based on `dt = .1, would be to average our two values .67 and .837 of x ', and use this average to recalculate our new x value.
In any case, continuing our original approximation, we can use our new point (.437, .40) to predict a new point:
Our equation is x ' = f(t, x) = x + t.
At (t, x) = (.40, .437) we have x ' = f( .40 , .437) = .837.
Using `dt = .1, our approximation is for the change in x is
`dx = x ' * `dt = .837 * .1 = .084,
where we have rounded off our `dx to two significant figures (more significant figures would be meaningless because of the error inherent in the approximation process).
Our new approximation is therefore
(t_new, x_new) = (.40 + .1, .437 + .084) = (.50, .53),
where again we have rounded to 2 significant figures.
We now have three points: (.37, .30), (.437, .40) and (.50, .53).
If you plot these points on a graph of x vs. t, you will see that they indicate a curve which is concave upward. This is what we would expect based on our calculations, since according to our calculations the values of x ' are positive and increasing (positive x ' means a positive slope, increasing x ' means increasing slope).
This curve is an approximate solution curve for our function.
Our curve is only approximate, since it is based on the assumption that x ' doesn't change between points.
Approximation errors accumulate with every step, multiplying in such a way that our curve varies more and more from the actual curve.
We could get a better approximation to the curve by using a smaller increment `dt (e.g., `dt = .01). It would take us 20 steps instead of 2 steps to get to the t = .50 point, but we would end up with a much more accurate solution curve.
Or after our first approximation of the new point, we could calculate the slope at the new point then re-predict that point based on the average of our two slopes. Using this method with `dt = .1 would give us a far more accurate solution curve in fewer steps than would be required with `dt = .01. There would be about twice as much calculation per step, but we would still be ahead.
Another way of analyzing solution curves is to use direction fields.
For any point (t, x) of the x vs. t plane, the equation x ' = x + t can be evaluated to obtain a value of the slope x '. Through any such point we can sketch a short line segment of the corresponding slope, and this segment will be the slope of a solution curve through that point.
If we do this for a grid of points, we get a direction field.
For the equation x ' = x + t, it is easy to evaluate x ' for the points on the grid
(0, 0), (0, 1/4), (0, 1/2), (0, 3/4), (0, 1)
(1/4, 0), (1/4, 1/4), (1/4, 1/2), (1/4, 3/4), (1/4, 1)
(1/2, 0), (1/2, 1/4), (1/2, 1/2), (1/2, 3/4), (1/2, 1)
(3/4, 0), (3/4, 1/4), (3/4, 1/2), (3/4, 3/4), (3/4, 1)
(1, 0), (1, 1/4), (1, 1/2), (1, 3/4), (1, 1).
The respective slopes range from 0 to 2. The respective slopes are
0, 1/4, 1/2, 3/4, 1
1/4, 1/2, 3/4, 1, 5/4
1/2, 3/4, 1, 5/4, 3/2
3/4, 1, 5/4, 3/2, 7/4
1, 5/4, 3/2, 7/4, 2
Plotted on a graph of x vs. t, we get something like the picture below:
All we have done is evaluate our function f(x, t) at a number of points within our region of interest.
It's easy to make a similar plot for any function f(x, t). All we need to do is plug in x and t values.
Having plotted our direction field, we can then sketch a solution curve starting from any point in the field.
The curve through (.30, .37) is approximated based on the hand-sketched direction field.
The t = .5 value of x predicted by the hand-sketched curve on the hand-sketched direction field is .63. Our calculated prediction was x = .53, and our approximations were seen to underestimate the actual changes in x, so a prediction of .63 might not be bad. However our calculated predication of the total change in x (which is .53 - .37 = .16) is probably not off by much more than 20%, and the sketch predicts a change of .63 - .37 = .26, so we can't claim a lot of accuracy for the hand sketch. The main value of the hand sketch is to provide a geometric picture of the behavior of this differential equation.
Direction fields are easy enough to sketch using a grid (though of course the process could be tedious). The process becomes much easier if we use isoclines:
f(x, t) = x + t has constant value c when x + t = c.
x + t = c when x = -t + c.
x = -t + c describes a straight line in the x vs. t plane, with slope -1 and y-intercept c.
It's easy to sketch these lines for a set of c values.
For c = 0, 1/2, 1, 3/2, 2 our lines would look something like the ones in the picture below:
The slope corresponding to each value of c is equal to c (this since x + t = c, and the slope is x ' = x + t). A number of segments of more or less appropriate slope have been sketched along each line. All the slopes along each line are the same, which makes it fairly easy to draw a decent slope field.
The figure isn't very well drawn; for example the c = 1/2 and c = 3/2 lines are badly misplaced, and the c = 1 line clearly missed the point (1, 0) on the t axis. The artist blames an unbuttoned sleeve for blocking his vision. Use of a straightedge and more care would have also been helpful.
The figure below depicts three solution curves. A solution curve can be sketched starting at any point, and moving either to the right or to the left. Once the slope field is drawn, the solution curves are fairly easy to sketch.
Had the equation been x ' = x + t^2 instead of x + t, the constant-slope condition would have been x + t^2 = c. This would give us x = -t^2 + c, which for each value of c is a parabola rather than a straight line.
In general, the equation x ' = f(x, t) yields constant-slope curves f(x, t) = c.
f(x, t) = x + t gives us a series of straight lines, along each of which the slope is constant.
f(x, t) = x + t^2 gives us a series of parabolas, along each of which the slope is constant.
In general a function f(x, t) gives us a series of curves f(x, t) = c, along each of which the slope is constant.
These curves are called isoclines ('iso' for 'one', 'cline' for inclination).
Spreadsheets and computer programs have obvious applications to the calculation schemes we have introduced here.
Let's consider for a moment the case of a second-order equation of the form
x '' = f(x, x', t).
We can still imagine a geometric picture depicting the behavior of the function f(x, x' t). However our picture graph would be in 3 dimensions, with an axis for each quantity x, x ' and t. Values of x '' would dictate changes in x ', not it x; and changes in x ' would then dictate changes in x. Our solution curves would become surfaces in a 3-dimensional space, our 3-dimensional space would no longer represent the 3-dimensional space we live in, and x vs. t solution curves would represent the intersections of these surfaces with the planes x ' = constant.
It's even more fun to try to imagine what happens when we go into higher dimensions.
However the scheme for numerically approximating the solution curves of a second-order equation isn't that much more complicated than the scheme we saw here.
There are three parts to this set of questions.
Part I, Part II and Part III follow below, in order. The links below might or might not be useful in helping you navigate.
Part I: The equation m x '' = - k x
Question: `q001. Show whether each of the following functions all satisfy the equation m x '' = -k x:
Your solution:
Confidence rating:
Given Solution:
If x = cos(t) then x ' = - sin(t) and x '' = - cos(t). Substituting the expressions for x and x '' into the equation we obtain
m * (-cos(t)) = - k * cos(t).
Dividing both sides by cos(t) we obtain m = k. If m = k, then the equation is satisfied. If m is not equal to k, it is not.
If x = sin(sqrt(k/m) * t) then x ' = sqrt(k / m) cos(sqrt(k/m) * t) and x '' = -k / m sin(sqrt(k/m) * t). Substituting this into the equation we have
m * (-k/m sin(sqrt(k/m) * t) ) = -k sin(sqrt(k/m) * t
Simplifying both sides we see that the equation is true.
The same procedure can and should be used to show that the third equation is true.
The question of the fourth equation is left to you.
Self-critique (if necessary):
Self-critique rating:
Question:
`q002. An incorrect integration of the equation x ' = 2 x + t yields x = x^2 + t^2 / 2. After all the integral of x is x^2 / 2 and the integral of t is t^2 / 2.
Show that substituting x^2 + t^2 / 2 (or, if you prefer to include an integration constant, x^2 + t^2 / 2 + c) for x in the equation x ' = 2 x + t does not lead to equality.
Explain what is wrong with the reasoning given above.
Your solution:
Confidence rating:
Given Solution:
The given function is a solution to the equation, provided its derivative x ' satisfies x ' = 2 x + t.
It would be tempting to say that the derivative of x^2 is 2 x, and the derivative of t^2 / 2 is t.
The problem with this is that the derivative of x^2 was taken with respect to x and the derivative of t^2 / 2 with respect to t.
We have to take both derivatives with respect to the same variable.
Similarly we can't integrate the expression 2 x + t by integrating the first term with respect to x and the second with respect to t.
Since in this context x ' represent the derivative of our solution function x with respect to t, the variable of integration therefore must be t.
We will soon see a method for solving this equation, but at this point we simply cannot integrate our as-yet-unknown x(t) function with respect to t.
Self-critique (if necessary):
Self-critique rating:
Question:
`q003. The general solution to the equation
m x '' = - k x
is of the form x(t) = A cos(omega * t + theta_0), where A, omega and theta_0 are constants. (There are reasons for using the symbols omega and theta_0, but for right now just treat these symbols as you would any other constant like b or c).
Find the general solution to the equation 5 x'' = - 2000 x:
Your solution:
Confidence rating:
Given Solution:
If x = A cos(omega * t + theta_0) then x ' = - omega A sin(omega * t + theta_0) and x '' = -omega^2 A cos(omega * t + theta_0).
Our equation therefore becomes
m * (-omega^2 A cos(omega * t + theta_0) ) = - k A cos(omega * t + theta_0).
Rearranging we obtain
-m omega^2 A cos(omega * t + theta_0) = -k A cos(omega * t + theta_0)
so that
-m omega^2 = - k
and
omega = sqrt(k/m).
Thus the constant omega is determined by the equation.
The constants A and theta_0 are not determined by the equation and can therefore take any values.
No matter what values we choose for A and theta_0, the equation will be satisfied as long as omega = sqrt(k / m).
Our second-order equation
m x '' = - k x
therefore has a general solution containing two arbitrary constants.
In the present equation m = 5 and k = 2000, so that omega = sqrt(k / m) = sqrt(2000 / 5) = sqrt(400) = 20.
Our solution x(t) = A cos(omega * t + theta_0) therefore becomes
x(t) = A cos(20 t + theta_0).
If theta_0 = 0 the function becomes x(t) = A cos( 20 t ). The graph of this function will be a 'cosine wave' with a 'peak' at the origin, and a period of pi / 10.
Self-critique (if necessary):
Self-critique rating:
Question:
`q004. In the preceding equation we found the general solution to the equation 5 x'' = - 2000 x. Assuming SI units, this solution applies to a simple harmonic oscillator of mass 5 kg, which when displaced to position x relative to equilibrium is subject to a net force F = - 2000 N / m * x. With these units, sqrt(k / m) has units of sqrt( (N / m) / kg), which reduce to radians / second. Our function x(t) describes the position of our oscillator relative to its equilibrium position.
Evaluate the constants A and theta_0 for each of the following situations:
Your solution:
Confidence rating:
Given Solution:
As seen in the preceding problem, a general solution to the equation is
x = A cos(omega * t + theta_0),
where omega = sqrt(k / m). For the current equation 5 x '' = -2000 x, this gives us omega = 20. In the current context omega = 20 radians / second.
So
x(t) = A cos( 20 rad / sec * t + theta_0 ).
Maximum displacement occurs at critical values of t, values at which x ' (t) = 0.
Taking the derivative of x(t) we obtain
x ' (t) = - 20 rad / sec * A sin( 20 rad/sec * t + theta_0).
The sine function is zero when its argument is an integer multiple of pi, i.e., when
20 rad/sec * t + theta_0 = n * pi, where n = 0, +-1, +-2, ... .
A second-derivative test shows that whenever n is an even number, our x(t) function has a negative second derivative and therefore a maximum value.
We can therefore pick any even number n and we will get a solution.
If maximum displacement occurs at t = 0 then we have
20 rad / sec * 0 + theta_0 = n * pi
so that
theta_0 = n * pi, where n can be any positive or negative even number.
We are free to choose any such value of n, so we make the simplest choice, n = 0. This results in theta_0 = 0.
Now if x = .3 when t = 0 we have
A cos(omega * 0 + theta_0) = .3
Self-critique (if necessary):
Self-critique rating:
Question:
`q005. Describe the motion of the oscillator in each of the situations of the preceding problem. SI units for position and velocity are respectively meters and meters / second.
Your solution:
Confidence rating:
Given Solution:
Self-critique (if necessary):
Self-critique rating:
Question:
Part II: Solutions of equations requiring only direct integration.
`q006. Find the general solution of the equation x ' = 2 t + 4, and find the particular solution of this equation if we know that x ( 0 ) = 3.
Your solution:
Confidence rating:
Given Solution:
Integrating both sides we obtain
x(t) = t^2 + 4 t + c,
where c is an arbitrary constant.
The condition x(0) = 3 becomes
x(0) = 0^2 + 4 * 0 + c = 3,
so that c = 3 and our particular solution is
x(t) = t^2 + 4 t + 3.
We check our solution.
Substituting x(t) = t^2 + 4 t + 3 back into the original equation:
(t^2 + 4 t + 3) ' = 2 t + 4 yields
2 t + 4 = 2 t + 4,
verifying the general solution.
The particular solution satisfies x(0) = 3:
x(0) = 0^2 + 4 * 0 + 3 = 3.
Self-critique (if necessary):
Self-critique rating:
Question:
`q007. Find the general solution of the equation x ' ' = 2 t - .5, and find the particular solution of this equation if we know that x ( 0 ) = 1, while x ' ( 0 ) = 7.
Your solution:
Confidence rating:
Given Solution:
Integrating both sides we obtain
x ' = t^2 - .5 t + c_1,
where c_1 is an arbitrary constant.
Integrating this equation we obtain
x = t^3 / 3 - .25 t^2 + c_1 * t + c_2,
where c_2 is an arbitrary constant.
Our general solution is thus
x(t) = t^3 / 3 - .25 t^2 + c_1 * t + c_2.
The condition x(0) = 1 becomes
x(0) = 0^3 / 3 - .25 * 0^2 + c_1 * 0 + c_2 = 1
so that c_2 = 1.
x ' (t) = t^2 - .5 t + c_1, so our second condition x ' (0) = 7 becomes
x ' (0) = 0^2 - .5 * 0 + c_1 = 7
so that c_1 = 7.
For these values of c_1 and c_2, our general solution x(t) = t^3 / 3 - .25 t^2 + c_1 * t + c_2 becomes the particular solution
x(t) = t^3 / 3 - .25 t^2 + 7 t + 1.
You should check to be sure this solution satisfies both the given equation and the initial conditions.
Self-critique (if necessary):
Self-critique rating:
Question:
`q008. Use the particular solution from the preceding problem to find x and x ' when t = 3. Interpret your results if x(t) represents the position of an object at clock time t, assuming SI units.
Your solution:
Confidence rating:
Given Solution:
Our solution was
x(t) = t^3 / 3 - .25 t^2 + 7 t + 1.
Thus
x ' (t) = t^2 - .5 t + 7.
When t = 3 we obtain
x(3) = 3^3 / 3 - .25 * 3^2 + 7 * 3 + 1 = 28.75
and
x ' (3) = 3^2 - .5 * 3 + 7 = 14.5.
A graph of x vs. t would therefore contain the point (3, 28.75), and the slope of the tangent line at that point would be 14.5.
x(t) would represent the position of an object. x(3) = 28.75 represents an object whose position with respect to the origin is 28.75 meters when the clock reads 3 seconds.
x ' (t) would represent the velocity of the object. x ' (3) = 14.5 indicates that the object is moving at 14.5 meters / second when the clock reads 3 seconds.
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Question:
`q009. The equation x '' = -F_frict / m - c / m * x ', where the derivative is understood to be with respect to t, is of at least one of the forms listed below. Which form(s) are appropriate to the equation?
Your solution:
Confidence rating:
Given Solution:
The right-hand side of the equation includes the function x ' but does not include the variable t or the function x.
So the right-hand side can be represented by any function which includes among its variables x '. That function may also include x and/or t as a variable.
The forms f(t) and f(x, t) fail to include x ', so cannot be used to represent this equation.
All the other forms do include x ' as a variable, and may therefore be used to represent the equation.
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Question:
`q010. If F_frict is zero, then the function x in the equation
x '' = -F_frict / m - c / m * x '
represents the position of an object of mass m, on which the net force is - c * x '.
Explain why the expression for the net force is -c * x '.
Explain what happens to the net force as the object speeds up.
Your solution:
Confidence rating:
Given Solution:
Newton's Second Law gives us the general equation
m x '' = F_net
so that
x '' = F_net / m.
It follows that
x '' = -F_frict / m - c / m * x '
represents an object on which the net force is -F_frict - c x '.
If F_frict = 0, then it follows that the net force is -c x '.
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Question:
`q011. We continue the preceding problem.
Your solution:
Confidence rating:
Given Solution:
If w(t) = x ' (t) then w ' (t) = (x ' (t) ) ' = x '' ( t ).
If x '' = - b / m * x ', then if w = x ' it follows that x '' = w ', so our equation becomes
w ' (t) = - b / m * w (t)
The derivative is with respect to t, so if we wish to integrate both sides we will get
w(t) = integral ( - b / m * w(t) dt),
The variable of integration is t, and we don't know enough about the function w(t) to perform the integration on the right-hand side.
[ Optional Preview:
There is a way around this, which provides a preview of a technique we will study soon. It isn't too hard to understand so here's a preview:
w ' (t) means dw / dt, where w is understood to be a function of t.
So our equation is dw/dt = -b / m * w.
It turns out that in this context we can sort of treat dw and dt as algebraic quantities, so we can rearrange this equation to read
dw / w = -b / m * dt.
Integrating both sides we get
integral (dw / w) = -b / m integral( dt )
so that
ln | w | = -b / m * t + c.
In exponential form this is
w = e^(-b / m * t + c).
There's more, but this is enough for now ... ].
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Question:
Part III: Direction fields and approximate solutions
`q012. Consider the equation x ' = (2 x - .5) * (t + 1). Suppose that x = .3 when t = .2.
If a solution curve passes through (t, x) = (.2, .3), then what is its slope at that point?
What is the equation of its tangent line at this point?
If we move along the tangent line from this point to the t = .4 point on the line, what will be the x coordinate of our new point?
If a solution curve passes through this new point, then what will be the slope at the this point, and what will be the equation of the new tangent line?
If we move along the new tangent line from this point to the t = .6 point, what will be the x coordinate of our new point?
Is is possible that both points lie on the same solution curve? If not, does each tangent line lie above or below the solution curve, and how much error do you estimate in the t = .4 and t = .5 values you found?
Your solution:
Confidence rating:
Given Solution:
At the point (.2, .3) in the (t, x) plane, our value of x ' is
x ' = (2 * .3 - .5) * (.2 + 1) = .12, approximately.
This therefore is the slope of any solution curve which passes through the point (.2, .3).
The equation of the tangent plane is therefore
x - .3 = .12 * (t - .2)
so that
x = .12 t - .24.
If we move from the t = .2 point to the t = .4 point our t coordinate changes by `dt = .2, so that our x coordinate changes by `dx = (slope * `dt) = .12 * .2 = .024. Our new x coordinate will therefore be .3 + .024 = .324.
This gives us the new point (.4, .324).
At this point we have
x ' = (2 * .324 - .5) * (.4 + 1) = .148 * 1.4 = .207.
If we move to the t = .6 point our change in t is `dt = .2. At slope .207 this would imply a change in x of `dx = slope * `dt = .207 * .2 = .041. Our new x coordinate will therefore be .324 + .041 = .365.
Our t = .6 point is therefore (.6, .365).
From our two calculated slopes, the second of which is significantly greater than the first, it appears that in this region of the x-t plane, as we move to the right the slope of our solution curve in fact increases. Our estimates were based on the assumption that the slope remains constant over each t interval. We conclude that our estimates of the changes in x are probably a somewhat low, so that our calculated points lie a little below the solution curve.
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Question:
`q013. Consider once more the equation x ' = (2 x - .5) * (t + 1).
Note on notation:
The points on the grid
(0, 0), (0, 1/4), (0, 1/2), (0, 3/4), (0, 1)
(1/4, 0), (1/4, 1/4), (1/4, 1/2), (1/4, 3/4), (1/4, 1)
(1/2, 0), (1/2, 1/4), (1/2, 1/2), (1/2, 3/4), (1/2, 1)
(3/4, 0), (3/4, 1/4), (3/4, 1/2), (3/4, 3/4), (3/4, 1)
(1, 0), (1, 1/4), (1, 1/2), (1, 3/4), (1, 1)
can be specified succinctly in set notation as
{ (t, x) | t = 0, 1/4, ..., 1, x = 0, 1/4, ..., 1}.
( A more standard notation would be { (i / 4, j / 4) | 0 <= i <= 4, 0 <= j <= 4 } )
Find the value of x ' at every point of this grid and sketch the corresponding direction field. To get you started the values corresponding to the first, second and last rows of the grid are
-.5, -.625, -.75, -.875, -1
0, 0, 0, 0, 0
...
...
1.5, 1.875, 2.25, 2.625, 3
So you will only need to calculate the values for the third and fourth rows of the grid.
Your solution:
Confidence rating:
Given Solution:
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Question:
`q014. We're not yet done with the equation x ' = (2 x - .5) * (t + 1).
x ' is the derivative of the x(t) function with respect to t, so this equation can be written as
dx / dt = (2 x - .5) * (t + 1).
Now, dx and dt are not algebraic quantities, so we can't multiply or divide both sides by dt or by dx. However let's pretend that they are algebraic quantities, and that we can. Note that dx is a single quantity, as is dt, and we can't divide the d's.
Your solution:
Confidence rating:
Given Solution:
The equation is easily rearranged into the form
dx / (2 x - .5) = (t + 1) dt.
Integrating the left-hand side we obtain 1/2 ln | 2 x - .5 |
Integrating the right-hand side we obtain t^2 / 2 + 4 t + c, where the integration constant c is regarded as a combination of the integration constants from the two sides.
Thus our equation becomes
1/2 ln | 2 x - 5 | = t^2 / 2 + t + c.
Multiplying both sides by 2, then taking the exponential function of both sides we get
exp( ln | 2 x - 5 | ) = exp( t^2 + 2 t + c ),
where as before c is an arbitrary constant.
Since the exponential and natural log are inverse functions the left-hand side becomes | 2 x + .5 |.
The right-hand side can be written e^c * e^(t^2 + 8 t), where c is still an arbitrary constant. e^c can therefore be any positive number, and we replace e^c with A, understanding that A is a positive constant.
Our equation becomes
| 2 x - .5 | = A e^(t^2 + 2 t).
For x > -.25, as is the case for our given value x = .3 when t = .2, we have
2 x - .5 = A e^(t^2 + 2 t)
so that
x = A e^(t^2 + 2 t) + .25.
Using x = .3 and t = .2 we find the value of A:
.05 = A e^(.2^2 + 2 * .2)
so that
A = .05 / e^(.44) = .03220, approx..
Our solution function is therefore
x(t) = .05 / e^(.44) * e^(t^2 + 2 t) + .25, or approximately
x(t) = .03220 e^(t^2 + 2 t) + .25
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Question:
`q015. OK, this time we are really going to be done with this equation. Again, x ' = (2 x - .5) * (t + 1)
Your solution:
Confidence rating:
Given Solution:
x ' = 1 when
(2 x - .5) * (t + 1) = 1.
Solving for x we obtain
x = 1/2 ( 1 / (t + 1) + .5) = 1 / (2(t + 1)) + .25.
The resulting curve is just the familiar curve x = 1 / t, vertically compressed by factor 2 then shifted -1 unit in the horizontal and .25 unit in the vertical direction, so its asymptotes are the lines t = -1 and x = .25. The t = 0 and t = 1 points are (0, .75) and (1, .5).
Similarly we find the curves corresponding to the other values of x ':
For x ' = 0 we get the horizontal line x = .25. Note that this line is the horizontal asymptote to the curve obtained in the preceding step.
For x ' = 2 we get the curve 1 / (t + 1) + .25, a curve with asymptotes at t = -1 and x = .25, including points (0, 1.25) and (1, .75).
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These questions can be challenging if you are rusty on first-year calculus and trigonometric functions.
Fortunately we won't be using these
functions a lot near the beginning of the course, but they become very important
later, and if you're rusty, or if you never really mastered these functions, you
want to get a good start now.
If you are rusty, try to answer the following without
the use of a calculator:
Quick refresher on trigonometric functions:
What are the maximum and minimum possible values of cos(theta)?
#$&*
For what values of theta does cos(theta)
take its maximum value, and for what values of theta is cos(theta) equal to
zero?
#$&*
For what values of t does y = cos(5 t) take its maximum value? For what t values
does this function take its minimum value?
#$&*
What is the maximum possible value of y = 4 cos(83 t)?
#$&*
Find at least one value of t for which y = cos(2 t + pi) takes the value zero.
#$&*
Find at least one value of t for which y = 3 cos(2 t + pi/4) takes its maximum
value.
#$&*
Find at least one value of t for which the derivative of y = cos(t) is zero.
#$&*
Find at least one value of t for which the derivative of y = cos(t) is zero, and
the second derivative is negative.
#$&*
Find the critical points of y = 4 sin(2 pi t + pi/2), then using a
second-derivative test find the values of t at which this function takes its
maximum and minimum values.
Feel free to submit a copy of these questions with your answers.