PERIMETER, AREA & VOLUME Rectangle ππ = 2ππ + 2π€π€ π΄π΄ = ππππ Square ππ = 4π π π΄π΄ = π π 2 Triangle P = add all sides 1 2 A = ππβ Parallelogram P= add all sides A = bh Trapezoid P = add all sides 1 A= (ππ1 + ππ2 )h 2 Circle πΆπΆ = ππππ = 2ππππ π΄π΄ = ππππ 2 Arc Length S = ππππ in radians ππ S= 180 ππππin degrees Circle Sector Area ππ 2 A= ππ ππππ 2 in degrees 360 A = ππ 2 in radians Rectangular solid S = 2ππππ + 2ππβ + 2π€π€β V = ππππβ Cube ππππ = 6π π 2 V = s3 Cylinder ππππ = 2ππππ 2 + 2ππππβ ππ = ππππ 2 β Cone ππππ = ππππππ + ππππ 2 ππ = 1 2 ππππ β 3 A = ππππ√ππ 2 + β2 Sphere SA = 4πr 2 4 V = πr 3 3 A = 4ππππ 2 Rectangle Pyramid SA = ππππ + 2ππππ + 2π€π€π€π€ 1 V = ππππβ 3 PYTHAGOREAN THEOREM leg2+ leg2 = hypotenuse2 DISTANCE FORMULA EXPONENT LAWS π₯π₯ 0 = 1 if x ≠ 0 1 ππ = οΏ½(π₯π₯2 − π₯π₯1 )2 + (π¦π¦2 − π¦π¦1 )2 π₯π₯ = π₯π₯ −ππ π₯π₯ 1 = ππ if x ≠ 0 π₯π₯ π₯π₯ ππ . π₯π₯ ππ = π₯π₯ ππ+ππ (π₯π₯ ππ )ππ = π₯π₯ ππ.ππ π₯π₯ ππ ÷ π₯π₯ ππ = π₯π₯ ππ (π₯π₯π₯π₯)ππ = π₯π₯ π¦π¦ π₯π₯ ππ π¦π¦ = π₯π₯ ππ−ππ if x ≠0 π₯π₯ ππ ππ ππ π₯π₯ ππ οΏ½ οΏ½ = π¦π¦ππ if y ≠0 ππ ππ ππ π₯π₯ = √π₯π₯ ππ if (a ≥ 0, m ≥0, n>0) PROPERTIES OF LOGARITHMS π¦π¦ = ππππππππ π₯π₯ ⇔ π₯π₯ = ππ π¦π¦ where a >0, a ≠ 0 ππ ππππππππ ππ = ππ ππππππππ (ππππ) = ππππππππ ππ + ππππππππ N ππ ππππππππ οΏ½ οΏ½ = ππππππππ ππ − ππππππππ ππ ππ ππππππππ ππ π₯π₯ = π₯π₯ππππππππ ππ ππππππππ ππ = ππππππππ ππ ππππππππ ππ SPECIAL PRODUCTS = ππππππππ ππππππππ = ππππππ ππππππ π₯π₯ 2 − π¦π¦ 2 = (π₯π₯ + π¦π¦)(π₯π₯ − π¦π¦) π₯π₯ 3 ± π¦π¦ 3 = (π₯π₯ ± π¦π¦)(π₯π₯ 2 β π₯π₯π₯π₯ + π¦π¦ 2 ) BINOMIAL THEOREM (π₯π₯ ± π¦π¦)2 = π₯π₯ 2 ± 2π₯π₯π₯π₯ + π¦π¦ 2 (π₯π₯ ± π¦π¦)3 = π₯π₯ 3 ± 3π₯π₯ 2 π¦π¦ + 3π₯π₯π¦π¦ 2 ± π¦π¦ 3 (π₯π₯ + π¦π¦)ππ = π₯π₯ ππ + πππ₯π₯ ππ−1 π¦π¦ ππ ππ(ππ−1) ππ−2 2 + π₯π₯ π¦π¦ + … + οΏ½ οΏ½ π₯π₯ ππ−ππ 2 ππ + … + πππππ¦π¦ ππ−1 + π¦π¦ ππ ππ(ππ−1)…(ππ−ππ+1) ππ where οΏ½ οΏ½ = ππ 1•2•3•…•ππ ππππππ sinππ = βπ¦π¦π¦π¦ ππππππ πππππππ½π½ = βπ¦π¦π¦π¦ ππππππ π‘π‘π‘π‘π‘π‘π½π½ = ππππππ ππππππ πππππππ½π½ = ππππππ βπ¦π¦π¦π¦ πππππππ½π½ = ππππππ βπ¦π¦π¦π¦ π π π π π π π½π½ = ππππππ π π π π π π π½π½ π‘π‘π‘π‘π‘π‘π½π½ = πππππππ½π½ πππππππ½π½ πππππππ½π½ = π π π π π π π½π½ 1 π π π π π π π½π½ = πππππππ½π½ 1 πππππππ½π½ = π π π π π π π½π½ 1 πππππππ½π½ = π π π π π π π½π½ 1 π π π π π π π½π½ = πππππππ½π½ 1 π‘π‘π‘π‘π‘π‘π½π½ = πππππππ½π½ 1 πππππππ½π½ = π‘π‘π‘π‘π‘π‘π½π½ π π π π π π 2 π½π½ + ππππππ 2 π½π½ = 1 π‘π‘π‘π‘π‘π‘2 π½π½ + 1 = π π π π π π 2 π½π½ ππππππ 2 π½π½ + 1 = ππππππ 2 π½π½ 1 − ππππππππππ π π π π π π 2 π½π½ = 2 1 + ππππππππππ 2 ππππππ π½π½ = 2 1 − ππππππππππ 2 π‘π‘π‘π‘π‘π‘ π½π½ = 1 + ππππππππππ π π π π π π (ππππ) = 2π π π π π π π½π½πππππππ½π½ ππππππ(ππππ) = ππππππ 2 π½π½ − π π π π π π 2 π½π½ = 2ππππππ 2 π½π½ − 1 = 1 − 2π π π π π π 2 π½π½ π‘π‘π‘π‘π‘π‘(ππππ) = 2π‘π‘π‘π‘π‘π‘π½π½ 1 − π‘π‘π‘π‘π‘π‘2 π½π½ Law of Sines π π π π π π πΆπΆ π π π π π π π·π· π π π π π π πΈπΈ = = ππ ππ ππ Law of Cosines ππ2 = ππ 2 + ππ 2 − 2ππππππππππππ ππ 2 = ππ2 + ππ 2 − 2ππππππππππππ ππ 2 = ππ2 + ππ 2 − 2ππππππππππππ Sum and Difference π π π π π π (πΆπΆ ± π·π·) = π π π π π π πΆπΆπππππππ·π· ± πππππππΆπΆπ π π π π π π·π· ππππππ(πΆπΆ ± π·π·) = πππππππΆπΆπππππππ·π· β π π π π π π πΆπΆπ π π π π π π·π· Sum to Product πΆπΆ + π·π· πΆπΆ − π·π· π π π π π π πΆπΆ + π π π π π π π·π· = 2π π π π π π οΏ½ οΏ½ ππππππ οΏ½ οΏ½ ππ ππ πΆπΆ − π·π· πΆπΆ + π·π· οΏ½ π π π π π π οΏ½ οΏ½ π π π π π π πΆπΆ − π π π π π π π·π· = 2ππππππ οΏ½ ππ ππ πΆπΆ − π·π· πΆπΆ + π·π· οΏ½ ππππππ οΏ½ οΏ½ πππππππΆπΆ + πππππππ·π· = 2ππππππ οΏ½ ππ ππ πΆπΆ + π·π· πΆπΆ − π·π· πππππππΆπΆ − πππππππ·π· = −2π π π π π π οΏ½ οΏ½ π π π π π π οΏ½ οΏ½ ππ ππ π‘π‘π‘π‘π‘π‘(πΆπΆ ± π·π·) = π‘π‘π‘π‘π‘π‘πΆπΆ ± π‘π‘π‘π‘π‘π‘π·π· 1 β π‘π‘π‘π‘π‘π‘πΆπΆπ‘π‘π‘π‘π‘π‘π·π· Product to Sum ππππππ(πΆπΆ − π·π·) − ππππππ(πΆπΆ + π·π·) 2 ππππππ(πΆπΆ − π·π·) + ππππππ(πΆπΆ + π·π·) πππππππΆπΆπππππππ·π· = 2 π π π π π π (πΆπΆ + π·π·) + π π π π π π (πΆπΆ − π·π·) π π π π π π πΆπΆπππππππ·π· = 2 π π π π π π (πΆπΆ + π·π·) − π π π π π π (πΆπΆ − π·π·) πππππππΆπΆπ π π π π π π·π· = 2 π π π π π π πΆπΆπ π π π π π π·π· = SANTA ANA COLLEGE 1530 West 17th Street, Santa Ana CA 92704 DERIVATIVES Definition: Derivative: ππ ′ (π₯π₯) = THE MATH CENTER www.sac.edu/MathCenter Room L-204 Phone: (714) 564-6678 Monday thru Thursday 9:00AM – 7:50PM Friday 10:00AM – 12:50PM Saturday 12:00PM – 4:00PM COMMON DERIVATIVES ππππππ ππ(π₯π₯+β)−ππ(π₯π₯) if this limit exists. β→0 β Applications: If π¦π¦ = ππ(π₯π₯) then, • ππ = ππ ′(ππ) is the slope of the tangent line to y=f(x) at x=a and the equation of the tangent line at π₯π₯ = ππ is given by π¦π¦ = ππ(ππ) + ππ ′(ππ)(π₯π₯ − ππ). • ππ′(ππ)is the instantaneous rate of change of ππ(π₯π₯)at π₯π₯ = ππ. • If ππ(π₯π₯)is the position of an object at time π₯π₯, then ππ ′(ππ) is the velocity of the object at π₯π₯ = ππ Critical points: π₯π₯ = ππ is the critical point of ππ(π₯π₯) = ππ provided either 1. ππ ′(ππ) = 0 or 2. ππ ′(ππ) does not exist. Increasing/Decreasing • If ππ ′(π₯π₯) > 0 for all x in an interval I, then f(x) is increasing on the interval I. • If ππ ′(π₯π₯) < 0 for all x in an interval I, then f(x) is decreasing on the interval I. • If ππ ′ (π₯π₯) = 0 for all x in an interval I, then f(x) is constant on the interval I. Concavity • If ππ ′′(π₯π₯) > 0 for all x in an interval I, then f(x) is concave up on the interval I. • If ππ ′′(π₯π₯) < 0 for all x in an interval I, then f(x) is concave down on the interval I. Inflection Points π₯π₯ = ππ is an inflection point of f(x)if the concavity changes at π₯π₯ = ππ. INTEGRATION ππ οΏ½ ππ(π₯π₯)ππππ = lim οΏ½ ππ(π₯π₯ππ∗ )βπ₯π₯ ππ ππ→∞ ππ=1 2) [ππ(π₯π₯) + ππ(π₯π₯)]’ = ππ’(π₯π₯) + ππ’(π₯π₯) 3) [ππ(π₯π₯)ππ(π₯π₯)]’ = ππ(π₯π₯)ππ’(π₯π₯) + ππ’(π₯π₯)ππ(π₯π₯) 4) [ππ(ππ(π₯π₯))]’ = ππ’(ππ(π₯π₯))ππ’(π₯π₯) 5) [ππππ(π₯π₯)]’ = ππππ’(π₯π₯) 6) [ππ(π₯π₯) – ππ(π₯π₯)]’ = ππ’(π₯π₯) – ππ’(π₯π₯) ππ(π₯π₯) ′ ππ(π₯π₯)ππ′(π₯π₯)− ππ(π₯π₯)ππ′ (π₯π₯) 7) οΏ½ = οΏ½ ππ(π₯π₯) [ππ(π₯π₯)]2 ππ π₯π₯ ππ ππ ππ 9) 11) [ππππ|ππ|]′ = (ππ−ππ) where βπ₯π₯ = ππ π₯π₯ ∫ ππ(π‘π‘)ππππ = ππ(π₯π₯)where ππ ≤ π₯π₯ ≤ ππ. ππππ ππ ππ Area: π΄π΄ = ∫ππ ππ(π₯π₯)ππππ Area between Curves: ππ INTEGRALS 1) ∫ π’π’ππ ππππ = 2) ∫ ππππ = −ππππππ ππ 17) = π π π π π π πππ‘π‘π‘π‘π‘π‘ππ 18) (ππππππππ)′ = −ππππππππππππππππ 19) (π π π π π π −1 ππ)′ = π’π’ π’π’ππ+1 ππ+1 22) (ππππππ −1 ππ) ′ = − 1 1+π₯π₯ 2 21) (π‘π‘π‘π‘π‘π‘−1 ππ)′ = ππ Cylinders V= ∫ππ ππππππππππππππππππππππππππ β βππππππβπ‘π‘ β π‘π‘βππππππππππππππ ππ Work: If a force of πΉπΉ(π₯π₯) moves an object in ππ ≤ π₯π₯ ≤ ππ, then the work done is W=∫ππ πΉπΉ(π₯π₯)ππππ ππ Average Function Value: The average value of ππ(π₯π₯) on ππ ≤ π₯π₯ ≤ ππ isππππππππππππππππ = ∫ ππ(π₯π₯)ππππ ππ−ππ ππ 1 1 1+π₯π₯ 2 + ππ, ππ ≠ −1 ∫ ππ π’π’ ππππ = ππ π’π’ + ππ ππ π’π’ +c ππππππ ∫ ππππππ ππππ = π’π’π’π’π’π’ππ − π’π’ + ππ 1 6) ∫ ππππ = ππππ|ππππππ| + ππ π’π’π’π’π’π’ππ 13) ∫ π π π π π π 2 ππ ππππ = π‘π‘π‘π‘π‘π‘ ππ + ππ 14) ∫ ππππππ 2 ππ ππππ = −ππππππππ + ππ 15) ∫ π π π π π π ππ π‘π‘π‘π‘π‘π‘ππ ππππ = π π π π π π ππ + ππ 16) ∫ ππππππππ ππππππππ ππππ = −ππππππππ + ππ 17) ∫ ππππ √ππ 2 −π’π’ = π π π π π π −1 ππππ + ππ, ππ > 0 2 1 |π₯π₯|√π₯π₯ 2 −1 24) (ππππππ −1 ππ)′ = − 12) ∫ ππππππππ ππππ = ππππ|ππππππππ − ππππππππ| + ππ Rings V= ∫ππ 2ππ(ππππππππππ ππ 2 − ππππππππππ ππ 2 ) 1 √1−π₯π₯ 2 11) ∫ π π π π π π ππ ππππ = ππππ|π π π π π π ππ + π‘π‘π‘π‘π‘π‘ππ| + ππ Volume of Revolution 1 √1−π₯π₯ 2 20) (ππππππ −1 ππ) ′ = − 10) ∫ ππππππππ ππππ = ππππ|π π π π π π ππ| + ππ Volumes: V= ∫ππ π΄π΄π΄π΄π΄π΄π΄π΄(π₯π₯) ππππ 2 (π π π π π π ππ)′ = ππππ|ππ| + ππ 4) ∫ ππ π’π’ ππππ = 5) 16) (ππππππππ)′ 9) ∫ π‘π‘π‘π‘π‘π‘ππ ππππ = ππππ|π π π π π π ππ| + ππ ππ “Success is the sum of small efforts, repeated day in and day out.” Robert Collier 1 3) 15) (π‘π‘π‘π‘π‘π‘ππ) ′ = π π π π π π 2 ππ 23) (π π π π π π −1 ππ)′ = π₯π₯π₯π₯π₯π₯ππ 25) (π π π π π π βππ)′ = coshππ 14) (ππππππππ)′ = −π π π π π π ππ 8) ∫ ππππππππ ππππ = π π π π π π ππ + ππ • π₯π₯ = ππ(π¦π¦); π΄π΄ = ∫ππ (ππππππβπ‘π‘ − ππππππππ ππππππππππππππ) ππππ ππ π₯π₯ 13) (π π π π π π ππ)′ = cosππ 7) ∫ π π π π π π ππ ππππ = −ππππππππ + ππ • π¦π¦ = ππ(π₯π₯); π΄π΄ = ∫ππ (π’π’π’π’π’π’π’π’π’π’ − ππππππππππ ππππππππππππππ)ππππ ππ 1 12) [ππππππππ ππ]′ = Part II: ππππ ∫ππ ππ(π₯π₯)ππππ = πΉπΉ(ππ) − πΉπΉ(ππ) where πΉπΉ(π₯π₯) is any anti-derivative of ππ(π₯π₯), i.e, a function such that πΉπΉ’ = ππ. Applications: ππ−1 (π₯π₯ )′ = πππ₯π₯ [ππ π₯π₯ ]′ = ππ π₯π₯ 10) [ππ π₯π₯ ]′ = ππ π₯π₯ ππππππ 8) Fundamental Theorem of Calculus: Suppose ππ(π₯π₯) is continuous on [ππ, ππ], then Part I: ππ(π₯π₯) =∫ππ ππ(π‘π‘)ππππ is also continuous on [ππ, ππ] and ππ′ (π₯π₯) = "Who has not been amazed to learn that the functionπ¦π¦ = πππ₯π₯ , like a phoenix rising from its own ashes, is its own derivative?" Francois le Lionnais 1) ππ’ = 0 Definition: Suppose ππ(π₯π₯) is continuous on [ππ, ππ]. Divide [ππ, ππ] into n subintervals of width βπ₯π₯ and choose π₯π₯ππ∗ from each interval. Then ∞ /ep2016 OPERATIONAL HOURS 1 |π₯π₯|οΏ½π₯π₯2 −1 26) (ππππππβππ)′ = π π π π π π βππ 27) (π‘π‘π‘π‘π‘π‘βππ) ′ = π π π π π π β2 ππ 28) (ππππππβππ)′ = −ππππππβ2 ππ 29) (π π π π π π βππ)′ = −π π π π π π βπππ‘π‘π‘π‘π‘π‘βππ 30) (ππππππππ)′ = −ππππππβππππππππβππ 1 31) (π π π π π π β−1 ππ)′ = √1+π₯π₯ 2 √π₯π₯ 2 −1 34) (ππππππβ−1 ππ)′ = 1 1−π₯π₯ 2 33) (π‘π‘π‘π‘π‘π‘β−1 ππ)′ = 35) (π π π π π π β−1 ππ)′ = − ππππ π’π’√π’π’2 −ππ2 1 1−π₯π₯ 2 1 |π₯π₯|√1−π₯π₯ 2 36) (ππππππβ−1 ππ)′ = − ππππ 1 ππ 18) ∫ 2 2 = π‘π‘π‘π‘π‘π‘−1 + ππ ππ ππ ππ +π’π’ 19) ∫ 1 32) (ππππππβ−1 ππ)′ = 1 |π₯π₯|οΏ½π₯π₯2 +1 = ππ1 π π π π π π −1 ππππ + ππ ππππ 1 ππ+ππ 20) ∫ ππ ππ = ππππ οΏ½ οΏ½+c 2ππ ππ−ππ ππ −ππ ππππ 1 ππ−ππ 21) ∫ ππ ππ = ππππ οΏ½ οΏ½+c 2ππ ππ+ππ ππ −ππ 22) ∫ π π π π π π −1 ππ ππππ = π’π’ π π π π π π −1 ππ + √1 − π’π’ 2 + ππ 23) ∫ ππππππ −1 ππ ππππ = π’π’ ππππππ −1 ππ + √1 − π’π’ 2 + ππ 1 24) ∫ π‘π‘π‘π‘π‘π‘−1 ππ ππππ = π’π’ tan−1 ππ − ln(1 + π’π’2 ) + ππ 25) ∫ π π π π π π βππ ππππ = ππππππβππ + ππ 26) ∫ ππππππβππ ππππ = π π π π π π βππ + ππ 2 27) ∫ π‘π‘π‘π‘π‘π‘βππ ππππ = ππππ(ππππππβππ) + ππ 28) ∫ ππππππβππ ππππ = ππππ|π π π π π π βππ| + ππ 29) ∫ π π π π π π βππ ππππ = π‘π‘π‘π‘π‘π‘−1 |π π π π π π βππ| + ππ 1 30) ∫ ππππππβππ ππππ = ππππ οΏ½π‘π‘π‘π‘π‘π‘β πποΏ½ + ππ 2 31) ∫ π π π π π π β2 ππ ππππ = π‘π‘π‘π‘π‘π‘β ππ + ππ 32) ∫ ππππππβ2 ππ ππππ = −ππππππβππ + ππ 33) ∫ π π π π π π βππ π‘π‘π‘π‘π‘π‘βππ ππππ = −π π π π π π βππ + ππ 34) ∫ ππππππβππ ππππππβππ ππππ = −ππππππβππ + ππ 35) ∫ π’π’π’π’π’π’ = π’π’π’π’ − ∫ π£π£π£π£π£π£ “Go down deep enough into anything and you will find mathematics.” Dean Schlicter