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Tangent Lines, Normal Lines, and Rectilinear Motion

Tangent Lines, Normal Lines, and Rectilinear Motion. David Smiley Dru Craft. Definition of Tangent Line:. The Linear function that best fits the graph of a function at the point of tangency. Definition of a Normal Line:. The negative reciprocal of a tangent line. Rectilinear Motion is.

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Tangent Lines, Normal Lines, and Rectilinear Motion

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  1. Tangent Lines, Normal Lines, and Rectilinear Motion David Smiley Dru Craft

  2. Definition of Tangent Line: • The Linear function that best fits the graph of a function at the point of tangency

  3. Definition of a Normal Line: • The negative reciprocal of a tangent line

  4. Rectilinear Motion is.. • The motion of a particle on a line

  5. Steps for solving a tangent line • Given the equation y = x² - 4x – 5 and the points (-2,7) • Find the equation of the tangent line.

  6. Step 1 • Given the equation y = x² - 4x – 5 and the points (-2,7) • Take a derivative • Y’ = 2x - 4

  7. Step 2 • Given the equation y = x² - 4x – 5 and the points (-2,7) • Take the derivative at a given point (put the x value into the derivative) • Y’ (-2) = 2(-2) – 4 = -8

  8. Step 3 • Given the equation y = x² - 4x – 5 and the points (-2,7) • Y value (plug the x value into the original problem to get y if the y value is not given) • Y(-2) = 7

  9. Example problem • Y= 2x – x³ and x= -2 • Find the derivitive at the given point and the y value.

  10. Solutions • Y’ = 2 – 3x² • Y’(-2) = -10 • Y(-2) = 4

  11. Take the same problem y = 2x-x^3 and put into the point slope formulaY’ = 2 – 3x²Y’(-2) = -10Y(-2) = 4

  12. Point Slope Formula • y-y1 = m(x-x1)

  13. Answer for y = 2x-x³ in point slope formula.. Y’ = 2 – 3x² Y’(-2) = -10Y(-2) = 4 • Answer: y+4 = -10(x+2)

  14. Take the same problem again y = 2x-x³ and continue to put into the slope intercept form. • y+4 = -10(x+2) into slope intercept… • Y= -10x - 24

  15. Try Me • Find the equation of the tangent line and put into slope intercept and point slope form. • Y=4x^3 - 3x – 1 at the point x=2

  16. Answers • Y’=12x² – 3 • Y’(2) = 45 • Y(2) = 25 • Point slope: y-25 = 45(x-2) • Slope Intercept: y = 45x-65

  17. Try Me • Find the equation of the tangent line and put in point slope and slope intercept form. • y = x³ – 3x at the point x=3

  18. Answers • Y’ = 3x² – 3 • Y’(3) = 24 • Y(3) = 18 • Point slope: y-18 = 24(x-3) • Slope intercept: y= 24x-54

  19. Normal lines • Negative reciprocal of tangent line • Tangent line y-4=-10(x+2) • Normal line of this would be.. • Y-4= 1/10(x+2)

  20. Try me Find the equation of the normal line given Y = 5-x at the point x = -3

  21. Answers Y = 5-x • Y= (5-x)^1/3 • Y’=1/3(5-x)^ -2/3 (-1) • Y’ (-3) = -1/12 • Y (-3) = 2

  22. Normal line answer • Y-2 = 12(x+3)

  23. Try me • Find the equation of the tangent line and the equation of the normal line and put both into slope intercept form • Y = X at the point x=8

  24. Y = X at the point x=8 • Y’ = 1/3(x)^ -2/3 • Y’ (8) = 1/12 • Y(8) = 2 • Tangent line: y = 1/12x – 4/3 • Normal line: y= -12x + 98

  25. Rectilinear Motion • Position: x(t) or s(t) • Velocity: x’ (t) or v(t) • acceleration: v’ (t) or a(t)

  26. Steps for solving a rectilinear motion problem • 1.) take a derivative • 2.) clean up the equation(must be a GCF) • 3.) draw a sign line

  27. Rectilinear motion example • X(t) = x³ - 2x² + x • X’(t) = 3x² – 4x + 1 • (3x-1) (x-1) = 0 • x=1/3 x=1

  28. Solution for example • Sign line x-3 - - - - - - .----------------------------------------------- x-1 - - - - - - - - -- - - .------------------------ _______________________ 0 + 1/3 - 1 + v(t)

  29. Try Me • V(t) = -1/3x³ – 3x² + 5x

  30. Solutions • V’(t) = x² – 6x + 5 • (x-1) (x-5) = 0 • X=1 x=5

  31. Sign line x-1 - - - - - - .----------------------------------------------- x-5 - - - - - - - - -- - - .------------------------ _______________________ 0 + 1 - 5 + a(t)

  32. Bibliography • http://www.classiccat.net/

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