Calculus, a technical project. And technicolor animation piece.

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Calculus, a technical project. And technicolor animation piece. By David M. Toner, with great respect

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By David M. Toner, with great respect. Calculus, a technical project. And technicolor animation piece. Power Rule- By the power rule, the derivative of x^n= nx^(n-1) 1. x^2= 2x 2.x^3= 3x^2. Exercise #1, Part 1. Quotient and Product rules- [(x^3-5)(x^2-1)]/(x^2+1) Becomes- - PowerPoint PPT Presentation

Transcript of Calculus, a technical project. And technicolor animation piece.

Page 1: Calculus, a technical project. And technicolor animation piece.

Calculus, a technical project. And technicolor animation piece.

By David M. Toner, with great respect

Page 2: Calculus, a technical project. And technicolor animation piece.

Exercise #1, Part 1

Power Rule-By the power rule, the derivative of x^n=

nx^(n-1) 1. x^2= 2x

2.x^3= 3x^2

Page 3: Calculus, a technical project. And technicolor animation piece.

Exercise #1, Part 2

Quotient and Product rules-[(x^3-5)(x^2-1)]/(x^2+1) Becomes-

(x^5-x^3-5x^2+5)/(x^2+1) Becomes-(5x^4-3x^2-10x)/(x^2+1) Becomes-

(x^2+1)(5x^4-3x^2-10x)-(2x)(x^5-x^3-5x^2+5)Final

Answer-[(3x^6+4x^4-3x^2-20x)/(x^2+1)^2]

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Exercise #2

The graph is not linear, and is infact quadratic. According to the 'rise over run'

approximation, using (2,4) and (3,9) the slope of this graph is 5/1, which is clearly not true.

The fault in this logic is that rise/run only works on linear graphs.

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Exercise 3

Our input is (2+h)^3-2^3/hThe limit from Exercise 1 for 2,

(3(2)^2)=12, which is what this equation implies, so the two are in perfect agreement.

Input -.01 -.1 .01 .1

Output 11.94 11.41 12.06 12.61

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Exercise 4, Part 1

To prove or disprove our third function, we've been given the following equation, [(x^3-5)

(x^2-1)]/x^2+1When x=2, y=1.8. When x=2.1, y=2.68

When placed into the equation to find the secant slope, it looks like this-

(2.68-2)/(2.1-2)=6.8

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Exercise 4 Part 2

We then place this into the equation...equation.

y=1.8+6.8(x-2)b,A(b)=2.1, 2.48

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Exercise 4 Part 3

According to the method I have been given, I should place SinX when x=0, as well as

Sinx=Pi/2, which make the following equations-Sin0, A(sin0)=0,5 and Sin(Pi/2),

A(Sin(Pi/2))=1,0

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Exercise 4, Part 4

The slope equation now flows as such, (0-5)/(1-0)=-5

This means the equation of the line is-5+-5(x-0)=5+-5x.

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Exercise 5

When give the equation, I find the derivative of the equation using the product rule-x(1-x) becomes x-x^2, which, using the power rule, becomes 1-2x. Using this, I can assume the

limit of this function will be .5. From here, I can plug in values near zero to

test the equation and my findings.

-.01 -.1 .01 .1

.51 .6 .49 .39

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Exercise 5, Part 2

Based on the information we found in the table and using the derivative rules(power

and product), the limit is .5, as I hypothesized earlier.