UNIT 1 ASSESSMENT
PROJECTBecca Ceremuga
AP Calculus Period 2
PART 1
LA QUEBRADA CLIFF DIVERS Acapulco, Mexico 35 meters (115 feet)
115 ft
Scale1 cm= 5.75
ft
Distance from Starting Point= 0 ft
Time= 0 sec
Distance from Starting Point= 12.5 ft
Scale1 cm= 5.75
ft
Time= 1 sec
102.5 ft
Scale1 cm= 5.75
ft
Distance from Starting Point= 35.8 ft
Time= 2 sec
79.2 ft
Scale1 cm= 5.75
ft
Distance from Starting Point= 51.6 ft
Time= 2.5 sec
63.4 ft
Scale1 cm= 5.75
ft
Distance from Starting Point=70.1 ft
Time= 3 sec
44.9 ft
Scale1 cm= 5.75
ft
Time= 3.5 sec
Distance from Starting Point=91.4 ft
23.6 ft
Scale1 cm= 5.75
ft
Time= 4 sec
Distance from Starting Point= 115 ft
0 ft
DISTANCE VS. TIME EQUATION
Data Points on the Video
Data from QuadraticRegression
Time (s) Height (ft)
0 115
1 100.6
2 84.8
2.5 58.9
Time (s) Height (ft)
0 115
1 102.5
2 79.2
2.5 63.4
3 44.9
3.5 23.6
4 0
General Equation d= ax2 +bx + c d= -5.51x2-6.74x+115 ft
4
115
Dis
tance
(f
t)
Time (seconds)
d= -5.51x2-6.74x+115
AVERAGE VELOCITYThe average velocity from 0 seconds to 4
seconds
v= d(tf)-d(ti)
tf- ti
v= d(4)- d(0) 4s- 0s
v= 0m–115ft 4s
v= -28.75ft/s
INSTANTANEOUS VELOCITYd= -5.51x2-6.74x+115The derivative of the distance formula=
velocityv= -11.02x-6.74 ft/s
Time (s) Velocity (ft/s)
0 -6.74
1 -17.77
2 -28.8
2.5 -34.31
3 -39.83
3.5 -45.34
4 -50.8550.85 ft/s= 35mph
d
vDis
tance
(f
t)
Time (seconds)
115
4
d= -5.51x2-6.74x+115 v= -11.02x-6.74
HIGHEST SPEEDDue to the acceleration of gravity, the diver’s velocity increases as he makes his descent. Therefore, the highest speed he travels is at the moment before he breaks the surface of the water.
-50.85 ft 1 m 15.5m
s 3.28 ft s
The diver never reaches a speed of 50
m/s.
ACCELERATIONv= -11.02x-6.74The derivative of the velocity formula= acceleration
a= -11.02 ft/s2
In reality, the acceleration of all objects in free fall have an acceleration of -9.8m/s2 (-32.2 ft/s2). However, because the video is in slow motion, the relationship between distance and time cannot be accurately found in a realistic manner. Also, the approximation error when finding the distance vs. time of the diver, contributes to the skewed acceleration.
Dis
tance
(f
t)
Time (seconds)
115
4
d
va
d= -5.51x2-6.74x+115 v= -11.02x-6.74 a= -11.02
PART 2
INSTANTANEOUS VELOCITY 1. Write the general equation
for the instantaneous rate of change of f(x) with respect to x at x=c
2. Determine the point at which the rate is to be found (c) and a point very close to c (x).
3. Find the values for f(x) and f(c) and substitute into the equation.
4. Calculate the rate of change at the specific point.
f’c= lim f(x)- f(c)
x-c
x c
f’c= lim f(2.01)- f(2)
(2.01-2)
x 2
f’c= lim 78.892 ft-79.18 ft .01 sx
2
f’c= lim -28.8 ft/sx 2
GRAPHICAL INTERPRETATION
Line of the tangent line at x=2
(y-y1)= m(x-x1) Point-slope Form(y-79.18)= -28.8 (x-2) Use the point: (2,78.18) and
the Instantaneous velocity: (-
28.8)y= -28.8x+136.78
d y
Time (seconds)
Dis
tance
(f
t)
2
78.18
y= -28.8x+136.78d= -5.51x2-6.74x+115
PART 3
PIECEWISE FUNCTION A function defined by different rules for
different intervals of its domain. Example:
f(x) -1.5x2-4x+10 if x≤ 22.5x-9 if 2≤x≥6
2.5x-6 if x>6
-1.5x2-4x+10 if x≤ 2
2.5x-9 if 2≤x≤6
2.5x-6 if x>6
f(x)
f(x)
62
REQUIREMENTS FOR CONTINUITY1. Value f(c) exists
*Filled in circle*0 cannot be in the denominator
2. lim(fx) exists (general limit)* lim(fx)= lim(fx)
3. f(c)= lim (fx)
x c
x c- x c+
x c
POINT OF DISCONTINUITYat x=6
1. f(x)=2.5x-9 if 2≤x≤6f(6)= 2.5(6)-9 *use the equation that has thef(6)=6 x value in its domain
2. lim(fx)= lim(fx) 2.5x-9=2.5x-6 *a general limit doesn’t exist 2.5(6)-9=2.5(6)-6 therefore #2 and #3 on the list 6= 2 of requirements cannot be met
x c- x c+
6
NO LIMIT AT X=6
CHANGING THE FUNCTION
Both equations need to have the same y value when x=6
#2 y= 2.5x-6 if x>6#1 y= 2.5x-9 if 2≤x≥6
y= 2.5x-b6= 2.5(6)-
b6=8-bb= 2
y=2.5x-9y= 2.5(6)-9y= 6
y= 2.5x-2
3. New Equation
1. Solve for the y value of #1
2. Use this y value and the x value to solve for b
LIMIT AT X=6
6
POINT OF NONDIFFERENTIABILITY
*Differentiable means continuous but continuous does not mean differentiable
at x=2
Left Side Slope f(x)= -1.5x2-4x+10lim f’(x)=-3x-4
lim f’(x)= -3(2)-4
lim f’(x)=-10
x 2-
x 2-
Right Side Slope f(x)= 2.5x-9lim f’(x)=2.5x 2+
x 2-
The slopes must be the same on the left and the right in order for a point to be differentiable.
2f’(x)= -10
f’(x)= 2.5
SLOPES DON’T MATCH
PART 4
End Behavior Model
FUNCTION APPROACHING INFINITY
f(x)=x3+5x-3
8
f(x)= x3lim +8x lim f(x) = 3
lim f(x)= 8
+8x
+8
x
lim f(x) = 3
lim f(x) = x3
lim f(x) = -
x
-
x
-
x
-
88
88
8
f(x)=x3+5x-3
FUNCTION APPROACHING 0
f(x)= x+2 x2-4
End Behavior Modellim f(x)= x = 1 x2 xlim f( )= 1
lim f( )= 0
8
8
f(x)= x+2 (x+2)(x-2)
f(x)= 1 (x-2) Vertical Asymptote
f(x)= x+2 x+2Removable Discontinuity
Horizontal Asymptote
+8
x +8
x +8
x
8
y= 0
x=2
x= -2
(same as)
8
x
-
Horizontal Asymptote at y=0Vertical Asymptote at x=2Removable Discontinuity at x=-2
f(x)= x+2
x2-4
FUNCTION WITH ASYMPTOTES
f(x)= x3+2x+3 4x3-4(
(
lim f(x)= x3 4x3
lim f(x)= 1 4
+8
x
x x
x
x
+8
x Horizonta
l Asymptote
y= 1 4
lim f(x)= 6 tiny+#
lim f(x)=
1 -1+
81 -
lim f(x)= 6 tiny-# lim f(x)= -
8
Vertical Asymptote
1+
x=1
(same as)
8
x
-
End Behavior Model
f(x)= x3+2x+3 4x3-4(
(
114
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