Kimberly Wathen
Chris Perkins
Jennifer Wright
MA 502 Project
I. Introduction
The scope of the module involves looking at the applications of graphs, limits, and derivatives to analyze and study the behavior of functions; more explicitly, functions that represent the position of an object relative to time. We will use various concepts learned in high school Calculus and their applications to physics. The lesson will be divided into three sections: graphs, limits, and derivatives. Each section will look at how its specific topic can be applied to physics. Positions, velocity, and acceleration can be evaluated by each of these three methods (graphs, limits, and derivatives). The connections we will explore play a large role in understanding upper level math and physics classes as well as applications to the real world. Each section should be used as supplemental information once the math components have been mastered. This particular module is to be used to learn how we can analyze the position functions using various forms of Calculus.
The intended audience for these lessons is Calculus and Physics students. These students would benefit from the exercises and "real" world examples. The necessary prerequisites for this lesson are Pre-Calculus as well as having been exposed to the various Calculus topics we discuss within the three sections. The students should have either already taken or are taking concurrently a Physics course.
II. Background
There were three major contributors to the study of functions. Isaac Newton played a major role in discovering information about functions in the 17th century. When he was attending college at Cambridge, a plague closed the school in 1665. He then returned home to Lincolnshire where he "laid the foundation for differential and integral calculus." He termed ‘method of fluxions’ which among other things, solved problems by finding "areas, tangents, the lengths of curves and the maxima and minima of functions" using derivatives. Eudoxus, Archimedes, Calvalieri, Fermat, and Barrow never explained limits precisely. Newton accomplished this phenomenon. However, he left the rigorous clarification to Cauchy.
Augustin Louis Cauchy, a 19th century French mathematician, made Newton’s notion of limits more explicit. He first defined the "continuity and derivative in terms of the limit." Second he defined limit as: "When the successive values attributed to a variable approach indefinitely a fixed value so as to end by differing from it by as little as one wishes, this last is called the limit of all others." Both Newton and Cauchy also made major contributions to physics.
Another great contributor was Gottfried Wilhelm von Leibniz, a 17th century mathematician from Germany. On November 21, 1675, Leibniz composed a manuscript with notation for integration and differentiation. He introduced the Power Rule in this manuscript. In 1676, Newton sent his results of differential calculus in a letter to Leibniz; however, Newton did not include his methods. The letter took a long time to get to Leibniz because it had to go through Oldenburg, another mathematician. Newton did not know that the letter took so long to get to him and thought that Leibniz was working on his reply for six weeks. In reality, Leibniz had not received the letter for six weeks. Upon receiving Newton’s letter, Leibniz quickly decided to publish his own methods. In the reply to Newton from Leibniz, Leibniz described his differential calculus. Newton then thought that Leibniz stole his methods and published them. Leibniz is credited with the methods of integration and differentiation independently of Newton.
A. Graphs: The word Tangent is derived from the Latin word tangens, which means "touching." So a tangent to a curve is a line that touches the curve. The slope of the tangent line at a point of a curve represents the slope of the curve at that point. Slope represents the rate of change so the slope at a particular point represents the rate of change at that point. We know that slope is commonly referred to as "rise over run". What this really means is ‘the change in the dependent variable over the change in the independent variable’ or how much the dependent variable changes per unit of change in the independent variable. If we have the graph of a function, and let’s say the function represents the position ‘s’ of an object (measured as positive or negative distance from a reference point) at time ‘t’, then any point on that graph of the function represents the position at a particular time. So the position (distance) is a function of time. Thus, time is the independent variable and position is the dependent variable.
So, what if we have a tangent line to a point on the curve of the function? What does this tangent represent? Again rise over run; change in dependent over change in independent; change in position (distance) over change in time. Wait a minute, doesn’t distance traveled divided by time elapsed equal the change in distance per unit of time. Yes it does, and another way to express this, is a concept called velocity (e.g., mph, m/sec, ft/sec, etc.). So here’s the idea: the slope of the tangent line at a given point of the position function is equal to the velocity at that particular time. In other words, tangent line at a given point represents the instantaneous velocity.
If we then constructed a new graph of the velocity function, based on the original function for position, and then took a tangent line at a given point, we would have the rate of change in velocity. How much the rate of change in position is changing per change in time or in other words how much the velocity changes per unit of time. So now we have change in position per change in time divided by the change in time. This tells us how fast the velocity increases or decreases. We refer to this rate of change in the velocity as "acceleration". So the slope of the tangent line at a given point of the graph of the velocity function is the instantaneous acceleration.
EXAMPLE #1
The graph of the position of a particle with respect to time is given.
Now if we examine the slope of the graph at each point along this line we will come up with the graph of the velocity of this particle with respect to time.
As you can see, the graph of the position has a slope of zero at the point (0,0). This is depicted on the graph above where the velocity is zero when the time is zero.
Below is the graph of the acceleration, which is a graph of all tangent lines to each point along the velocity graph.
Again, you can see that based on the graph of the velocity, which is a linear graph, that the tangent to that graph (the velocity graph) has no slope. Therefore, the graph of the acceleration is a horizontal line.
B. Limits: Earlier, when we were looking at rates of change, we examined them by graphing tangent lines at particular points. We also discussed that rates of change were measured in time intervals. But what if we wanted to look at the instantaneous rates of change without graphing them? If our position function is s(t) and we evaluate the position function on a time interval from time=t to time=t+h, then we can get the change in position from s(t+h)-s(t) (where s(t) is the position at time ‘t’ and s(t+h) is the position at time ‘t+h’.) So the average velocity over this time interval would be [s(t+h)-s(t)/h.]
Now suppose we compute the average velocities over shorter and shorter time intervals, [t,t+h]. So we are letting h approach zero. Now we can define instantaneous velocity at time=t to be the limit of these average velocities.
v(t)= 
The same logic can be applied to the acceleration.
a(t)=
EXAMPLE #2
Suppose a ball is dropped of the upper observation deck of the CN Tower in Toronto, 450 meters above the ground.
a) What is the velocity of the ball after 5 seconds?
b) How fast is the ball traveling when it hits the ground?
SOLUTION
In trying to solve this problem we use the fact, discovered by Galileo almost four centuries ago, that the distance fallen by any freely falling body is proportional to the square of the time it has been falling. If the distance fallen after t seconds is denoted by s(t) and measured in meters, then Galileo’s law is expressed by the equation
s(t)= 4.9t2
Use this fact to find the velocity v(t) after t seconds:
v(t) = 
= (4.9(t+h)2–4.9t2) / h
= [4.9(t2+2th+h2–t2)] / h = [4.9(2th+h2)] / h
= [4.9(2t+h)] / h = 9.8t
a) The velocity after 5 seconds is v(5)=(9.8)(5)= 49 m/s.
b) Since the observation deck is 450m above the ground, the ball will hit the ground at the time t1 when
s(t1)=450, that is
4.9t12 = 450
This gives t1 2 = 
The velocity of the ball as it hits the grounds is therefore
v (t1) = 9.8 t1
= 9.8 
. 94 m/s
EXAMPLE #3
If an arrow is shot upward on the moon with a velocity of 58 m/s, its height (in meters) after t seconds is given by s=58t-0.83t2.
a) Find the velocity of the arrow after 1 second.
b) Find the velocity of the arrow when t=t1.
c) When will the arrow hit the moon?
d) With what velocity will the arrow hit the moon?
SOLUTION
v(t) = 
[s(t+h)-s(t)] / h = [58(t+h)-0.83(t+h)2-(58t-0.83t2)] / h
= [58t+58h-0.83(t2+2th+h2)-58t+0.83t2] / h
= [58h-1.66th-0.83h2] / h = [58-1.66t-0.83h] = 58-1.66t
58-1.66(1)=56.34 m/s.
0 = t(58-0.83t)
t = 0 and 58-0.83t = 0; t = 
Therefore, t = 0 and t = 69.88 seconds. This implies that when we start the arrow is on the moon and then after 69.88 seconds, the arrow falls back to the moon after we shoot it upward.
d) The velocity the arrow is traveling when it hits the moon is v(t) = 58-1.66(58/.63) = -58 m/s.
C. Derivatives: Another way of expressing the limit in these cases dealing with rates of change is called the derivative. We define the derivative of f(a)=f(x) as:
s’(t)= 
We can now pull in many of the commonly used tools of Calculus we often employ to make solving derivatives easier:
The Power Rule
The Product Rule
The Quotient Rule
The Chain Rule
The Derivatives of Trig Functions
Implicit Differentiation
Now we define s(t). . . . the position equation where s is a function of time.
s’(t) = v(t). . . . . . the velocity function is the derivative of the position function.
s’’(t) = v’(t) = a(t). . . . . the velocity function is the derivative of the position function.
Example #4
The position of a particle is given by the equation
s(t)= f(t)= t3–6t2+9t
where t is measured in seconds and s in meters.
a) Find the velocity at time t.
b) What is the velocity after 2 seconds? After 4 seconds?
c) When is the particle at rest?
d) When is the particle moving in the positive direction?
e) Find the total distance traveled by the particle during the first five seconds.
SOLUTION
a) The velocity function is the derivative of the position function.
s(t)= f(t)= t3–6t2+9t
v(t)= s’(t)= 3t2–12t+9
b) The velocity after 2 seconds means the instantaneous velocity when t=2, that is v(2)= 3(2)2–12(2)+9=-3 m/s. The velocity after 4 seconds is v(4)=3(4)2–12(4)+9=9 m/s.
c) The particle is at rest when v(t) = 0, that is,
3t2–12t+9=3(t–1)(t–3)=0
and this is true when t=1 or t=3. Thus the particle is at rest after 1 second and after 3 seconds.
d) The particle moves in the positive direction when v(t)>0, that is, 3t2–12t+9=3(t–1)(t–3)>0
This inequality is true when both factors are positive or when both factors are negative. Thus the particle moves in the positive direction in the time intervals t<1 and t>3. It moves in the negative direction when 1<t<3.
e) The distance traveled in the first second is
½ f(1)–f(0)½ = ½ 4-0½ = 4 m
From t=1 to t=3 the distance traveled is
* f(3)–f(1)* = * 0-4* = 4 m
From t=3 to t=5 the distance traveled is
* f(5)-f(3)* = * 20-0* = 20 m
The total distance is 4 + 4 + 20 = 28 m.
Example #5
If a ball is thrown into the air with a velocity of 40 ft/s, its height (in feet) after t seconds is given by
y=40t-16t2. Find the velocity when t=2.
SOLUTION
The formula for the position of the ball is s(t)=40t-16t2. By taking the derivative, we get the velocity:
s’(t)=v(t)=40-32t. Since we know the formula for velocity, we can find the velocity at a certain time.
At t=2, v(t)=40-32(2)= -24 ft/s.
Limits:
The height s of a ball thrown straight up with an initial speed of 80 ft/sec from a rooftop 96 feet high is
s=s(t)= -16t2+80t+96 where t is the elapsed time that the ball is in the air. The ball misses the rooftop on its way down and eventually strikes the ground.
a. When does the ball strike the ground?
SOLUTION
0 = -16t2+80t+96
0 = -16(t2-5t-6)
0 = -16(t-6)(t+1)
The solutions to the equation is t=6 and t=-1. We disregard t=-1 because we cannot have negative time. Therefore, the ball will strike the ground at 6 seconds.
96 = -16t2+80t+96
0 = -16(t2-5t)
0 = t(t-5)
The solutions to the equation is t=0 and t=5. We will reject the solution t=0 because that is where the ball is before it is thrown. Therefore, the ball will pass the rooftop at 5 seconds.
v(t)=s’(t)= 
= [–16(t+h)2+80(t+h)+96 –(-16t2+80t+96)] / h
= [–16t2-32th-16h2+80t+80h+96+16t2-80t-96] / h
= –32th-16h2+80h = –32t-16h+80 = -32t+80
Therefore, v(t)=-32t+80 = –16(2t-5). So, at t=2, v(2)=-16(2(2)-5)=16 ft/sec.
Derivatives:
Two people are playing catch by throwing a football back and forth to each other. The position function the ball is traveling is s(t)=(-1/3)t2–t+6.
SOLUTION
The velocity function is s’(t)=v(t)=(-2/3)x-1. At t=2, v(2)=(-2/3)(2)-1 = -7/3 ft/sec. The velocity is negative because the ball is on its way down. It is moving in a negative direction. Therefore, we can simply say that the velocity of the ball is 7/3 ft/sec.
s(t)=0=(-1/3)t2-t+6
0=(-1/3)(t2+3t-18)
0=(-1/3)(t+6)(t-3)
When the ball is at t=-6 and t=3, the height of the ball is equal to 0. Since, time cannot equal a negative number; we can reject that answer. Therefore, at 3 seconds, the ball will strike the ground provided that there is no interference with the trajectory.
Graphs:
Below is a graph of the position of a car relative to time. Time is the independent variable and position (measured in meters) is the dependent variable. Based on the graph below, draw the graph of the velocity verses time.
Solution:
Below is the graph of the velocity of the car’s journey.
Exercise:
This is a hands-on activity for the students to be able to physically do an application. The students will roll a ball down an inclined plane to gather data concerning time versus distance. An inclined plane is used because free falling objects create too much error.
The students will be given the formula for a position of a ball after time t and will use derivatives to find the formulas for velocity and acceleration. The students will conduct an experiment where they roll a ball down an inclined plane to see if what they find corresponds to what the formulas give them.
Materials needed:
timer
ball
yard stick
something to prop the yard stick up
The yardstick needs to be propped up by an object like a book for example. The angle 2 will be defined as the angle of inclination. For best results, make the inclination small. Let the ball roll down the yard stick and time how long it takes to get to the bottom.
Position
1. The function for the position of the ball at time t is given by:
position = s(t)=gt2sin2 (where g is gravity, in inches g = 192) = 192t2sin2
So first, 2 needs to be found.
Find 2 using sin 2 = length of opposite/length of hypotenuse
2. After measuring the distance of the inclined plane, time how long (in seconds) it takes for the ball to roll down the inclined plane. Do this three times.
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3. Now use the formula for position to find if the experiment was accurate. Insert the distance for s(t) and solve for t.
Position = s(t) = 192t2sin2
Is your answer close to what you found in the experiment? If not, what could have happened? Some explanations could be friction and human error.
1. Velocity is equal to the derivative of the position function.
Position = s(t) = 192t2 sin2
Velocity = v(t) = s`(t) =
Find v(t).
2. Using the fact that velocity = v(t) = s`(t), calculate what the velocity of the ball was when it was rolled down the inclined plane.
3. Prove velocity can also be defined as v2 = 2 g sin2 d (where d is position).
Hint: v = g t sin2 d
(g t sin2 )2 = 2 g sin2 (1/2 g t2sin2 )
4. Is this constant or always increasing? Why?
1. Acceleration is equal to the derivative of the velocity function.
Position = s(t) = 192t2 sin2
Velocity = v(t) = s`(t)
Acceleration = a(t) = v`(t) = s``(t)
Find a(t).
2. Calculate what the acceleration of the ball was when it was rolled down the inclined plane.
3. What if the ball was rolled down a plane that was 120 inches long, what would the acceleration be?
4. Why is acceleration always constant?
What if 2 increased or decreased? How would this change velocity and acceleration?
If two objects of different weight were dropped off a building, which would hit the ground first?
Works Cited
Calinger, Ronald. "Mathematics Department Information Service", http://euler.ciens.ucv.ve/English/mathematics/leibniz, access date: 11-13-98.
Knight, Randall D. "Physics: A Contemporary Perspective", Vol. One, Addison-Wesley, Reading: 1997. Chapter 8: p.268-287.
O’Connor, John J. and Robertson, Edmund F. "The MacTutor History of Mathematics Archive", University of St. Andrews, Scotland: Jan 1997, http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Cauchy.html, access date: 11-13-98.
O’Connor, John J. and Robertson, Edmund F. "The MacTutor History of Mathematics Archive", University of St. Andrews, Scotland: Oct 1998, http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Leibniz.html, access date: 11-13-98.
O’Connor, John J. and Robertson, Edmund F. "The MacTutor History of Mathematics Archive", University of St. Andrews, Scotland: Feb 1997, http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Newton.html, access date: 11-13-98.
Schlissel, Arthur. "Mathematics Department Information Service", http://euler.ciens.ucv.ve/English/mathematics/cauchy.html, access date: 11-13-98.
Schuster, J.A. "Mathematics Department Information Service", http://euler.ciens.ucv.ve/English/mathematics/newton, access date: 11-13-98.
Stewart, James. "Calculus," 2nd ed. Belmont: Brooks/Cole Publishing, 1991. Section 1: p. 57-105.
Stewart, James. "Calculus," 3rd ed. Pacific Groove: Brooks/Cole Publishing, 1995. Section 1-2: p. 46-175.
Sullivan, Michael and Sullivan III, Michael. "Precalculus: Graphing and Data Analysis", Prentice Hall, Saddle River: 1998. Section 12.4: p.951-961.
Wachsmuth, Bert G. and Golba, Paul. "Interactive Real Analysis", An Interactive Textbook, Seton Hall University: 1996, http://pirate.shu.edu/~wachsmut/reals/history/cauchy.html, access date: 11-13-98.