AP Calculus Summer Project Solutions
- Show all work on separate paper.
- Write your name in pen at the top of each page.
- Use graph paper for all graphs. Do all examples.
- Prepare to be tested on all topics on the first day of class.
- If you have any questions during the summer, email the instructor at w at least 48 hours for a response.
Topic: Slope, Lines, and Linear Equations
1. Find slope from two given points. Example: (4 , –7) and (–5 , 8)
Slope (m) of the line through (x1,y1) and (x2,y2) is m =
Slope of the line through (4 , –7) and (–5 , 8) is m = = =
2. Write the slope and an equation for a horizontal line through (4, 7). m = 0
y = 7
3. Write the slope and an equation for a vertical line through (-3 , 5).
no slope or undefined slope ; equation: x = –3
Point-Slope Form: y – y1 = m(x – x1)
4. Use point-slope form to write an equation for the line through (6 , –5) with
slope 2.
y – (–5) = 2(x – 6)
y + 5 = 2x – 12
y = 2x – 17
5. Use point-slope form to write an equation for the line through (4 , –7) and (–5 , 8).
First find the slope: m =
Then choose either point and use point-slope form: y – y1 = m(x – x1)
y – 8 = ( x – (–5) )
y – 8 = ( x + 5)
y – 8 =
y = or 5x + 3y = –1
6. Convert from point slope form to slope-intercept form: y – 7 =
Distribute and simplify: y – 7 =
y =
7. Define and illustrate a tangent line.
A tangent line is a line that touches a curve at exactly one point.
8. Define and illustrate a secant line.
A secant line is a line that crosses a curve at two points.
9. Define “average rate of change” algebraically and graphically.
The ARC on [ a , b ] is the slope of the secant line through (a , f (a)) and (b , f (b)).
Algebraically: ARC = or
Geometrically: Average rate of change is slope of a secant line.
10. Define “instantaneous rate of change” algebraically and graphically.
Algebraically: IRC is the derivative or the limit as h approaches zero of the DQ.
Geometrically: IRC is the instantaneous slope of a tangent line
Do the following AP released free response questions found online at ap central.
- 2010 FRQ #2(a)ARC = hundred entries per hour
- 2011 FRQ #2(a)ARC@ 3.5 = ° Celsius/minute
13. 2008 FRQ #2(a)ARC@(5.5) =
14. 2005 FRQ #3(a)ARC@(7) = ° Celsius/minute
Topic: Functions
Graph the 15 parent functions listed below. Use graph paper. State the domain and
range for each.
DomainRange
15. f (x) = x2all Ry ≥ 0
16. f (x) = x3all Rall R
17. f (x) =x ≥ 0y ≥ 0
18. f (x) = all Rall R
19. f (x) = |x|all Ry ≥ 0
20. f (x) = [x]all Rall integers
21. f (x) = sin(x)all R[ –1 , 1 ]
22. f (x) = cos(x)all R[ –1 , 1 ]
23.f (x) = tan(x)all R except odd multiples of all R
24. f (x) =all R except x = 0all R except y = 0
Topic: Semi-Circles
25. Write the general function for a semi-circle: f (x) = , where r is the radius.
26. Give examples of three semi-circle functions and graphs. (Answers may vary.)
y = ; y = ; y = ; y =
Be able to instantly recognize the equation and graph of a semi-circle!
Topic: Special Right Triangles
- Describe and illustrate the ratios for the sides of a 30º–60º–90ºtriangle.
In a 30º-60º-90º triangle, the hypotenuse is double the length of the short side
and the longer leg is times the length of the short side.
The ratio of the sides is 1 : 2 :(short leg : hypotenuse : long leg)
- Describe and illustrate the ratios for the sides of a 45º–45º–90º right triangle.
In a 45º-45º-90º triangle, the hypotenuse is times the length of either leg.
The ratio of the sides is 1 : 1 :(leg : leg : hypotenuse)
- Label all three sides for 30º–60º–90º triangles with hypotenuse of length
1 , 2 , 5 , 8 , 10, and x.
Short Leg / Long Leg / Hypotenuse/ / 1
1 / / 2
/ / 5
4 / 4 / 8
5 / 5 / 10
/ / x
30. Label all three sides for 45º–45º–90º triangles with hypotenuse of length
1 , 2 , 5 , 8 , 10, and x.
Leg / Hypotenuseor / 1
/ 2
or / 5
4 / 8
10 / 10
or / x
Topic: Trigonometric Ratios
31. Know sine, cosine, and tangent of the following angles to automaticity.
θ =
If necessary, make 48 flashcards. Identify all 48 trigonometric ratios from randomly-ordered flashcards to 99% accuracy in four minutes or less.
Topic: Seven Deadly Trig Identities
Prove these seven identities and their corollaries and memorize the results.
32. Pythagorean Identity #1: sin2θ + cos2θ = 1
Proof: From a right triangle diagram with acute angle θ, adjacent leg labeled x,
opposite leg labeled y, and hypotenuse h:
x2 + y2 = h2 by the Pythagorean Theorem
Divide through by h2:
cos2θ + sin2θ = 1 or sin2θ+ cos2θ = 1
Remember: “This sign costs one dollar.”
33. Pythagorean Identity #2: tan2θ + 1 = sec2θ
sin2θ + cos2θ= 1
Divide through by sin2θ:
Remember: “One vampire got caught in a casket.” or
You cannot buy just “one cotton ball at Costco”.
34. Pythagorean Identity #3: 1 + cot2θ = csc2θ
sin2θ +cos2θ = 1
Divide through by cos2θ:
Remember: You should only “tan for one second”.
Note: You need to prove the addition and subtraction identities before the double angle identities. I use a geometric proof starting with a rectangle ABCD. Right triangle ∆AED is inscribed in the rectangle with right angle AEF and hypotenuse AF = 1.
35. Sine Addition and Sine Subtraction: sin(α ± β) = sin α cos β ± cos α sin β
36. Cosine Addition and Cosine Subtraction: cos(α ± β) = cos α cos β ± sin α sin β
Next we can prove these:
37. Sine Double Angle Identity: sin 2θ = 2sinθ cosθ
Start with sin 2Ө= sin(Ө + Ө)
= sin Ө cos Ө + cos Ө sin Ө
= 2 sin Ө cos Ө
38. Cosine Double Angle Identities
(a) cos 2θ = cos2θ – sin2θ
Start with cos 2Ө= cos(Ө + Ө)
= cosӨ cos Ө – sin Ө sin Ө
= cos2Ө – sin2Ө
(b) cos 2θ = 2cos2θ – 1
Start with cos 2Ө = cos2Ө – sin2Ө
Use the first Pythagorean Identity: sin2Ө = 1 – cos2Ө
Substitute: cos 2Ө = cos2Ө – (1 – cos2Ө)
cos 2Ө = cos2Ө – 1 + cos2Ө
cos 2Ө = 2cos2Ө – 1
(c): cos 2θ = 1 – 2sin2θ
Start with cos 2Ө = cos2Ө – sin2Ө
Use the second Pythagorean Identity: cos2Ө = 1 – sin2Ө
Substitute: cos 2Ө = (1 – sin2Ө) – sin2Ө
cos 2Ө = 1 – 2sin2Ө
Topic: Sinusoidal Waves
Know the amplitude, period, horizontal shift, and vertical shift for trigonometric functions in the following forms. f (x) = a sin b(x – c) + d and f (x) = a cos b(x – c) + d
Graph and label these sinusoidal waves. Use a separate graph for each.
39. f (x) = 3 sin 2 + 1 40.f (x) = –2 cos 3 – 3
Topic: Absolute Value Functions
41. Define the absolute value function, f (x) = | x | as a piecewise function.
Algebraically: | x | = x when x ≥ 0 and | x | = –x when x < 0.
Graphically: | x | is the distance from 0 to x on the number line.
Graph and label the following absolute value functions on four separate graphs.
42.f (x) = | x + 3 | – 5 V-shape with vertex at ( –3 , –5 )
43. f (x) = 2| x | – 4 Skinny V-shape with vertex at ( 0 , –4 )
44. f (x) = –| x – 2 | + 1 A-shape with vertex at ( 2 , 1 )
45. f (x) = | 3 – x |V-shape with vertex at ( 3 , 0 )
Topic: Greatest Integer Function
46. Define the greatest integer function.
[[ x ]] is the greatest integer that is less than or equal to x. It is a rounding DOWN function.
Graph and label the following on four separate graphs. These are all “stair step” functions with an open circle on the left and a closed circle on the right.
47. f (x) = 2[x]49. f (x) = [x + 3]
48. f (x) = –[x]50. f (x) = [0.5x]
Topic: Piecewise Functions Graph the following piecewise functions.
51. f (x) =52. f (x) =
53. f (x) =
Topic: Exponential Functions
Graph. Label the y-intercept, the horizontal asymptote, and one anchor point.
Function GraphHADomainRangeAnchor Point
54. f (x) = 2xincreasing for all Ry = 0all Ry > 0(0 , 1)
55. f (x) = exincreasing for all Ry = 0all Ry > 0(0 , 1)
56. f (x) = 10xincreasing for all R y = 0all Ry > 0(0 , 1)
57. f (x) = ex+2– 4 same as ex shifted y = –4all Ry > –4(0, –3)
left 2 and down 4
58. f (x) = –5xdecreasing for all Ry = 0all Ry < 0(0 , –1)
59. f (x) = 4x + 3increasing for all Ry = 3all Ry > 3(0 , 4)
60. f (x) = 10x+2 + 3same as 10x shifted
left 2 and up 3y = 3all Ry > 3(0 , 4)
31. f (x) = log(x – 3) same as log(x) shifted right 3x > 3all R
32. f (x) = log(x + 5) – 1 same as ln(x) shifted left 5 and down 1x > –5all
Topic: Logarithmic Functions
61. Graph f (x) = ln(x) and g(x) = ex on one graph.
Label the x- and y-intercepts and one anchor point on each graph.
( 1 , 0 ) and (e , 1) are on f (x) = ln(x)
( 0 , 1 ) and (1 , e) are on g(x) = ex
The y-intercept for g(x) = exis y = 1.
The x-intercept for f (x) = ln(x) is ( 1 , 0 ).
62. Graph f (x) = log(x) and g(x) = 10x on one graph.
Label the x- and y-intercepts and one anchor point on each graph.
The y-intercept for g(x) = 10xis y = 1.
( 0 , 1 ) and (1 , 10) are on g(x) = 10x
The x-intercept for f (x) = log(x) is ( 1 , 0 ).
( 1 , 0 ) and (10 , 1) are on f (x) = log(x)
Topic: Transformations of Logarithmic Functions
Graph. Label the x-intercept, the vertical asymptote (VA) and one anchor point. Use four separate graphs.
- f (x) = ln(x – 1) + 2VA: x = 1Anchor Point: ( 2 , –3)
- f (x) = ln(x – 3) + 5VA: x = 3Anchor Point: ( 4 , 5 )
- f (x) = ln(x + 2) + 4VA: x = –2Anchor Point: (–1 , 4)
- f (x) = ln(x + 4) – 2VA: x = –4Anchor Point: ( –3 , –2)
Topic: Solving Logarithmic Equations
67. ln(x – 2) + 4 = ln x
4 = ln x – ln(x – 2)
4 = ln
e4 =
e4 (x – 2) = x
xe4 – 2e4 = x
xe4 – x = 2e4
x( e4 – 1 ) = 2e4
x =
68. 3(x+1) = 7(x–2)
ln 3(x+1) = ln 7(x–2)
( x + 1) ln 3 = ( x – 2 ) ln 7
x ln 3 + ln 3 = x ln 7 – 2 ln 7
ln 3 + 2 ln 7 = x ln 7 – x ln 3
ln 3 + 2 ln 7 = x ( ln 7 – ln 3 )
= x
= x
= x
69. log5(x–2) = log5x + log57
log5(x–2) – log5x = log57
log5 = log57
= 7
x – 2 = 7x
–2 = 6x
= x
70. ex(ex–4 ) = 1
ex–4 =
ex–4= e-x
x – 4 = –x
–4 = –2x
2 = x
Topic: Inverse Functions
- What is inverse function notation for f (x)? f -1(x)
- How do you find the inverse of f (x) algebraically? Switch x and y, then solve for x.
73. How do you find the inverse of f (x) graphically?
Reflect the graph across the diagonal line y = x.
74. Give 3 examples of points on a function and points on the inverse.
( 5 , –2 ) (–2 , 5 )
( 3 , 4 ) ( 4 , 3 )
(–7 , –10 ) (–10 , –7 )
75. If f (x) and g(x) are inverses, what is f (g(x))? What is g(f (x))?
f (g(x)) = x
g(f (x)) = x
76. Give 3 examples of slopes of functions and their inverses at corresponding points.
Slope of Function at ( x , y ) / Slope of Inverse at ( y , x )5 / 1/5
–3 / –⅓
½ / 2
77. Inverse functions are reflections across what line? y = x
78. The slopes of inverse functions are reciprocals of each other at corresponding points reflected across the line y = x.
Topic: Polynomials and Rational Expressions
79. X-Intercepts occur when the numerator of a rational expression equals zero.
80. Vertical Asymptotes occur when the denominator of a rational expression equals zero.
81. Horizontal Asymptotes occur when:
the degree of the numerator ≤ the degree of the denominator.
82. Holes occur when there is a:
zero of both the numerator and denominator with multiplicity ≥ in the numerator.
Non-Vertical Asymptotes:In a rational expression, let n = the degree of the numerator and
d = the degree of the denominator. Describe the end behavior.
83. If n < d, then the horizontal asymptote is y = 0 (the x-axis).
84. If n = d, then the horizontal asymptote is y =
where LCON is the leading coefficient of the numerator and
LCOD is the leading coefficient of the denominator.
85. If n > d, and n = d + 1, then the non-vertical asymptote is a line.
86. If n > d, and n = d + 2, then the non-vertical asymptote is a parabola.
87. If n > d, and n = d + 3, then the non-vertical asymptote is a cubic.
Find the x-intercepts, vertical asymptotes, horizontal asymptotes, and holes.
Graph each function separately on graph paper. Describe the end behavior.
Optional: Find and .
88.
HA: y = 0
VA: none
Holes: none
x-intercepts: x = 0 and x = –7
= 0 and = 0
89. =
HA: none, but there is a parabolic asymptote at approx. y = –2
VA: none
Holes: x = 0
x-intercepts: none
= -2and = -2
90. =
HA: none, but there is a parabolic asymptote at approx. y = x2
VA: none
Holes: x = 0
x-intercepts: x =
= ∞and = ∞
91. =
HA: y = 1
VA: none
Holes: none
x-intercepts: x = 4 and x = –3
= 1 and = 1
Topic: Area Formulas for Basic Geometric Shapes
92. Area of a Square with side s.A = s2
93. Area of a Semi-Circle with diameter DA = or A =
94. Area of an Isosceles Right Triangle with leg x.A =
95. Area of an Isosceles Right Triangle with Hypotenuse h. A =
96. Area of a TrapezoidA =
97. Area of a Rectangle with width x and length equal to three times the width. A = 3x2
98. Area of an Equilateral Triangle with side s. A =
Do the following AP released free response questions at
99. 2011B #6: Find the area of the triangle on [–2π , 4π].A = 6π2
100. 2011 #4: Find the area between the graph and the x-axis on the intervals
(a) [–4 , –3]:A = units2
(b) [–3 , 0]:A = units2
(c) [0 , 1.5]:A = units2
101. 2010 #5: Find the area between the graph and the x-axis on the intervals
[–7, –2]:A = units2
[–2 , 2]: A = units2
[2 , 4.5]:A = units2
[4.5, 5]:A = units2
102. 2010B #4: Find the area of the three trapezoids on [0 , 18].
1st Trapezoid on [0 , 9]:A = = 140 units2
2nd Trapezoid on [9 , 15]:A = = 50 units2
3rd Trapezoid on [15 , 18]:A = = 25 units2
Topic: Difference Quotient and Derivative
- Find the difference quotient.
- Find the derivative, f '(x). Use the limit of the difference quotient as h approaches 0.
103. f (x) = –x – 4
DQ =
DQ =
DQ =
DQ = = –1f '(x) =
104. f (x) = x2
DQ =
DQ =
DQ =
DQ =
DQ = 2x + hf '(x) =
105. f (x) = x3 + x
DQ =
DQ =
DQ =
DQ =
DQ = f '(x) =
DQ = 3x2 + 3xh + h2 + 1f '(x) = 3x2 + 1
106. f (x) = 4x2 – 3x – 2
DQ =
DQ =
DQ =
DQ =
DQ = f '(x) =
DQ = 8x + 4h – 3 f '(x) = 8x – 3
107. f (x) =
DQ =
DQ =
DQ = .
DQ = f '(x) =
DQ = f '(x) =
DQ = f '(x) =
108. f (x) =
DQ =
DQ =
DQ =
DQ =
DQ = f '(x) =
DQ = f '(x) =
109. f (x) = sin(x)
DQ =
DQ =
DQ = You can STOP here.
We will find the derivative, f '(x) = cos(x) soon.
110. f (x) = cos(x)
DQ =
DQ =
DQ = You can STOP here.
We will find the derivative, f '(x) = –sin(x) soon.