Tangent Lines, Normal Lines, and Rectilinear Motion

Tangent Lines, Normal Lines, and Rectilinear Motion

David SmileyDru 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..

• The motion of a particle on a line

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.

Step 1


• Take a derivative• Y’ = 2x - 4

Step 2


• Take the derivative at a given point (put the x value into the derivative)

• Y’ (-2) = 2(-2) – 4 = -8

Step 3


• Y value (plug the x value into the original problem to get y if the y value is not given)

• Y(-2) = 7

Example problem

• Y= 2x – x³ and x= -2

• Find the derivitive at the given point and the y value.

Solutions

• Y’ = 2 – 3x²• Y’(-2) = -10

• Y(-2) = 4

Take the same problem y = 2x-x^3 and put into the point slope formula

Y’ = 2 – 3x²Y’(-2) = -10

Y(-2) = 4

Point Slope Formula

• y-y1 = m(x-x1)

Answer for y = 2x-x³ in point slope formula..

Y’ = 2 – 3x² Y’(-2) = -10

Y(-2) = 4

• Answer: y+4 = -10(x+2)

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

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

Answers

• Y’=12x² – 3• Y’(2) = 45• Y(2) = 25

• Point slope: y-25 = 45(x-2)• Slope Intercept: y = 45x-65

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

Answers

• Y’ = 3x² – 3• Y’(3) = 24• Y(3) = 18

• Point slope: y-18 = 24(x-3)• Slope intercept: y= 24x-54

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)

Try me

Find the equation of the normal line given Y = 5-x at the point x = -3

Answers

Y = 5-x

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

Normal line answer

• Y-2 = 12(x+3)

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

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

Rectilinear Motion

• Position: x(t) or s(t)

• Velocity: x’ (t) or v(t)

• acceleration: v’ (t) or a(t)

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

Rectilinear motion example

• X(t) = x³ - 2x² + x

• X’(t) = 3x² – 4x + 1

• (3x-1) (x-1) = 0

• x=1/3 x=1

Solution for example• Sign line

x-3 - - - - - - .-----------------------------------------------

x-1 - - - - - - - - - - - - .------------------------

_______________________ 0 + 1/3 - 1 + v(t)

Try Me

• V(t) = -1/3x³ – 3x² + 5x

Solutions

• V’(t) = x² – 6x + 5

• (x-1) (x-5) = 0

• X=1 x=5

Sign line

x-1 - - - - - - .-----------------------------------------------

x-5 - - - - - - - - - - - - .------------------------

_______________________ 0 + 1 - 5 + a(t)

Bibliography

• http://www.classiccat.net/

Tangent Lines, Normal Lines, and Rectilinear Motion

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