Tuesday, November 20, 2012

Derivatives, Velocity, and Acceleration

Calculus can be viewed as the study of rates of change of quantities.  The most familiar rate of change in our ordinary experiences is velocity as the rate of change of position.  As we ride our skateboards, bicycles, and cars, we understand that high velocities mean our position is changing quite rapidly; and we understand that when our velocity is zero, we are standing still.

So imagine the experience of a perfect rocket-car that experiences no friction and has no brakes.  The only way to change its velocity is with rocket blasters that are installed on either end of the vehicle.  Consider the following trajectory, illustrated as an animation, and repeating in a loop.  A timer (16 seconds) is also shown to provide a measurable sense of time.

The following table describes when and which direction the rockets are firing.

Time Interval Direction of Rocket Magnitude
[0,1) None 0
(1,3) Right Moderate
(3,5) None 0
(5,7) Left Large
(7,8) None 0
(8,12) Right Small
(12,16] None 0

In addition, the rocket-car is driving along a track which has positions marked so that we can think of the position as a variable (in meters).  That is, we can think of examining the relation between the variables of time and position.  The following table provides the position of the rocket-car as recorded at each second.

 time (s) position (m)  time (s) position (m) 0 1 2 3 4 5 6 7 0.0 0.0 1.0 4.0 8.0 12.0 14.0 12.0 8 9 10 11 12 13 14 15 8.0 4.5 2.0 0.5 0.0 0.0 0.0 0.0

A table is only useful for a coarse overview of the relation between the variables.  For example, we can see that between t=2 s and t=3 s, the position went from x=1.0 m to x=4.0 m.  We can compute an average velocity over this time interval:

So the average velocity on this interval was 3.0 m/s.  However, this was during one of the intervals the rocket was firing.  It would have been going slower at the beginning of the interval and faster at the end.  The table does not give enough information for us to estimate the actual velocities.

Another representation of the relation between time and position is with a graph.

The derivative defines a new variable, dx/dt.  Although this looks like a fraction, we should really think of the entire symbol as the name of our new variable, the derivative.  This variable measures the rate of change of position x with respect to time t.  On the graph, this corresponds to the slope of the graph at each point.

For example, if I were to consider the point at t=2 s and x=1 m, we might draw the tangent line and measure the slope.  This slope is the derivative, which for this point corresponds to dx/dt=2.0 m/s.  Notice that the derivative is a velocity, which is precisely what the derivative measures for position with respect to time.  So we will use the variables v and dx/dt interchangeably.

We could compute the derivative for every point of the original relation.  This new variable could be added to our table.

 time (s) position (m)  velocity (m/s)  time (s) position (m)  velocity (m/s) 0 1 2 3 4 5 6 7 0 0 1 4 8 12 14 12 0 0 2 4 4 4 0 -4 8 9 10 11 12 13 14 15 8 4.5 2 0.5 0 0 0 0 -4 -3 -2 -1 0 0 0 0

But it would be better, just as with our original relation, to consider the relation as a graph.  The figure below illustrates both graphs one above the other.

Notice that the velocity graph itself also has a slope at (nearly) every point.  The slope of a velocity graph is also a rate of change, measuring how the rate of change of velocity (m/s) with respect to time (s).  Consequently, the derivative dv/dt, which is called acceleration has units (m/s)/s, or more directly m/s2. (If velocity was measured in miles per hour, then acceleration might be measured as mph/s.)  We can use the variable name a=dv/dt.

Because our velocity is defined as piecewise linear, the acceleration will be piecewise constant.  On intervals where the velocity was constant, the acceleration will be zero.

Notice that the acceleration is directly related to our original discussion of the rocket-blasters.  When the rockets are blasting to the right, the acceleration is positive; when the rockets are blasting to the left, the acceleration is negative.  In fact, this is exactly the idea behind Newton's second law of mechanics, F=ma.  The rocket's thrust corresponds to the force F.  Newton's law simply states that the acceleration is proportional to the force.  The mass, which measures inertia, provides the proportionality constant. If we had the exact same rockets and the car had twice the mass, then the acceleration would be cut in half.

Because our rocket-car does not have brakes, the only way to slow down is to turn on the rockets in the opposite direction of motion.  In the language of derivatives, the acceleration must have the opposite sign from the velocity.  If the acceleration is the same sign as the velocity, then the effect of acceleration is an increase in the speed (the magnitude of velocity).

We close this reading by considering the effect of acceleration on the graph of position.  Slowing down corresponds to making the slope closer to horizontal.  Speeding up corresponds to making the slope steeper.  So, let us look at the original graph of position as a function of time, but marking the graph with different colors, depending on how the car is accelerating.
When the velocity and acceleration are opposite, I have marked the graph in red.  Notice that this is when the graph is becoming closer to horizontal (left-to-right).  When the velocity and acceleration are the same direction, I have marked the graph in green.  Notice that this is when the graph is becoming steeper (left-to-right).  When there is no acceleration, I have marked the graph in purple (straight lines).

To visualize this relative to the original animation, I have color-coded the timer and have placed a colored-flag on the car.  Try to connect the information in the graph to the visualization of the moving rocket-car.

Concavity is the word that describes how a graph bends.  When the second derivative (acceleration) is positive, the graph will be concave up.  On our graph, this corresponds to the first green segment, ∈ (1,3), and the last red segment,  ∈ (8,12).  When the second derivative is negative, the graph will be concave down.  This corresponds to the middle red and green segments,  ∈ (5,7).