Download Rectilinear Motion and Tangent Lines

Rectilinear Motion
and Tangent
Lines
Katie Faith
Laura Woodlee
Let’s
Review
some
Calculus!
ARF
What is Rectilinear Motion?
• Definition: It is any motion along a straight line.
• Application to Calculus:
– In the AP Calculus AB course, this method is used
to solve problems dealing with particle motion.
Three equations are found and utilized to find the
position, velocity, and acceleration of the particle
at different points of time.
Rectilinear Motion
(Continued)
• The position of the particle is found using the
position function, which is the original
function. This is usually written as x(t) or p(t)
.It describes the particle’s location at a
specific point in relation to time (t). If given
velocity, this function is found by taking the
integral of the velocity function and then
plugging in a known point to find the
constant (C).
Rectilinear Motion
(Continued)
• The velocity of the particle is found by taking
the derivative of the original function
(position function). It can also be found by
taking the integral of the acceleration
function and then plugging in a known point
to find the value of the constant (C). The
function is written as v(t) . The velocity
describes the movement of the particle,
either left, right, or stopped. This direction is
determined using a sign line.
STOP!!! SIGN LINE REVIEW!!!
•
In this application of calculus, sign lines are used to show when the velocity of the particle
is moving to the right (positively), moving to the left (negatively), or changing direction (the
point when it switches from positive to negative).
• To make a sign line, first the zeroes of the function must be found. For example:
v(t)= t²-3t + 2
v(t)= (t-1)(t-2)
Dots represent the
• Then the sign line is set up:
zeroes (this is
where it is at rest)
+
+
The zeroes are
Things to
written on the
(t-1)
remember:
Side
(t-2)
Negative (dotted)
V(t)
1
2
a time line with the numbers is written on the bottom
• This sign line shows that the function is increasing from negative infinity
to 1, resting at 1, decreasing from 1 to 2, resting at 2, and increasing from
2 to infinity.
lines are always to
the left of the zero
(point) unless there
is a negative in
front of t
Positive (solid) lines
are always to the
right unless there is
a negative in front
of t
Rectilinear Motion
(Continued)
• The acceleration of the particle is found by
taking the second derivative of the position
function, or the first derivative of the velocity
function. It is written as a(t). The acceleration
of the particle describes the absolute value of
distance that the particle travels during a
certain interval of time. The zeroes of this
function describe when the particle is at rest.
Example Problem 1
1) 1983 AB 2: A particle moves along the x-axis
so that at time t its position is given by:
x(t)= t³ - 6t²+ 9t + 11
a) What is the velocity of the particle at t=0?
b) During what time intervals is the particle
moving to the left?
c) What is the total distance traveled by the
particle from t=0 to t=5?
Step-by-Step Solution 1
a)To solve, you must first find the formula for velocity, which is done by taking the
derivative of the position formula. Using chain rule, this would be:
v(t)= 3t² - 12t + 9
To find the velocity when t=0, plug in 0 for t:
v(0)= 3(0)²- 12 (0) + 9
v(0)=9
b)To find the time intervals that the particle is moving to the left, you must make a
sign line of the velocity function. First find the zeroes of the velocity function:
v(t)= 3t² - 12t +9
v(t)= 3(t-3)(t-1)
Then make the sign line: +
+
3
(t-3)
(t-1)
V(t)
1
3
The sign line shows the particle is moving to the left from 1 to 3.
Step-by-Step Solution 1
c)To find the total distance traveled, you will be looking at the position
that the particle is at different times. The sign line, in the last part of
the problem, showed that the particle was at rest at t=1, and changed
direction at t=3. These are two important points that the position must
be found at, along with the endpoints, which are t=0 and t=5.
First we must find the position of the particle at these critical
points using the position formula: p(t)= t³ - 6t² + 9t + 11
Then find the total distance traveled at each interval and add to find
total distance:
p(0)=11
4
p(1)=15 +
58
p(3)=-43 + 186
p(5)=143
698 units total
Example Problem 2
2) 1982 AB 1: A particle moves along the x-axis in such
a way that its acceleration at time t for t > 0 is
given by:
a(t)= 3/t²
When t=1, the position of the particle is 6 and the
velocity is 2.
a) Write an equation for the velocity, v(t), of the
particle for all t>0.
b) Write an equation for the position, x(t), of the
particle for all t>0.
c) Find the position of the particle when t=e.
Step-by-Step Solution 2
a)
To find the equation for velocity, you must first
take the integral of the acceleration equation and
then plug in the known point v(1) = 2 to find the
value for C:
a(t)= 3
Final Velocity Equation:
t²
v(t)= -6 + C
v(t)= -6 + 8
t
t
2= -6 + C
C= 8
1
Step-by-Step Solution 2
b) To find the equation for position, you must take the integral of
the velocity equation and then plug in the known point x(1)=6
to find the value for C:
v(t) = -6 + 8
t
x(t) = -6ln t + 8t + C
Final Position Equation:
6 = -6ln1 + 8(1) + C
6 = -6(0) + 8 + C
6=8+C
C = -2
x(t) = -6lnt + 8t - 2
c) To find the position when t =e, plug in e for t in the position
formula:
x(e) = -6lne + 8e –2
= -6 + 8e –2
= 8e – 8 or 8(e – 1)
Try Me Problem
3) 1987 AB 1: A particle moves along the x-axis so that
its acceleration at any time t is given by a(t)= 6t –
18. At time t=0 the velocity of the particle is
v(0)=24, and at time t=1, its position is x(1)= 20.
a) Write an expression for the velocity v(t) of the
particle at any time t.
b) For what values of t is the particle at rest?
c) Write an expression for the position x(t) of that
particle at any time t.
d) Find the total distance traveled by the particle from
t=1 to t=3.
Solution
a)
b)
c)
d)
v(t) = 3t² - 18t + 24
t = 2, t = 4
x(t) = t³ - 9t² + 24t + 4
x(1) = 20
4
x(2) = 24
2
x(3) = 22 6 total units
Work
(If Needed)
What are Tangent Lines?
• Definition: A line that corresponds to a
function and only intersects it at one specific
point.
• General Equation: y-y1=m(x-x1)
Tangent Lines
(Continued)
• To solve tangent line problems, certain components
must be given or solved for:
– The original function
– The derivative of that function
– X-coordinate (x1)
– Y-coordinate(y1)
– The solution of the derivative by plugging in the
original x- and y- coordinates
Once these components are solved for, plug them
into the general equation.
Example Problem 1
• 1978 AB 1: given the function f defined
by f(x)= x³ - x² - 4x + 4. Write an
equation of the line tangent to the
graph of f at x= -1
Step-by-Step Solution 2
• To find the equation of the line tangent, you must
first find the y- coordinate and the slope:
f(x)= x3-x2-4x+4
f(-1) = (-1)³ - (-1)² - 4(-1) + 4
f(-1) = -1 – 1 + 4 +4
When the elements are
f(-1) = 6
plugged into the general
f’(x) = 3x² - 2x – 4
equation, the tangent line is:
y – 6 = 1 ( x + 1)
f’(-1) = 3(-1)² - 2(-1) –4
or
f’(-1) = 3 + 2 –4 = 1
y=x+7
Example Problem 2
• 1985 AB 1: Let f be the function given by :
f(x)= 2x – 5
x² - 4
Write an equation for the line tangent to the
graph of f at the point (0, f(0)).
Step-by-Step Solution 2
• To find the equation of the line tangent, you must first find the
y- coordinate and the slope:
f(x)= 2x – 5
When the elements are
x² - 4
plugged into the general
equation, the tangent line is:
f(0) = 2(0) –5
y-(5/4) = -8x
(0)² - 4
or
f(0) = 5
y = -8x + (5/4)
4
f’(x) = (x² -4)(2) – (2t – 5)(2t)
(t² -4)²
f’(0) = (0² - 4)(2) – (2(0) – 5)(2(0))
(0² - 4)²
f’(0) = (-4)(2) = -8
16
Try Me Problem
• 1986 AB 1: Let f be defined by:
f(x) = 7 – 15x + 9x² - x³
Write an equation of the line tangent to the
graph of f at x=2.
Solution
y- 5 = 9(x – 2) or
y =9x –13 or
y – 9x = -13
Work
(If Needed)
What are Normal Lines?
• Normal lines are similar to tangent lines
except their slope is perpendicular to
the tangent lines’ slope.
• The general equation is the same:
y-y1=m(x-x1)
except the slope is the opposite reciprocal
Normal Lines
(Continued)
• To solve normal line problems, certain
components must be given or solved for:
–
–
–
–
–
The original function
The derivative of that function
X-coordinate (x1)
Y-coordinate(y1)
The solution of the derivative by plugging in the
original x- and y- coordinates, then finding the
opposite reciprocal of the solution
STOP!!! WHAT DOES
PERPENDICULAR MEAN?!?!?
• A perpendicular line has the opposite
reciprocal slope of the line it is perpendicular
to. This means it crosses the line at a 90º
angle.
• Opposite reciprocal means that the slope is
“flipped” and the sign is changed. For
example:
1) m= -2
m=½
2) m= 7
m= -12
12
7
Example Problem
• Find the equation of the normal line to
the curve:
f(x)=x2-2x+1
at the point x=3.
Step-by-Step Solution
•
To find the equation of the line normal to the graph, you must first find the
y- coordinate and the slope:
f(x)=x2-2x+1
f(3)=(3)2-2(3)+1
f(3)=4
f’(x)=2x-2
f’(3)=2(3)-2
f’(3)=4
This is the slope of the tangent line and needs to be made
perpendicular for the line to be normal to the graph.
m= -¼
When the elements are plugged into the general equation, the normal line is:
y-4=-¼(x-3)
y-4=-¼x+(3/4)
y=-¼x+19/4
Try Me Problem
• Find the equation of the normal line to
the graph of:
f(x)=(x2+3)½
at the point (−1, 2).
Solution
y-2=2(x+1) or
2x-y=-4
Work
(If Needed)
Bibliography
• "Rectilinear motion." Connexions - Sharing Knowledge and Building
Communities. N.p., n.d. Web. 1 Mar. 2011.
<http://cnx.org/content/m13612/latest/>..
• Picture. 1 Mar. 2011.
<http://www.analyzemath.com/calculus/Problems/tangent_2.gif>.
• Picture. 2 Mar. 2011. <http://www.capecodshops.com/id524/ImgUpload/P_507686_1284420.JPG>.
• "Tangents and Normals." Series Math Study. N.p., n.d. Web. 6 Mar.
2011. <http://www.seriesmathstudy.com/tangentexample.htm>.
•KAYE AUTREY
© Laura Woodlee and Katie Faith, 3/7/11