The Basics of Physics with Calculus & Systems of Equations

Mathy AP Physics C

Pythagoras started it all…6th Century Pythagoras first got interested in music when

he was walking past a forge and heard that the sounds of the blacksmiths' hammers sounded good together. He talked to the blacksmiths and found out that this was because the anvils they were using were scaled down copies of each other: one full size, one half size, and one two thirds size. He got an idea that maybe a simple fraction relationship in the size of two instruments is what makes them sound good together.

This idea led to the belief that mathematics was tied to the physical world.

Galileo Galilei Galileo too grew up around music as

his father was a musician. His father, however, felt that the Pythagorean Harmonies were too simple for the now emerging Italian Renaissance.

Galileo, like his father, was dealing with

this same idea as well. His studies with kinematics (motion) were too abstract and too simple to explain complicated motion.

Differential Calculus – More sophisticated! 25 years later Isaac Newton and Gottfried Leibniz

developed a sophisticated language of numbers and symbols called Calculus based on work. Newton began his work first but it was Leibniz who first published his findings. Both led the other towards accusations of plagiarism.

What is calculus? Calculus is simply very advanced algebra and

geometry that has been tweaked to solve more sophisticated problems.

Question: How much energy does the man use to push the crate up the incline?

The “regular” way For the straight incline, the man

pushes with an unchanging force, and the crate goes up the incline at an unchanging speed. With some simple physics formulas and regular math (including algebra and trig), you can compute how many calories of energy are required to push the crate up the incline.

What is calculus? It is a mathematical way to express something that is ……CHANGING! It could be anything??

But here is the cool part:

Calculus allows you to ZOOM in on a small part of the problem and apply the “regular” math tools.

We already know How Graphs show change

Calculus is about “rates of change”. n  A RATE is anything divided by time. n  CHANGE is expressed by using the Greek letter, Delta, Δ.

The Derivative…aka….The SLOPE! Since we are dealing with quantities that are changing it may be

useful to define WHAT that change actually represents.

Suppose an eccentric pet ant is constrained to move in one dimension. The graph of his displacement as a function of time is shown below.

t

x(t)

t + Δt

x(t +Δt)

A

B

At time t, the ant is located at Point A. While there, its position coordinate is x(t). At time (t+ Δt), the ant is located at Point B. While there, its position coordinate is x(t + Δt)

The secant line and the slope

t

x(t)

t + Δt

x(t +Δt)

A

B

Suppose a secant line is drawn between points A and B. Note: The slope of the secant line is equal to the rise over the run.

The “Tangent” line READ THIS CAREFULLY!

If we hold POINT A fixed while allowing Δt to become very small. Point B approaches Point A and the secant approaches the TANGENT to the curve at POINT A.

t

x(t)

t + Δt

x(t +Δt)

A

B

We are basically ZOOMING in at point A where upon inspection the line “APPEARS” straight. Thus the secant line becomes a TANGENT LINE.

A

A

The derivative Mathematically, we just found the slope!

line tangent of slope)()(lim

linesecant of slope)()(

0

12

12

=Δ

−Δ+

=Δ

−Δ+=

−

−=

→Δ ttxttx

ttxttx

xxyyslope

t

Lim stand for "LIMIT" and it shows the delta t approaches zero. As this happens the top numerator approaches a finite #.

This is what a derivative is. A derivative yields a NEW function that defines the rate of change of the original function with respect to one of its variables. In the above example we see, the rate of change of "x" with respect to time.

Fundamental Theorem : The Derivative In most Physics books, the derivative is written

like this:

Mathematicians treat dx/dt as a SINGLE SYMBOL which means find the derivative. It is simply a mathematical operation.

The bottom line: The derivative is the slope of the line tangent to a point on a curve.

Easier Theorem : The Derivative Since the derivative is the slope of the line

tangent to a point on a curve…… velocity is the derivative of a position function x(t) in the first order acceleration is the second order derivative of a position function x(t) The relationship described here can be summarized in the following diagram

position velocity acceleration

€

v =ΔxΔt#

$ %

&

' ( small

= limΔt→0ΔxΔt

=dxdt

€

a =ΔvΔt#

$ %

&

' ( small

= limΔt→0ΔvΔt

=dvdt

=d2xdt 2

differentiate differentiate

integrate integrate

€

a(t)

€

Δv(t)

€

Δx(t)

Differentiation has specific Rules ! Consider the function x(t) = 3t +2 What is the time rate of change of the function? That is, what is the NEW

FUNCTION that defines how x(t) changes as t changes. This is actually very easy! The entire equation is linear and looks like y =

mx + b . Thus we know from the beginning that the slope (the derivative) of this is equal to 3.

Nevertheless: We will follow through by using the definition of the derivative…The fundamental Theorem of Calculus!!!

We didn't even need to INVOKE the limit because the delta t's cancel out. Regardless, we see that we get a constant.

Let’s cheat…..Forgive us Mrs. H ! But we can bypass the F.T.C…..

“Secret Weapon”

Advanced Position Functions.

This function gives us our DISPLACEMENT (x) changing as a function of

TIME (t). However ,the rate at which displacement changes is also called VELOCITY, thus if we use our derivative we can find out how fast the object is traveling at t = 2 second. �

�We will use a special rule that will differentiate like this: �

Core Problem 1.0 A particle is has a position function as described by x(t) = 6t2 +2t - 4 . Find its velocity at t =2 s and its acceleration !

€

x(t) = 2t 3 + t 2 − 4

v(t) =dxdt

= 6t 2 + 2t − 4

v(2) = 28 m /s

How did you do that math trick ?

Evaluating Derivatives: Power Rule

€

If x(t) = atn

Then v(t) =dxdt

= natn−1

My high school Calculus teacher would say “Can you differentiate a nat……..anyone ?”

Look back at how power rule worked on the last problem

What happened to the “4”??? Well, since it doesn’t change the actual slope…we treat it like “b” in y=mx+b and we drop it out. Sometimes it is called C in calculus (constant)

€

x(t) = 2t 3 + t 2 − 4

v(t) =dxdt

= (3)(2)t 2 + (2)t1

v(2) = 28 m /s

Practice

€

1. If x(t) = 20t 3,thendxdt

=

2. If x(t) = 5t 3 + 6t + 7,thendxdt

=

3. If x(t) = 2t 6 + 7t 4 + 4t + 2,thendxdt

=

What if the exponents are negative or aren’t whole integers ? How do we subtract 1 then ?

We use more little math rules concerning exponents!

€

b−p =1bp

bpr = bpr

€

t12 = t2

t−1 =1t

t−2 =1t 2

t−12 =

1

t12

t−32 =

1

t32

€

v(t) =

€

v(t) =

€

v(t) =

Evaluating Derivatives Again.

The rate at which velocity changes is also called Acceleration, thus if

we use our velocity derivative we can find out how fast the object’s rate of change is changing at t = 2 seconds. �

�"" "We will now diferentiate the velocity function once: �

A particle is has a position function as described by x(t) = 6t2 +2t -4 . Find its acceleration at 2 secs.

Oh….that’s the power rule again!

€

v(t) = 6t 2 + 2t

a(t) =d2xdt 2

=12t + 2

a(2) = 26m /s2

Practice What if the exponents are negative or aren’t whole integers ? How do we subtract 1 then ?

We use more little math rules concerning exponents!

€

b−p =1bp

bpr = bpr

€

t12 = t2

t−1 =1t

t−2 =1t 2

t−12 =

1

t12

t−32 =

1

t32

€

1. If v(t) = 60t 2,thendvdt

= a(t) =

2. If v(t) =15t 2 + 6 ,thendvdt

= a(t) =

3. If v(t) =12t 5 + 28t 3 + 4,thendvdt

= a(t) =

Putting it together. Core Problem 2.0 A particle is has a position function as described by x(t) = 4t2 -6t -40 . Find : a)  Its velocity at any time:

b)  Its acceleration at any time :

c)  The time at which the object is at the origin:

d)  When does the object change direction ?

Common calculus derivative rules