Day one of acceleration
This past summer I read Steven Strogatz's Infinite Powers: How Calculus Reveals the Secrets of the Universe. I was reminded that the earliest steps towards calculus were taken by those who were trying to solve physical science problems. And their approaches were both algebraic and geometric in their reasoning. I realized I could try retooling my approach to acceleration from a more Galilean approach, to not provide the quadratic regression from the get go on day one, but to edge into this relationship more carefully. To build more slowly from the material of the second week, and to not presume that a parabolic curve is the automatic result of a constant increase in speed.
I began class on Monday by sketching a time versus distance graph onto the board with a speed of one meter per second and the week ago Monday RipStik speed of 1.67 meters per second. I reminded the class that we had found slope was speed, and that increased speed led to an increased slope, a steeper slope.
I then sketched a time versus velocity graph with both v = 1.0 m/s and v = 1.67 m/s on the graph. Both have a slope of zero on this graph: there is no change in the velocity with respect to time. I noted that the equation was v = constant value, such as v = 1 or v = 1.67. There is no time t in the equation - no x-value.
I noted that the two graphs both meant that distance increased as the product of the velocity and the time.
I then sketched a time versus velocity graph further to the right, leaving space on the board to the left. On this graph I drew a steadily increasing velocity, a linearly increasing velocity. I noted that the slope was no longer zero: the slope is now a constant value, representing an increasing velocity. I noted that the equation would now be "v = mt" where v is the y-value and t is the x-value.
I then moved back to the left and drew a blank time versus distance graph. I put a question mark in the blank graph. I asked what the time versus distance graph might look like for the constantly increasing velocity. I asked students to sketch this on a piece of scratch paper.
Some students seemed unable to imagine what might happen and stared at the blank graph.
This student opted for what appeared to my eye to be a translated cubic, although I suspect the student did not think of the curve in this manner.
This student said that a "smiley" curve would be the result. From his description I do not think the student thought of this as a parabolic curve, just as a smiley.
A neighboring student opted to directly disagree and drew a frownie curve as he put it.
One student was spot on with the shape, but I asked the name of the shape and the student shrugged - they did not know if the shape had a name.
And therein lies the rub with providing a parabola: what guarantee does the pre-calculus student have that the curve must be a second degree curve, the result of a quadratic equation? I am to try to tackle this algebraically and geometrically, without calculus, on Wednesday in class.
At this point I suggested that we go outside and I would attempt to make the RipStik go faster and faster at a steady rate of increase in speed. Note that I had not used the word acceleration and would not use the word acceleration during class on Monday. I only ever spoke of increasing my speed at a steady rate of increase.
Based on prior terms, I laid down lines at 1, 2, 3, 5, 7, and 9 meters. Four timers were deployed and asked to time a whole run - I wanted a timing sequence off of a single timer. My notes from last term suggest I used median times on three watches, but start time desynchronization and reaction time errors seem likely to plague that data.
With the students timing I could focus on increasing my speed. Two runs were done, the second showing a slightly higher acceleration.
As a student read off the times, I entered the data into Desmos on my Motorola Moto G7. After both the first and second run I showed the class the result, walking up and down the sidewalk. The two curves both showed the same pattern of curving upwards. The student who produced the smiley face quickly noted that he was half right. I withheld that he was actually more than half right, there is another half to this curve.
The above data is from the first run. Later I used what appears to be a new capability in Desmos to do arithmetic inside a table to generate a table of estimated mark-to-mark speeds.
The numerator changes from one to two in the second column as the timed distances increased from one meter intervals to two meter intervals.
Note that the in the second equation I did not choose to run a linear regression of the second table. Doing so generates a line that fits the speed data better with a slightly lower slope. I opted instead to use the acceleration from the quadratic to generate the time versus velocity line.
I have made this choice because this will eventually be the path forward: the slope of the quadratic will be the speed, and that is the blue line above. The difference has to do with those being average speeds over discrete distances and not instantaneous speeds at a point. The velocity data is essentially an average speed over an interval during which the speed is increasing. This will be less than the instantaneous speed at the end of the interval. The blue line is the derivative, the data points are discrete average speeds at locations in time to the "left" of where they appear above. This is something I will likely quietly steer clear of in class on Wednesday.
My intent on Wednesday is to start with that constantly increasing velocity with respect to time line from Monday and try to work geometrically backwards to what the time versus distance graph should look like. My intent is to take a velocity that increases at one meter per second per second, a slope of one.
To calculate the distance I will need to work my way around to why the area under the line is the distance, that ½×base×height produces the distance. That will not make sense as a starting point. So I believe I will need to argue that my speed is increasing along that one second interval.
While at the end of the one second I have reached one meter per second, that was not my speed for the whole second. I have to consider my average speed for the one second, and that happened by at halfway through the second, and 0.5 seconds. At that time I was traveling at 0.5 meters per second. Times one second and the distance is one half of a meter. Which is the area of the triangle.
If I build similar triangles out at two, three, and four seconds, and calculate average velocities multiplied by the duration I get a sequence such as:
0.5 × 1 = 0.5
1.0 × 2 = 2
1.5 × 3 = 4.5
2.0 × 4 = 8
If one were to treat 0.5 as the base unit of one, then the numbers double to the more familiar 1, 4, 9, 16. Perhaps I should consider use of a slope of 2 meters per second and see if the more familiar numbers would spill out directly. Note that the first number above is indeed ½×base of the triangle. Plotting the numbers above, even at a slope of one, will yield that same curve that was generated on Monday. And that was the intent - to mimic the curve from Monday. To get stuck at about the same place Galileo did 500 years ago, just before the dawn of calculus. We have a relationship that is going up quadratically, but little clue as to why nor what exactly the slope formula might be.
At this point I suspect I should just introduce the v = at formula as the derivative of d=½at² and suggest that while I cannot prove this with Newton's fluxions or Leibniz's calculus, this is where the above is heading.
Time permitting I might still get in the "arc" of acceleration run on the RipStik to wrap up the class.
Post-script: the two acceleration runs with their accelerations
The above chart is also available online. Thus my acceleration on Monday was roughly 0.29 m/s² and 0.42 m/s².
I began class on Monday by sketching a time versus distance graph onto the board with a speed of one meter per second and the week ago Monday RipStik speed of 1.67 meters per second. I reminded the class that we had found slope was speed, and that increased speed led to an increased slope, a steeper slope.
I then sketched a time versus velocity graph with both v = 1.0 m/s and v = 1.67 m/s on the graph. Both have a slope of zero on this graph: there is no change in the velocity with respect to time. I noted that the equation was v = constant value, such as v = 1 or v = 1.67. There is no time t in the equation - no x-value.
I noted that the two graphs both meant that distance increased as the product of the velocity and the time.
I then sketched a time versus velocity graph further to the right, leaving space on the board to the left. On this graph I drew a steadily increasing velocity, a linearly increasing velocity. I noted that the slope was no longer zero: the slope is now a constant value, representing an increasing velocity. I noted that the equation would now be "v = mt" where v is the y-value and t is the x-value.
I then moved back to the left and drew a blank time versus distance graph. I put a question mark in the blank graph. I asked what the time versus distance graph might look like for the constantly increasing velocity. I asked students to sketch this on a piece of scratch paper.
Some students seemed unable to imagine what might happen and stared at the blank graph.
This student opted for what appeared to my eye to be a translated cubic, although I suspect the student did not think of the curve in this manner.
This student said that a "smiley" curve would be the result. From his description I do not think the student thought of this as a parabolic curve, just as a smiley.
A neighboring student opted to directly disagree and drew a frownie curve as he put it.
One student was spot on with the shape, but I asked the name of the shape and the student shrugged - they did not know if the shape had a name.
And therein lies the rub with providing a parabola: what guarantee does the pre-calculus student have that the curve must be a second degree curve, the result of a quadratic equation? I am to try to tackle this algebraically and geometrically, without calculus, on Wednesday in class.
At this point I suggested that we go outside and I would attempt to make the RipStik go faster and faster at a steady rate of increase in speed. Note that I had not used the word acceleration and would not use the word acceleration during class on Monday. I only ever spoke of increasing my speed at a steady rate of increase.
Based on prior terms, I laid down lines at 1, 2, 3, 5, 7, and 9 meters. Four timers were deployed and asked to time a whole run - I wanted a timing sequence off of a single timer. My notes from last term suggest I used median times on three watches, but start time desynchronization and reaction time errors seem likely to plague that data.
With the students timing I could focus on increasing my speed. Two runs were done, the second showing a slightly higher acceleration.
As a student read off the times, I entered the data into Desmos on my Motorola Moto G7. After both the first and second run I showed the class the result, walking up and down the sidewalk. The two curves both showed the same pattern of curving upwards. The student who produced the smiley face quickly noted that he was half right. I withheld that he was actually more than half right, there is another half to this curve.
Note that the in the second equation I did not choose to run a linear regression of the second table. Doing so generates a line that fits the speed data better with a slightly lower slope. I opted instead to use the acceleration from the quadratic to generate the time versus velocity line.
I have made this choice because this will eventually be the path forward: the slope of the quadratic will be the speed, and that is the blue line above. The difference has to do with those being average speeds over discrete distances and not instantaneous speeds at a point. The velocity data is essentially an average speed over an interval during which the speed is increasing. This will be less than the instantaneous speed at the end of the interval. The blue line is the derivative, the data points are discrete average speeds at locations in time to the "left" of where they appear above. This is something I will likely quietly steer clear of in class on Wednesday.
My intent on Wednesday is to start with that constantly increasing velocity with respect to time line from Monday and try to work geometrically backwards to what the time versus distance graph should look like. My intent is to take a velocity that increases at one meter per second per second, a slope of one.
To calculate the distance I will need to work my way around to why the area under the line is the distance, that ½×base×height produces the distance. That will not make sense as a starting point. So I believe I will need to argue that my speed is increasing along that one second interval.
While at the end of the one second I have reached one meter per second, that was not my speed for the whole second. I have to consider my average speed for the one second, and that happened by at halfway through the second, and 0.5 seconds. At that time I was traveling at 0.5 meters per second. Times one second and the distance is one half of a meter. Which is the area of the triangle.
If I build similar triangles out at two, three, and four seconds, and calculate average velocities multiplied by the duration I get a sequence such as:
0.5 × 1 = 0.5
1.0 × 2 = 2
1.5 × 3 = 4.5
2.0 × 4 = 8
If one were to treat 0.5 as the base unit of one, then the numbers double to the more familiar 1, 4, 9, 16. Perhaps I should consider use of a slope of 2 meters per second and see if the more familiar numbers would spill out directly. Note that the first number above is indeed ½×base of the triangle. Plotting the numbers above, even at a slope of one, will yield that same curve that was generated on Monday. And that was the intent - to mimic the curve from Monday. To get stuck at about the same place Galileo did 500 years ago, just before the dawn of calculus. We have a relationship that is going up quadratically, but little clue as to why nor what exactly the slope formula might be.
At this point I suspect I should just introduce the v = at formula as the derivative of d=½at² and suggest that while I cannot prove this with Newton's fluxions or Leibniz's calculus, this is where the above is heading.
Time permitting I might still get in the "arc" of acceleration run on the RipStik to wrap up the class.
Post-script: the two acceleration runs with their accelerations
The above chart is also available online. Thus my acceleration on Monday was roughly 0.29 m/s² and 0.42 m/s².
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