Download Tangent Lines, Normal Lines, and Rectilinear Motion

Tangent Lines, Normal Lines,
and Rectilinear Motion
Conner Moon
Evan Haight
2nd period
Mrs. Autrey
Tangent Lines
• The definition of a tangent line is a line
which intersects a curve at a point where
the slope of both the curve and the line are
equal. In other words, it is a line that
touches a curve at one point without
crossing over.
Three Things Needed to Find the
equation of a tangent line
1. Derivative
2. Derivative at the point
3. Y-value at the point
Derivative
• The first step of finding the equation of a
tangent line is to take the derivative of the
original equation.
y  x  3x  2
2
y '  2x  3
Derivative at the point
• The second step is to find the derivative at
the point. For this example we will give
the point x=2.
y '  2x  3
y '  2(2)  3
Cont…
y' 1
After finding the derivative at the point, your
answer will be the slope of the tangent line.
Y-value at the point
• Now go back to the original equation and
plug in your x-value given earlier, x=2, to
find the y-value.
y  x  3x  2
2
y  (2)  3(2)  2
2
y 0
Now, Slope Intercept Formula
• After finding the slope of the tangent line,
and your y-value, we plug in the those
points to find the equation of the tangent
line, shown below.
y  y1  m( x  x1 )
y  0  1( x  2)
y  x2
Example 1
Here we will find the equation of the tangent line at
the point x=-2.
y  x  4x  5
2
Walk Through
y  x  4x  5
2
y '  2x  4
y '  2(2)  4
y '  8
y  (2)2  4(2)  5
y 7
y  y1  m( x  x1 )
y  7  8( x  2)
Solution
y  8 x  9
Try Me!!!!
• Find the equation of the line tangent at the point
x=-1
y  3  2x
How did you do?
y  2 x  6
Normal Line
• The definition of as normal line is a line
that is perpendicular to the tangent line at
the point of tangency. In other words it is a
line that intersects the tangent line with a
slope that is the negative reciprocal of the
slope of said tangent line.
Things Needed to Find Equation of
Normal Line
• Original Equation of the Tangent line
- Slope of Tangent line
• Negative Reciprocal
Negative Reciprocal
• Ok this sounds a lot harder than it actually
is. So if the slope of the tangent line is
y’=4, then the slope of the normal line, the
negative reciprocal of the slope of the
tangent line, is y’=-1/4.
Example 1
• Here we will find the equation of the normal line at the
point x=-2
y  x  4x  5
2
Walk Through
y  x  4x  5
2
y '  2x  4
y '  2(2)  4
y '  8
This is the part where we take
the negative reciprocal of the
tangent line, sooo, the slope
of this normal line is 1/8.
y  (2)2  4(2)  5
y 7
y  y1  m( x  x1 )
y  7  1 / 8( x  2)
Solution
1
29
y  x
8
4
Example 2
• Here we will find the equation of the
normal line at the point y=-1
Try Me!!!!
y  3  2x
How did you do?
1
9
y  x
2
2
Rectilinear Motion
• In this case when we talk about rectilinear
motion lets think of it as the motion of a
particle, and the motion of this particle is
illustrated by a given expression. Using
this expression and derivatives we will be
able to calculate the position, velocity, and
acceleration of said particle.
Stages of Rectilinear Motion
• X(t)=position
• V(t) / X’(t)=velocity
• A(t) / X’’(t)=acceleration
Explanation of a Sign Line
• Staying with the particle idea, a sign line
will visually show us the positive or
negative value of a group of numbers,
using this information we can determine if
the particle is moving left or right, has a
positive or negative velocity, or if the
acceleration of the particle is positive or
negative.
Example 1
• Find when the particle, expressed by the
equation x(t )  t 3  3t 2  3t  2 , is moving to
the right.
Walkthrough
x(t )  t 3  3t 2  3t  2
x '(t )  3t 2  6t  3
x '(t )  3(t  1)(t  1)
Solution
• As the sign line shows the particle is
moving right from  ,  .
Try Me!!!!
• Find when the particle, expressed by the
equation x(t )  (t  2)2 (t  6), is moving to
the left.
How did you do??
(2,5)
Now you know how to do
tangent and normal lines and
rectilinear motion.
The end.