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

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Presentation on theme: "Tangent Lines, Normal Lines, and Rectilinear Motion David Smiley Dru Craft."— Presentation transcript:

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 formula Y’ = 2 – 3x² Y’(-2) = -10 Y(-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) = -10 Y(-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’ = 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 x _______________________ 0 + 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 x _______________________ a(t)

32 Bibliography


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