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by david m toner with great respect n.
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Calculus, a technical project. And technicolor animation piece. PowerPoint Presentation
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Calculus, a technical project. And technicolor animation piece.

Calculus, a technical project. And technicolor animation piece.

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Calculus, a technical project. And technicolor animation piece.

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

  2. 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

  3. 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] Exercise #1, Part 2

  4. 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. Exercise #2

  5. Our input is (2+h)^3-2^3/h The limit from Exercise 1 for 2, (3(2)^2)=12, which is what this equation implies, so the two are in perfect agreement. Exercise 3

  6. To prove or disprove our third function, we've been given the following equation, [(x^3-5)(x^2-1)]/x^2+1 When 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 Exercise 4, Part 1

  7. We then place this into the equation...equation. y=1.8+6.8(x-2) b,A(b)=2.1, 2.48 Exercise 4 Part 2

  8. 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 Exercise 4 Part 3

  9. 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. Exercise 4, Part 4

  10. 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. Exercise 5

  11. 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. Exercise 5, Part 2