A_slope_rule_is_a_differential_equation

A rule that specifies the slope at every point of a region of the plane is a differential equation.

A slope in the y vs. t plane specifies a ratio which we interpret as rise / run.  A smooth curve in the y vs. t plane will at every one of its points have a slope, which is the slope of the tangent line at the point.  If y is a function of t, then we can denote the derivative of y with respect to t as y ', and identify the value of y ' with the slope of the tangent line.  This point could be made more specific, but for now we'll leave it at that.

If f(t, y) is a function of t and y, then y ' = f(t, y) is a rule which associates to every point (t, y) a slope y '.

If you can evaluate f(t, y) at a number of representative points in a region, you can use these values to construct a good picture of how slopes behave in that region.

y ' = f(t, y) is a rule for finding the slopes, so it is a differential equation.

Many differential equations are of this type, but there are also other types of differential equations, as we will see later.

Use_the_slope_rule_to_sketch_approximate_curves

If we have a picture representing the slopes y ' = f(t, y) in a region of the y vs. t plane, then we can start from any point of the plane and sketch a good approximation to a curve through that point, satisfying the condition that the slope at any point of the curve is f(t, y).

Using the rule we can construct an approximate curve starting at any point

We can use the rule itself to approximate the slope at a given point, and construct a short segment having that slope.  Having constructed that segment, we can then use the point at either of its ends to evaluate the function and sketch a new segment.  By repeating this process we can construct a good approximation to the curve through that point.

The_equation y ' = t / 2

We can evaluate the function f(t, y) = t / 2 at the points (0, 0), (1, 0), (2, 0), (0, 1), (1, 1), (2, 1), (0, 2), (1, 2) and (2, 2), then sketch the corresponding slope segments at these points.

Using this picture we can sketch a solution curve starting at any selected point in the interval.

We can also solve the equation y ' = t / 2.  We get y = t^2 / 4 + c, where c can be any constant number. 

If we substitute y = t^2 / 4 + c into the equation y ' = t / 2, we get (t^2 / 4 + c) ' = t / 2.  Since (t^2 / 4 + c) ' is in fact equal to t / 2, we see that the function y(t) = t^2 / 4 + c is a solution to the equation y ' = t / 2.

Check_the solution_against_the approximate_curve

We can compare our solution y(t) = t^2 / 4 + c to our approximate curve.

We first find that the y intercept of our solution curve is about (0, .3).

Thus our solution y(t) = t^2 / 4 + c should match our approximate curve, provided the graph of y(t) vs. t passes through the point (0, .3).

This will be the case if

y(0) = .3.

Since y(t) = t^2 / 4 + c, our condition dicates that

y(0) = 0^2 / 4 +  c

must be equal to .3.

It follows that the value c = .3 yields a solution curve through (0, .3).  Our solution is thus

y(t) = t^2  4 + 0.3.

If we graph this curve we find that it matches our previously constructed curve pretty well.

The rule_y ' = t / 2 was_a_simple_function_of_t and was therefore_easily_integrated

It was easy to integrate the y ' = t / 2 equation, because it's easy to find an antiderivative for t / 2.

This isn't always the case.  Our f(t, y) function doesn't have to be a function of t alone, and even if it is, we can't always integrate it.  For example y ' = e^(t^2) is a perfectly good equation, but we can't integrate e^(t^2) with respect to t.

In our original picture, y ' = y / 5 + t / 4.  It was easy to construct a slope picture for this equation, but we can't simply integrate y / 5 + t / 4 with respect to t.  This is because the function y(t) itself, the function we are trying to find, is itself a function of t.  We don't know the function, so we can't simply integrate it with respect to t.

It isn't too difficult to solve this equation, as we will see fairly soon, but it can't be solved by direct integration with respect to t.

sketch_curves_from_slope_rule

We use pretend 'buttons' to execute an unspecified rule to evaluate slope and/or reciprocals of slopes, and construct solution curves through various points..

something_that_should_have_bothered_you

When we draw straight line segments, they don't account for continuously changing values of the slope.

We therefore get a little bit of an error each time we sketch a segment.

So the curves we sketch in this manner are only approximations to the actual solution curves.

We can improve on the accuracy of our approximation by improving our estimates of the slopes, using a trick you don't have to remember right now.

fundamental_theorem_of_calculus

y ' (t) is the derivative with respect to t of the function y(t).  So the function y(t) is an antiderivative of the function y ' (t). 

Any function y(t) + c is an antiderivative of the function y ' (t).

So if we integrate both sides of an equation of the form

y ' = f(t)

we get

y + c = integral( f(t) dt)

so that y (t) is of the form

y(t) = integral ( f(t) dt) + c,

where again c is an arbirary constant function.

From the Fundamental Theorem of Calculus we know that the integration and differentiation are inverse operations, and in particular that for a well-behaved function h(t) the indefinite integral of h ' (t) is h(t) + c, where c is a constant (here the ' is taken to indicate differentiation with respect to the independent variable, in this case with respect to t).  So for example if h ' (t) = t^2, it follows that h(t) = t^3 / 3 + c.  We can easily verify that for this function h(t), no matter what the value of the constant c, we have h ' (t) = t^2.  Put in more succinct notation:  integral( h ' (t) dt) = h(t) + c. 

The solution of the equation h ' (t) = t^2 in the preceding example is solved by integrating both sides with respect to t.  We start with the equation:

h ' (t) = t^2.

We integrate both sides:

integral( h ' (t) dt ) = integral ( t^2 dt).

We get

h(t) = t^3 / 3 + c

and our function h(t) is the solution of our equation.

If we are given the equation dx / dt = 3 t, a solution to the equation is a function x(t) whose derivative with respect to t is 3 t. 

It is easy to see that any function x(t) = 3 t^2 / 2 + c satisfies this equation. 

The equation dx/dt = 3 t is called a differential equation. 

This particular differential equation is easily solved by simply integrating both sides. 

Most differential equations are not so easily solved. 

You may if you wish try to solve the equation dx/dt = 3 t + x.  We will soon learn how to solve this equation, but until we have developed the right set of tricks the solution will probably elude you.

As the ball coasts down the incline its velocity v changes with respect to clock time at a constant rate. 

The rate of change of velocity with respect to clock time is the derivative dv / dt, and the mathematical expression of this condition is dv/dt = constant. 

If we let letter 'a' stand for the constant we have dv / dt = a. 

Letting ' indicate the derivative with respect to t, this equation can be written v ' = a. 

The notations v ' and dv / dt are interchangeable; sometimes one notation is more convenient, sometimes the other.

ball_on_ramp_in_water

If two steel balls roll down two ramps in equal time intervals, then if one ramp is submerged in water the two balls will not roll down in equal time intervals.  The presence of the water impedes the motion of the ball on the submerged ramp.

modeling_ball_on_ramp_with_drag

If the v vs. t graph of the ball on the non-submerged incline is a straight line, then the slope picture on a v vs. t graph consists of a series of parallel lines, leading to straight-line solution curves..

The submerged ball has to move water out of the way more and more quickly as it speeds up, so the faster it goes the less quickly it speeds up.  As a result, the greater the value of v the less the slope of the v vs. t graph.  This leads to solution curves which are concave downward.

modeling_with_linear_drag_force

As the ball coasts down the incline it speeds up.  If the water wasn't there it would speed up at a constant rate.  However the presence of the water has its effect:  the faster the ball moves, the more the water resists it, so the ball speeds up more and more slowly. 

In some circumstances the resisting effect of the water causes the ball's speed to change at the rate a - c v, where a and c are positive constant numbers. 

You can see that the expression a - c v is equal to a when v = 0.  However as v increases the value of (a - c v) decreases, and when v gets large enough the value of (a - c v) becomes zero. 

The resulting differential equation would be

dv / dt = a - c v;

or alternatively

v ' = a - c v. 

As we will see before long, this equation isn't particularly difficult to solve.  However it won't work to just integrate both sides with respect to t.  Since you don't yet know the function v(t), you can't hope to evaluate integral ( (a - c v(t) ) dt ).

pendulum_model

The further we pull a pendulum back from its equilibrium position, the greater the force tending to pull it back toward equilibrium.

The greater the force, the more quickly the speed of the pendulum changes.

Using v for the velocity, the rate of change of the velocity of the pendulum is v '.

So if v is the velocity of the pendulum, and x is its position, we have

v ' = - k / m * x.

Not the velocity of the pendulum is the rate at which its position changes with respect to time.  So v = x ', and v ' = (x ' ) ' = x ''.

Our equation v ' = - k / m * x therefore becomes

x '' = - k / m * x.

The solution is therefore a function x(t) whose second derivative is a negative multiple of itself.

brief_physics_of_pendulum

This analysis is slightly simplified, so don't carry it with you into your physics or engineering class.  For those classes you need to work out the vectors carefully.  However for small displacements from equilibrium, this explanation is very nearly accurate.

The force of gravity on a pendulum mass, hanging at equilibrium, causes the string from which the mass is suspended to stretcch slightly.  The stretching continues until the tension in the string is equal and opposite to the force exerted by gravity, so that the pendulum mass remains in equilibrium.

If the mass is pulled slightly back in the horizontal direction, the tension in the string does not change significantly, but that tension is no longer directed upward.  The tension in the string tends to pull its ends together, so that the tension now pulls the mass in the direction of the string.  Gravity continues to pull the mass downward, and the tension force remains mostly upward, but the tension also now has a component perpendicular to the direction of the gravitational force.  This horizontal component of the tension tends to pull the mass back toward equilibrium.

It isn't difficult to show that the horizontal component of the tension force is proportional to its displacement from the equilibrium position, so that the net force on the pendulum is F_net = - k x.

The net force on an object is the product of its mass and acceleration:  F_net = m a, where a = v ' = x '' is the derivative of the velocity function and the second derivative of the position function.

Thus we have

m a = - k x

which we can express as

m x '' = - k x,

leading us to the previouisly stated equation

x '' = - k x.

pendulum_in_water

A pendulum oscillating in air loses energy very slowly, maintaining the original amplitude of its oscillation through several oscillations.

A pendulum oscillating in water loses energy much more quickly, with the amplitude of its oscillations clearly decreasing with every cycle.

pendulum_in_changing_water_depth

A pendulum initially oscillating while only partly submerged, and continuing to oscillate while water depth increases, loses energy more quickly than in air, as would be expected.

This situation is, however, much more complex to model than any of the preceding situations.

equation_of_pendulum_in_water

A pendulum swinging in water still experiences a restoring force of the form F_drag = - k x, though due to buoyant forces the value of k is different than in air.

The pendulum also experiences a drag force due to the water.  This drag force depends on how fast the pendulum is moving through the water, and it in the direction opposite the motion.

For certain shapes and velocities, the drag force is simply proportional to the velocity v of the pendulum.  Such a force is equal to - c v, where c is a constant number that depends on the shape and size of the actual pendulum mass.  The negative sign occurs because the force is in the direction opposite the velocity.

This leads to the equation m a = - k x - c v, which since v = x ' and a = x '' can be written

x '' = -k/m * x - c/m * x '.

As we will see, this equation is not very difficult to solve.

Now for some speeds, the drag force is more of the form F_drag = - k v^2.  This leads to the equation

x '' = - k / m * x - c / m * (x ' )^2.

Except for the square on the x ', this equation looks a lot like the first.  However that square on the x ' makes a big difference in the difficulty of solution.  This equation is very difficult, perhaps impossible, to solve.

We can say that the first equation, with just multiples of first powers of x, x ' and x '', is linear.  The second equation, with the square on the x ', is not linear.

Nonlinear equations are usually very challenging and often impossible to solve.

The equation x ' = - k / m * x - c / m * x ' is of the form x ' = f ( x, x '), where f(x, x ') = - k / m * x - c / v * x '.

This function contains only first powers of x and x ', so we call it a linear function.

Equations of the form x '' = f( x, x ' ) are generally pretty easy to solve.

The equation x ' = - k / m * x - c / m * (x ' )^2 is of the form x ' = f ( x, x '), where f(x, x ') = - k / m * x - c / v * (x '.)^2.

In this case the function f(x, x ') contains a second power of x ' and is therefore not linear, making it difficult or impossible to solve.

equation_pendulum_in_rising_water

If a pendulum swings in water which is rising, so that more and more of the pendulum is submerged, then the drag force will again depend on the pendulum's velocity v. 

Whereas before, for some range of velocities, we expected that the drag force would be of the form F_drag = - c v, in this case the value of c would expected to increase with the rising water.  So we can't talk about a constant value of c, but would instead expect that the c in this case would be a function of time.  This function would depend on how fast the water is rising, and on the shape of the pendulum's mass.

So we might have a drag force of the form

F_drag = -c(t) * v.

This would lead the an equation of the form

x '' = -k/m * x - c(t) / m * x '

(you can ignore the following comment, which isn't necessary for the point being made here, but for accuracy we should say that fact, k would also be a function of t, since the increasing buoyant.force on the pendulum would change the tension in the string and affect the horizontal component of tension).

The right-hand side of the equation now includes t as a variable, which introduced additional complications.  The right-hand side being a function of x, x ' and t, we could now write the equation as

x '' = f(t, x, x ')

form_of_differential_equation

A differential equation might be in any of a variety of forms, depending on the order of the highest derivative (x ' is the first-order derivative, x '' is second-order, etc.), whether t is involved explicitly, and other characteristics.  Some possible forms are

x ' (t) = f(t), for example x ' = t / 2

x ' (t) = f(t, x), for example x ' = 2 x

x '' (t) = f(t), for example x '' = 4 t^3

x ''(t) = f(t, x), for example x '' = 3 cos(t^2) + 2 x

x ''(t) = f(x, x ' ), for example x '' = 3 x ' + 5 x

x '' ( t) = f(t, x '), for example x '' = 5 t^2 x '

x '' ( t) = f(t, x, x '), for example x '' = 2 t x + (x ' )^2.

Some of these equations are linear, involving only first powers of x and its derivatives (if t is present, the power of t has no effect on the linearity of the equation).

Some are first-order, involving only x and x ' , possibly along with t.

Some are second-order, involving x '' and lower powers of x and/or x '.

forms_of_some_differential_equations

The form x ' = f(t) is easily solved if f(t) can be easily integrated.

The form x ' = f(x) is easily solved if 1 / f(x) can be easily integrated.

An equation of the form x ' = f(t, x) might be realatively easy to solve, or it might be difficult or even impossible.

An equation of the form x '' = f(t) is easily solved if f(t) can be easily integrated twice with respect to t.

An equation of the form x '' = f(x, x ') is not difficult to solve if f(x, x ') is a linear function of x and x '.

Some equations of the other forms can be tricked into giving up their solutions, and some can't.

A wooden peg rotating around a vertical circle can provide a good model for way the position of the pendulum changes in time.

By rotating a vertical ring 30 degrees at a time and aligning it with a metal angle, we can see that the horizontal position of a point on the ring changes by different amounts, depending on the initial horizontal position of the mark.

By marking the ring at 30 degree intervals, and at 45 degree intervals, an lying it flat beneath the camera, we can see how the horizontal positions change as the position on the ring changes.

By placing the ring on a set of x-y axes, with the origin at the center of the ring, we can trace the circle, the mark positions around the circle at 30 and 45 degree intervals (starting from the positive x axis).

Projecting these points onto a horizontal axis (parallel to the x axis) we see how the x coordinate of the point changes with angle, and we gain information about how the position of a pendulum might change with time.

It should be clear that the position of a pendulum which is moving in conjunction with a point moving around the circle will change more quickly when the pendulum is close to the center than when it is close to the extreme points of its motion.

Done very accurately our picture will exhibit right-left symmetry.

circular_model_of_sine_and_cosine_4

A closer look reveals how a pendulum moving in conjunction with a point moving around the circle, starting from the extreme rightmost point of its motion, will speed up as it approaches the equilibrium position at the midpoint of its motion, then slow as it approaches the leftmost point.

We label the horizontal line onto which we have projected the points on our circle as the x axis, abandoning our original x axis, placing the origin of the x axis below the center of the circle..  We do this only for convenience in labeling.

We assume that the circle has a radius of 1 unit, so that the x coordinates of our various points lie between -1 and 1.

We proceed to estimate the x coordinates of our various points.

The x-axiis points corresponding to 0, 30, 60, 90, 120, 150, 180, 210, 240, 270, 300, 330 and 360 degrees are estimated to lie at respective coordinates x = 1, .9, .5, 0, -.5, -.9, -1, -.9, -.5, 0, .5, .9 and 1.

The exact x coordinates corresponding to these points, on an ideal and accurately divided circle, are called the cosines of the angles 0 deg, 30 deg, 60 deg, 90 deg, 120 deg, etc..  So for example the cosine of 120 deg would, according to our estimate, be -.5.  Our estimate for the cosine of 30 degree is .9.  (As we will see, and as you may in fact already know, the exact value of cosine(30 deg) is sqrt(3) / 2, approximately .866).

Our circle has been divided into 360 degrees.  We can also divide the circle into 2 pi radians.  Since 30 degrees is 1/12 of the 360 degrees into which the circle has been divided, it follows that 30 degrees should also be 1/12 of 2 pi  radians.  Since 1/12 of 2 pi radians is pi / 6 radians, it follows that 30 degrees is equivalent to pi / 6 radians.

We can label our point in increments of pi / 6, obtaining a degree label and a radian label for each point.

circular_model_of_sine_and_cosine_6

A circle of radius 1 is divided into equal increments of 30 degrees or pi/6 radians.  The points are labeled according to their angles, and the horizontal and vertical projections of the labeled points are indicated.

The sine of an angle is the y coordinate the corresponding projection, and the cosine is the x coordinate.

This is the same as saying that the x coordinates of the points we have labeled on the circle are the cosines of those angles, and the y coordinates are the sines.

We reason out the exact values of the sines and cosines of all angles which are multiples of 30 degrees or pi/6 radians.  This is based on the '30-60-90' right triangle, which is half of an equilateral triangle, and the '45-45-90' triangle, which has two equal legs. 

We use the Pythagorean Theorem with the characteristics of these triangles and conclude that the cosine and sine of 30 degrees are respectively sqrt(3) / 2 and 1/2, while the cosine and sine of 45 degrees or pi/4 radians are both equal to sqrt(2) / 2.