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