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# AP Calculus: A Guide to Survive - PowerPoint PPT Presentation

AP Calculus: A Guide to Survive. By: Florin Gjergjaj and Neomis Rodriguez. Table of Contents. The Masterminded Authors Chapter 1: Limits and Continuity Chapter 2: Derivatives Chapter 3: Antiderivatives Chapter 4: Application Problem. The Masterminded Authors.

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### AP Calculus: A Guide to Survive

By: Florin Gjergjaj and Neomis Rodriguez

• The Masterminded Authors

• Chapter 1: Limits and Continuity

• Chapter 2: Derivatives

• Chapter 3: Antiderivatives

• Chapter 4: Application Problem

Florin is known for having a difficult last name. He is a student at the High School for Environmental Studies. He defeated the formidable course known as AP Calculus AB, and he mastered all of the concepts. His academic goal is to conquer the BC course and master the concepts just as well. His long term goal is to go to NYU and achieve a medical career.

Neomis is known for having a difficult first name (if you think it’s easy, you’re probably pronouncing it wrong). She is currently a student at HSES with several AP classes. Her academic goals include mastering the Spanish language and pursuing a scientific career.

A limit is the calculation of f(x) as x approaches a. The three criteria for a limit to exist are:

1. Lim f(x) exists (right)

xa+

2. Lim f(x) exists (left)

xa-

3. Lim f(x) = Lim f(x) = L

xa+ xa-

Limits can be found three ways: Numerically, Algebraically and Graphically

Algebraically has three steps:

• Factor

• Simplify

• Substitute

Graphically requires tracing from the right and the left. The f(x) must be the same number for both for the limit to exist.

Numerically requires a table with x and f(x) values. It shows f(x) as x approaches a, from both the right and left.

AlgebraicallyExample: Lim x – 2 x2 x² – x – 2 (x- 2) (factor the denominator out and cross out like terms)(x-2) (x+1) 1 (Simplify completely and then substitute 2 for x) (x+1) 1 = 1/ 3 ((2) + 1)Numerically requires plugging in x values that approach a to find f(x). Graphically requires tracing the graph from the left and right as x approaches a (requires the same value in order for the limit to exist)

Continuity (at a point and an open interval) requires no holes (limits), gaps, jumps and must be unbroken.

The criteria for Continuity:

• 1. F(c) is defined

• 2. Lim f(x) exists

xc

• 3. Lim f(x) = f(c)

xc

• If it is not unbroken, it is discontinuous. There are two types of discontinuity:

• 1. Removable (with a limit)

• 2. Non-removable (without a limit)

• Continuous Discontinuous

A derivative is the slope of a function at a point. Some derivative notations include: f ‘(x), dy/dx,  y’,  d/dx f(x) or dx [y].

The definition of tangent line with slope m is calculated using the formula

Lim f(c + ∆ x) – f(c) = m

∆ x0 ∆ x

The definition of a derivative using limit process is calculated using the formula

F ‘(x) = Lim f (x + ∆ x) – f(x)

∆ x  0   ∆ x

The steps to solve are:

1. Substitute (x + ∆ x) for each x variable

2. Cross out opposite signs

3. Substitute for ∆ x

• Example:

f(x) = x2 + 1 (0,1)

Lim (x + ∆x)2 + 1- (x2 + 1)

∆x 0 ∆x

Lim x2 + 2∆x2 + ∆x2+ 1– x2– 1

∆x  0 ∆ x

Lim 2∆x2 + ∆x2

∆x  0∆ x

Lim ∆x (2x + ∆x)

∆x 0 ∆ x

Lim (2x + ∆x)

∆x  0

(2x + (0)) = 2x

1. Constant Rule d/dx [c] = 0

*The derivative of variables such as x and y is 1.*

2. Power Rule d/dx [xn] = nxn-1

This means that you move the exponent to the front and subtract 1.

Ex. y = x3 is 3x2

3. Sine and Cosine

d/dx [sinx] = cosx

d/dx [cosx] = -sinx

d/dx [tanx] = sec2x

d/dx [secx] = secxtanx

d/dx [cscx] = - cscxcot

d/dx [cotx] = - csc2x

Anything with a c is negative. *

4. Product Rule d/dx [ f(x) g(x)] = f ‘(x) g(x) + f(x) g’(x)

Ex. (3x +1) * (5 + 4x)* f ‘(x) = 3 and g‘(x) = 4*

f(x) g(x)

3 * (5 + 4x) + (3x +1) * 4

f ‘(x) g(x) + f(x) * g ‘(x)

15 + 12x +12x + 4 = 24x + 19

5. Quotient Rule d/dx [ f(x)/ g(x)] = g (x) f ‘(x) – f(x) g’(x) [g(x)]2

* Formemorization: lo di hi – hi di lo over lo lo*

Ex. 5x – 2* f ‘(x) = 5 and g ‘(x) = 2x

x2 + 1

[x2 + 1 * 5] – [5x – 2 * 2x] = [5x2 + 5] – [10x2 – 4x]

(x2 + 1)2 (x2 + 1)2

= - 5x2– 4x + 5

(x2 + 1)2

6. Chain Rule d/dx = f ‘ [g(x)] * g’ (x)

This means that you multiply the original function by the derivative of the inside function.

Ex. (2x2 + 5)7 (Power rule: move 7 to front and subtract 1.)

7(2x2 + 5)6 * 4x (Derivative of the inside is multiplied, inside left intact)

28x(2x2 + 5)6

When you implicitly differentiate yvariables require dy/dx.

Ex. d/dx [y3] = 3y2dy/dx

Guidelines

1. Differentiate both sides with respect to x

2. Collect all terms involving dy/dx

3. Factor dy/dx to the left side

4. Solve for dy/dx

Ex.    y3+ y2– 5y – x2= - 4

3y2dy/dx + 2y dy/dx – 5 dy/dx – 2x = 0

dy/dx (3y2+ 2y – 5) = 2x

dy/dx = 2x

3y2 + 2y – 5

Rate = distance  Average Velocity = ∆d

time             ∆t

S (t) = Position Function

S ‘(t) = Velocity (Find the derivative)

V ‘(t) = Acceleration (Find the second derivative)

A(t) = S ‘’(t)

Ex.  S (t) = 3x2+ 3

V(t) = 6x

A(t) = 6

Antiderivatives vs. DerivativesDefinite Integrals vs. Indefinite

Antiderivative: a function F is an antiderivative of f on an interval I if F’(x) = f(x) for all x in I.

An antiderivative is truly the inverse operation of the derivative.

A definite integral expresses the difference between the upper and lower limits of the function. Example:

An indefinite integral does not include the limits of the function. Example:

Accumulation functions shows how you can figure out the velocity and position of a function when you’re given the acceleration. You can find the velocity by finding the antiderivative of the acceleration. Then, to find the position of the object you have to find the antiderivative of the velocity.

Ex. a(t) = 6 sec/min2, x(0) = 2 and v(0) = 5

v(t) =

v(t) = 6t + c (plug in initial condition to solve for c)

s(t) =

s(t) = 6t + 5t + c (plug in initial condition again)

s(t) = 6t+5t+2

How Calculus Relates to Physics

Calculus can be applied to Physics in terms of velocity, acceleration and position.

Ex. A model rocket is fired vertically upward from rest. Its acceleration for the first three seconds is a(t) = 60t at which time the fuel is exhausted and it becomes a freely “falling” body. Fourteen seconds later, the rocket’s parachute opens, and the (downward) velocity slows linearly to -18ft/second in 5 seconds. The rocket then “floats” to the ground at that rate.

At what time does the rocket reach its maximum height?

The rocket accelerates for the first three seconds. Hence, the rocket reaches its maximum height at t = 3 seconds.

What is the height?

A(t) = 60t therefore to find the velocity you find the antiderivative of a(t). V(t) = 30t2 +c. To find the height you must find the antiderivative of the velocity D(t) = 10t2 +C. Using three seconds as your maximum height, D(3) = 270 meters.