Find the tangent line

Report
Topanga’s Real-World Applicability……………………………. Pg.3
Hyde’s Helpful Mathematician’s Contributions………….. Pg.4
Angela’s Analytical Examples…………………………............ Pg.5-8
Midge’s Multiple Choice Example…………………………….. Pg.9-10
Cory’s Conceptual Example………………………………………. Pg.11
Foreman’s Arc Length with Graphing Calculator………. Pg.12-13
Mr. Feeny’s Free Response Example………………………... Pg.14-16
Sabrina’s Works Cited………………………………………….…….Pg.17
How many times have you seen a cat walk in
the path of x²? Almost never. Cats (like Salem)
pounce, jump, and walk in circles. However,
there are no functions to describe Salem’s
movement. By incorporating time into a
function, parametric equations allow us to
describe these paths of movements that we
could not describe before. One function relies
on time, t, to describe the movement with
respect to x, and the other function relies on
time, as well, to describe the movement with
respect to y. Together, they give you graphs that
can describe movement we could not describe
before like the movement of the eight ball as
Cory and Shawn play pool or the path of
Forman’s smoke.
Mustapha Amrani is a professor at the University of Cambridge
and is affiliated with the Isaac Newton Institute. He studies
parametric equations in detail so as to find real-world applications
for the concept and he gives talks on his findings. Amrani has
discovered a use of parametric equations that was previously
unknown-- he has begun to incorporate parametric equations and
semi-parametric equations into electricity
consumption forecasting. For utilities it is necessary to have a good
idea of the future's electricity consumption and Armani has found
that through the use of parametric equations in his work he is able
to better predict the consumption trends that will occur thereby
aiding the utilities. Since Armani uses primarily semi-parametric
structures he is able to create an even more accurate prediction as
his models create a complex but interesting trade-off between
previous accurate forecasts and new adaptations to different data
sets. Although these contributions are still gaining credibility and
usage in the real-world, Armani has proven how parametric
equations can be utilized with a multitude of different
geographical climates and horizons in order to try to accurately
predict the electricity consumption of the future.
First, let’s learn how to eliminate the parameter.
Eliminating the parameter allows for you to go from parametric
equations to polynomial equations. Ultimately we are
eliminating time, t, from the equation.
x (t ) 
t
3
t3
So, when told to eliminate the parameter, look at the x and-y functions and decide in which equation would it be
easier to solve for time, t. Although both of the equations
above are quite messy, the y-equation seems to be easier
to solve for time, t.
2 y 
1
t
2
2
4 2y  t
2
4 2y  t
The last step is to solve for time,
t, in the y-equation and plug it
in for time, t, in the x-equation.
You just eliminated the
parameter! Wasn’t that easy?
Graph and find the equation of the tangent line to
the graph at t=1 if x(t)=2t²-4 and y(t)=t²-3t+1
First, let’s graph the parametric equation!
Graphing parametric equations are quite different from regular
equations. Because the equations are given with respect to time, we
are actually calculating positions at certain times. So, you would want
to create a table of values like the one to the bottom-left and graph
the points by plugging time into the x- and y-functions. For example,
at t=0, the x-value is -4 and the y-value is 1. So, the first point will be (4,1). Continue this with the rest of the points. Also, one thing greatly
different from regular graphs is direction. So add an arrow to specify
what direction you are going.
t
0
x(t)
-4
y(t)
1
.5
1
-3.5
-2
-.25
-1
1.5
2
.5
4
-1.25
-1
Now, it’s time to find the tangent line!
When finding tangent lines, it’s not too different from regular
equations. First, find the derivatives of both x- and y- functions.
x ' (t )  4 t
y ' (t )  2 t  3
Next, it’s time to find the slope of the tangent line. But, how do we
do this when we have two functions!?!? You simply divide the
derivative of the y-function by the derivative of the x-function.
dy / dt
dx / dt

y ' (1)

x ' (1)
2 (1)  3

1
4 (1)
4
Now that we have the slope of the tangent line, it’s time to find
the x- and y-coordinates. This step is a bit more straight forward.
Just plug in t=1 into the functions!
x (1)  2 (1)
y (1)  (1)
2
2
 4  2
 3 (1)  1   1
Finally, it’s time to put it all together and graph the tangent line!
y 1 
1
( x  2)
4
y 1 
1
( x  2)
4
Now you know how to graph parametric equations and find
their tangent lines!
At which time, t, does the graph of   = ( ) and   =
 have a vertical tangent line on the interval − ≤  ≤ ?
a. 0
b. 
c. 2
d. 0, 
e. 0, , 2
To find a vertical tangent line, we must solve for when the
derivative of the x-function is equal to zero. This will give us
the time at which there is no movement in the x-direction
and thus a vertical tangent line!
x ' ( t )   4 t sin t
0   4 t sin t
t0
2
2


sin t
2
 0
t 
2

So what will make this equation equal to zero? Obviously when
t is equal to zero, the whole function will be zero so that is one
option. When sine is zero, the whole function will be zero as
well. At 0 and , sine is equal to zero but because we already
have 0, we can consider .
t
t0
t

We now have the two times at which there is no movement in
the x-direction. So we’re done now right!? Well not quite. We
have to test these times in the derivative of the y-function and
make sure it does not equal zero. Why may you ask? Well, what’s
happening to the graph when both the derivative of x and y are
zero? That’s right, nothing. There’s no movement, and therefore,
there wouldn’t be a vertical tangent line at that moment.
y ( t )  sin 2 t
y ' ( t )  2 cos 2 t
y ' ( 0 )  2 cos( 2 ( 0 ))
y ' (  )  2 cos 2 
2
Now that we have established that
there is still movement at those times,
we can officially say we have gotten
our answer! The times at which a
vertical tangent line occurs are at 0
and . Letter A doesn’t work because
it only has 0, Letter B only has ,
Letter C doesn’t have either, and Letter
E has too many answers. The only
answer that matches our choice is
Letter D!
  1 . 840
What is the concavity of the graph of   =  + −
and   =  −  at time, t=-3?
Once you hear the word concavity, immediately think the second
derivative! Wait, but how do we take the second derivative of a
parametric function? First we find the first derivative which we already
learned how to do. Differentiate the x- and y-functions then set dy over
dx. This is the first derivative. Now differentiate the first derivative to
get the second derivative, but wait a second! The second derivative is in
terms of dt! It should be in terms of dx²! To solve this problem just
divide the second derivative we just found by dx. Then evaluate at t=-3!
dx
 22 t  e
dy
t
dt
dt
dy / dt

dx / dt
d
d
dy

dx
 3
3
22 t  e
t
 dy 
t
t
t
 dx 

22 t  e 0  3 22  e   3 22  e 




t 2
t 2
dt
( 22 t  e )
( 22 t  e )
 dy 
 dx 


2
dt
dx
dx
dt
2
2
2

 3 22  e
t
t
2
( 22 t  e )
 3 ( 22  e )

1
22 t  e
t
3
d y
dx

d y

t  3
( 22 (  3 )  e )
3 3
0
Because the second derivative evaluated at t=-3 is
greater than zero than the graph is concave up!
If the eight ball Eric hit is described in inches by the

functions   =  +  and   =  − , what is the

distance the eight ball traveled between time t=1 and t=3?
To find the distance the eight ball traveled, we will want to use the arc
length equation.

′()2 + ′()2 
 ℎ =

So first, we need to find the derivatives of the x- and y-functions.
x ' (t ) 
8
t
3
1
y ' ( t )  21 t
2
Now just plug in the derivatives into the arc length equation
and solve it by using your graphing calculator!
3

1
2
8

2

1

21
t
 3

 t



2
dt
The eight ball that Eric hit traveled a total distance of
182.138 inches!
Sabrina is practicing spells with her aunts Zelda and Hilda when-OH
NO!--she accidentally hits Salem with a jumping jinx! Salem begins
to jump uncontrollably with an initial velocity of 40 ft/sec at angle
of 75° to the ground. Despite his efforts to acquiesce to the 32ft/sec acceleration of gravity he is unable to stop jumping. If
Salem was initially resting 3ft above the ground on the arm of the
couch, create X(t) and Y(t).
To begin to find the position functions you must utilize the general
equations:
 =  cos
V =  sin
First focus on finding X(t). Substitute the information from the
problem into the general equation to get the velocity function for X.
V() = 4075°
Next, integrate the equation to get the position function for X.
()  = 4075°
() = 4075° 
Now that you have found X(t), focus on finding Y(t). Integrate the
acceleration function to get the velocity of Y.
()  =
−32 
V  = −32 + 
Add the constant of integration that you get from the general
equation for the velocity of Y.
  = −32 + 4075°
Integrate the velocity function to get the position function of Y.
   =
−32 + 4075°
  = −16 2 + 4075°  + 
Add the initial condition.
  = −16 2 + 4075°  + 3
The final answer is:
  = 4075° 
  = −16 2 + 4075°  + 3
Eliminate the parameter:
t
2) x ( t ) 
1)  = 4 + 1
  = 2 − 1
t 1
y (t )  t  1
3
3)   =  2 + 1
y (t ) 
3t
2
t 1
4
Graph the following parametric equations:
t 3
2
4) x ( t )  t
6) x ( t ) 
5) x ( t )  t 2  3 t
y (t )  2 t  1
y (t )  t  2 t  2
3
2
t 3
3
t  4t
4
y (t ) 
3
Find 
7) x ( t )  t 2  3 t
y (t )  2 t  3
3
:
8)
x (t ) 
1
t t
4
2
2
y ( t )  ln( t  2 )
2
1
3
Find the tangent line:
9) At t=2:
10) At t=1:
x (t )  t  4 t  6
x ( t )  4 t  3t  5
3
2
y (t ) 
1
y ( t )  7 t  10 t  2
2
t 5
2
4
11) At t=3:
x ( t )  ln( 2 t  3 t  5 )
2
y (t )  t  5t  7
3
12) If  = sin  and  = −cos , then
2
 2
(A) − 3
(D) 3
2
(B)−3
3
(C) 0

6
at t= equals
(E) 3
13) What is the length of the path described by the
parametric equations   = 2  − 5 and   = 3 2 + 6,
where 0 ≤  ≤ 5?
(A)309.122
(B)36.993
(C)325.687 (D)39.046 (E) 369.826
14) The velocity vector of a particle moving in the xy-plane
is given by   =< 4 2 − 3 + 1, 3 − 2 >. At t=0,
the particle is at the point (0,-1). What is the position
vector at t=3?
(A) < 27,8 >
(B)< 8,27 >
(C)< 6.5,25.5 >
(D)< 25.5,6.5 >
(E) < 25.5,7.5 >
15)At what time(s) t does the graph of   = 5 2 − 2
and   = 3 + 1 have a vertical tangent line?
(A) 0 & 2/5 (B) None of the above (C) 0, 2/5, 1/5 (D) 1/5 (E)-1/3
16) Find the speed of a particle traveling along the path of
1
  = 4 3 − 3 + 1 and   =  2 +  at t=2.
2
(A) 7.348
(B)27.295
(C)45.100
(D)2601
(E)93.134
Cory and Topanga’s relationship has had up and downs
and everything in between over the years. The path of
their love is modeled by the velocity vector

−1

−1
< = sin  +  , = 18 2 cos( 3 ) − 2 >

4

for 0 ≤  ≤ 3 and at t=0 the position of their love is
(3,0).
a) Find the speed of the particle and the acceleration
vector at t=2.
b) Find the x-and y-values of their love at t=2.
c) Find the equation for the line tangent at t=2.
d) Find the length of their love from t=0 to t=3.
C
P
A
E
L
R
C
M
U
I
L
T
A
T
T
E
O
D
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Multiple-Choice & Free-Response Questions in Preparation for the AP
Calculus (BC) Examination by David Lederman

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