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..
• 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
• Given the equation y = x² - 4x – 5 and the
points (-2,7)
• Take a derivative
• Y’ = 2x - 4
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
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
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 – 3x²
Y’(-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=
•
•
•
•
Y= (5-x)^1/3
Y’=1/3(5-x)^ -2/3 (-1)
Y’ (-3) = -1/12
Y (-3) = 2
5-x
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/