Kinematics with Calculus: dervitives

AP Physics C

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” wayFor 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.

The “calculus” wayFor the curving incline, on the other

hand, things are constantly changing. The steepness of the incline is changing — and not just in increments like it’s one steepness for the first 10 feet then a different steepness for the next 10 feet — it’s constantly changing. And the man pushes with a constantly changing force — the steeper the incline, the harder the push. As a result, the amount of energy expended is also changing, not every second or every thousandth of a second, but constantly changing from one moment to the next. That’s what makes it a calculus problem.

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.

“Regular” math vs. “Calculus”

“Regular” math vs. “Calculus”

“Regular” math vs. “Calculus”

Learn the lingo!Calculus is about “rates of change”.

A RATE is anything divided by time.

CHANGE is expressed by using the Greek letter, Delta, D.

For example: Average SPEED is simply the “RATE at which DISTANCE changes”.

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 + Dt

x(t +Dt)

A

B

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

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

The secant line and the slope

t

x(t)

t + Dt

x(t +Dt)

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” lineREAD THIS CAREFULLY!

If we hold POINT A fixed while allowing Dt 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 + Dt

x(t +Dt)

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 derivativeMathematically, we just found the slope!

line tangent of slope)()(

lim

linesecant of slope)()(

0

12

12

t

txttx

t

txttx

xx

yyslope

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.

The derivativeIn the beginning of math 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.

Derivative exampleConsider the function x(t) = 3t +2What 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

We didn't even need to INVOKE the limit because the delta t's cancel out.

Regardless, we see that we get a constant.

ExampleConsider the function x(t) = kt3, where k =

proportionality constant equal to one in this case..

What happened to all the delta t's ? They went to ZERO when we invoked the limit!

What does this all mean?

The MEANING?

For example, if t = 2 seconds, using x(t) = kt3=(1)(2)3= 8 meters.

The derivative, however, tell us how our DISPLACEMENT (x) changes as a function of TIME (t). 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. Since dx/dt = 3kt2=3(1)(2)2= 12 m/s

23

3)(

ktdt

ktd

THERE IS A PATTERN HERE!!!! Now if I had done the previous example with kt2, I would have gotten 2t1

• Now if I had done the above example with kt4, I would have gotten 4t3

• Now if I had done the above example with kt5, I would have gotten 5t4

So the fast an easy way to find a derivative is……Practice : Find the

displacement an speed at 3 seconds for each motion equation:A> x(t) = 5t + 3 x(3) = 18 meters dx/dt = 5 meters/sec = velocity

B> x(t) = 5t2 x(3) = 45 meters (freefall) dx/dt = 10 t = instantaneous velocity = v(t) = 10 (3) = 30 m/s = v(3)

C> x(t) = 5t3

x(3) =5 (t)3 = 125

v= dx/dt = 15x2 = 15(3)2

=135

This shortcut for finding derivatives is called The Power Rule

. )(then

constant, a is where)( If

1

n

n

nCtdt

dxtv

CCttx

Calculate the instantaneous velocity as a function of time and at the specific time indicated for each of the following:

Practice Problems

s 4 3)(

s 3 5)(

s 3 4)(

s 2 3)(

2

4

tttx

ttx

tttx

tttx

To find the derivative of a polynomial, we just take the derivative of each term.

Derivative of a Polynomial

dt

dB

dt

dA

dt

dx

tBAx

of functions are B andA where

Find the derivatives of the following polynomials:

Derivative of a Polynomial

constants. are and , where)(

s 2 6523)(

002

21

00

23

avxattvxtx

tttttx

Find the instantaneous velocity function for each equation, then evaluate it at t = 2 sec.

Exercise 1: From Summer Assignment

23

3

4

23 .3

2 .2

5 .1

ttx

tx

tx

32

1

32 .5

1 Use:Hint

4 .4

ttx

ttt

x

Average acceleration is the rate of change of velocity with respect to time.

Just as with velocity, the average acceleration depends on the time interval chosen to calculate it.

Average Acceleration

t

vaAV

The instantaneous acceleration at time t is the average acceleration calculated over a very small interval, where our two endpoints are very close to time t, so that ∆t is very close to 0 .

This quantity is called the derivative of velocity with respect to time.

Instantaneous acceleration

t

va

t

0

limdt

dva

Instantaneous acceleration:Example 1

tdt

dxvtx 3 then 5.1 If 2

333*1 011 ttdt

dva

Since a is the derivative of a derivative, it is sometimes called a second derivative.

The “2” exponents do not refer to squares. They just mean to take the derivative twice.

Second derivative

dtdtdx

d

dt

dva

2

2

dt

xda

Sample problem 1: A particle travels from A to B following the function x(t) = 3.0 – 6t + 3t2.

a) What are the functions for velocity and acceleration as a function of time?

b) What is the instantaneous velocity at 6 seconds?

c) What is the initial velocity?

Calculate the instantaneous velocity as a function of time and at the specific time indicated for each of the following:

Practice Problems

s 4 3)(

s 3 5)(

s 3 4)(

s 2 3)(

2

4

tttx

ttx

tttx

tttx

Shorthand: dots for time derivatives

Position is xVelocity is ẋAcceleration is Ẍ

Instantaneous Acceleration

?)(,/]ˆ4ˆ21[)(2

2 tasmjt

ittv

Find the acceleration equation if the velocity of an object is defined as:

dt

jt

itd

dt

dva

)ˆ4ˆ21( 2

2

ssmjt

it //]ˆ8ˆ42[3

a = dv/dt = d2x/dt2

v x

Find the instantaneous acceleration at 2.45 seconds

Other common calculus derivative rules on reference tables that we will use:

Now lets do some AP kinematics and force questions that require

derivatives

Calculus Kinematics Practice

A force problem related to acceleration: