1 / 11

Calculus, a technical project. And technicolor animation piece.

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-

charity-herrera
Télécharger la présentation

Calculus, a technical project. And technicolor animation piece.

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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

More Related