Math Formulas Percentage and interest rate πΎπΎ = πΎπΎ0 · (1 + ππ)ππ πΎπΎ0 : Initial value n: number of interest terms πΎπΎ: Future value (after n interest terms) r: interest rate (growth rate) ππ = 1 + ππ a: growth factor. Factorization and Quadratic Identities π₯π₯ · ππ + π₯π₯ · ππ = π₯π₯ · (ππ + ππ) (ππ ± ππ)2 = ππ2 + ππ 2 ± 2ππ · ππ Fraction rules Exponent rules ππ2 − ππ 2 = (ππ − ππ) · (ππ + ππ) ππ ππ ππ · = ππ·ππ ππ ππ/ππ ππ/ππ = ππ·ππ ππ·ππ ππ ππ/ππ = ππ·ππ ππ ππππ · ππ π π = ππππ+π π ππππ = ππππ−π π ππ π π (ππππ )π π = ππππ·π π (ππ · ππ)ππ = ππππ · ππ ππ ππ ππ ππππ οΏ½ οΏ½ = ππ ππ ππ ππ0 = 1 , π π ππ−ππ = ππ ≠ 0 1 ππππ √ππππ = ππ ππ οΏ½ οΏ½ π π Make a Good Start Course, SDU-S, DWN, 2020 1 ππ ππ ππ ππ · = ππ·ππ ππ·ππ ππ/ππ ππ = ππ ππ·ππ Logarithm rules The common logarithm: log(ππ π₯π₯ ) = π₯π₯ · log (ππ) log(ππ · ππ) = log(ππ) + log(ππ) ππ log οΏ½ οΏ½ = log(ππ) − log (ππ) ππ log(10) = 1 Similar rules apply to the natural logarithm (and all other logarithm functions) ln(ππ π₯π₯ ) = π₯π₯ · ln (ππ) ln(ππ · ππ) = ln(ππ) + ln(ππ) ππ ln οΏ½ οΏ½ = ln(ππ) − ln (ππ) ππ ln(ππ) = 1 Graph of the natural logarithm (base e) Only defined for x>0 Vertical asymptote at x=0 Intersects the x-axis at x=1 ln(1) = 0 Heads towards infinity ln(π₯π₯) → ∞ for π₯π₯ → ∞ Intersects the point (e,1) Graph of the common logarithm (base 10) Only defined for x>0 Vertical asymptote at x=0 Intersects the x-axis at x=1 log(1) = 0 Heads towards infinity log(π₯π₯) → ∞ for π₯π₯ → ∞ Intersects the point (10,1) Make a Good Start Course, SDU-S, DWN, 2020 2 Functions, graphs and tangents Linear function Definition ππ(π₯π₯) = ππ · π₯π₯ + ππ The graph of a linear function is a straight line: π¦π¦ = ππ · π₯π₯ + ππ a: the slope of the line b: 2nd coordinate of the point where the graph intersects with the y-axis. The slope can be calculated by use of two points (π₯π₯1 , π¦π¦1 ) og (π₯π₯2 , π¦π¦2 ) on the graph: ππ = Δy y2 − y1 = Δx x2 − x1 The equation of a tangent t to the graph of the function ππ at (π₯π₯0 , ππ(π₯π₯0 )): π¦π¦ = ππ´(π₯π₯0 ) · (π₯π₯ − π₯π₯0 ) + ππ(π₯π₯0 ) The equation of a straight line can also be expressed ππ · (π₯π₯ − π₯π₯0 ) + ππ · (π¦π¦ − π¦π¦0 ) = 0 ππ where οΏ½ οΏ½ is a vector that is perpendicular to the line and (π₯π₯0 , π¦π¦0 ) is a point on the line. ππ To fit a linear model to a data set you use linear regression analysis. Make a Good Start Course, SDU-S, DWN, 2020 3 Exponential function Definition: ππ(π₯π₯) = ππ · ππ π₯π₯ , ππ > 0 , ππ > 0 , π₯π₯ ∈ β a: growth factor (a>1) a: decay factor (0<a<1) b: second coordinate of the point where the graph intersects with the y-axis. ππ = 1 + ππ Growth constant Decay constant ππ = ππ2 = ππ½ = π₯π₯2 −π₯π₯1 οΏ½ π¦π¦2 π¦π¦1 log(2) ln(2) = log(ππ) ln(ππ) 1 log οΏ½2οΏ½ log(ππ) = 1 ln οΏ½2οΏ½ ln(ππ) Exponential growth A certain change of x produces a certain %-change of the range element regardless of the initial value of x: f ( x + h) = a h ⋅ f ( x ) To fit an exponential model to a data set you use exponential regression analysis. Make a Good Start Course, SDU-S, DWN, 2020 4 Power function Definition: ππ(π₯π₯) = ππ · π₯π₯ ππ , ππ > 0 , ππ ∈ β , π₯π₯ > 0 Determination of the exponent a using 2 points (π₯π₯1 , π¦π¦1 ) og (π₯π₯2 , π¦π¦2 ) on the graph: ππ = log(π¦π¦2 ) − log(π¦π¦1 ) log(π₯π₯2 ) − log(π₯π₯1 ) ππ = ln(π¦π¦2 ) − ln(π¦π¦1 ) ln(π₯π₯2 ) − ln(π₯π₯1 ) When the value of x increases by a certain percentage the range element of a power function will also change by a certain percentage regardless of the initial value of x: f (h ⋅ x) = h a ⋅ f ( x) To fit a power model to a data set you use power regression analysis. Make a Good Start Course, SDU-S, DWN, 2020 5 Polynomial Degree n polynomial ππ(π₯π₯) = ππππ · π₯π₯ ππ + ππππ−1 · π₯π₯ ππ−1 + β― + ππ1 · π₯π₯ + ππ0 Degree 1 polynomial Degree 2 polynomial ππ(π₯π₯) = ππ · π₯π₯ + ππ ππ(π₯π₯) = ππ · π₯π₯ 2 + ππ · π₯π₯ + ππ Roots (= zeros) of the degree 2 polynomial (if any) π₯π₯ = −ππ ± √ππ 2ππ ππ = ππ 2 − 4ππ · ππ When π₯π₯1 and π₯π₯2 are roots (could be alike) in a degree 2 polynomial it can be factorized ππ(π₯π₯) = ππ(π₯π₯ − π₯π₯1 )(π₯π₯ − π₯π₯2 ) Any polynomial ππ with one or more roots can be factorized: ππ(π₯π₯) = (π₯π₯ − π₯π₯1 )ππ(π₯π₯) where π₯π₯1 is a root of ππ and q(x) is a polynomial with or without any roots. Make a Good Start Course, SDU-S, DWN, 2020 6 Trigonometric functions The unit circle The length x is equal to the angle v in radians π₯π₯ = π£π£ = Definition of cosine og sine Degrees Sin 30° 45° 60° 90° 0 1 2 √2 2 √3 2 1 √3 3 1 √3 - 1 Cos 0 Tan 180 π₯π₯ ππ 0° 0 Radians ππ π£π£ 180 ππ 6 √3 2 ππ 4 √2 2 ππ 3 1 2 ππ 2 0 sin2 (π₯π₯) + cos2 (π₯π₯) = 1 Cosine Cosine is periodic: Cosine is even: Sine Sine is periodic: Sine is odd: Harmonic function 7 cos(π₯π₯ + 2ππ) = cos (π₯π₯) cos(−π₯π₯) = cos (π₯π₯) cos(ππ − π₯π₯) = −cos (π₯π₯) sin(π₯π₯ + 2ππ) = sin (π₯π₯) sin(−π₯π₯) = −sin (π₯π₯) sin(ππ − π₯π₯) = sin (π₯π₯) f (t ) = A ⋅ sin(ω ⋅ t + Ο ) + k A : amplitude ω : angular frequency Ο : initial phase k : vertical shift T= 2π ω Make a Good Start Course, SDU-S, DWN, 2020 Ο : phase shift (to the left) ω Differential calculus The derivative of f at x = x 0 : ππ(π₯π₯0 +β)−ππ(π₯π₯0 ) β β→0 ππ ′ (π₯π₯0 ) = lim Differentiation rules f ´(x) = lim x → x0 f ( x) − f ( x 0 ) x − x0 ′ οΏ½ππ · ππ(π₯π₯)οΏ½ = ππ · ππ ′ (π₯π₯) ′ οΏ½ππ(π₯π₯) ± ππ(π₯π₯)οΏ½ = ππ ′ (π₯π₯) ± ππ′ (π₯π₯) ′ οΏ½ππ(π₯π₯) · ππ(π₯π₯)οΏ½ = ππ ′ (π₯π₯) · ππ(π₯π₯) + ππ(π₯π₯) · ππ′(π₯π₯) ′ οΏ½πποΏ½ππ(π₯π₯)οΏ½οΏ½ = ππ′(ππ(π₯π₯)) · ππ′(π₯π₯) Function The derivative ln(x) 1 π₯π₯ ππ π₯π₯ 8 ππ π₯π₯ ππ ππ·π₯π₯ ππ · ππ ππ·π₯π₯ ππ π₯π₯ ππ π₯π₯ · ln (ππ) π₯π₯ ππ ππ · π₯π₯ ππ−1 √π₯π₯ 1 2√π₯π₯ 1 π₯π₯ − cos (π₯π₯) sin (π₯π₯) k (constant) 1 π₯π₯ 2 − sin(π₯π₯) cos(π₯π₯) 0 Make a Good Start Course, SDU-S, DWN, 2020 Integral calculus Function Indefinite integral (antiderivative) (+ an arbitrary constant) ππ π₯π₯ ππ π₯π₯ ππ ππ·π₯π₯ 1 ππ π₯π₯ ππ 1 π₯π₯ ππ+1 ππ+1 √π₯π₯ 2 π₯π₯√π₯π₯ 3 ππ π₯π₯ 1 π₯π₯ cos (π₯π₯) sin (π₯π₯) k (constant) · ππ ππ·π₯π₯ ππ π₯π₯ ln (ππ) ln|x| sin (π₯π₯) −cos (π₯π₯) k· π₯π₯ Integration rules 9 F (x) is an antiderivative of f Indefinite integrals ∫ ππ(π₯π₯) ππππ = πΉπΉ(π₯π₯) + ππ ∫ ππ · ππ(π₯π₯) ππππ = ππ · ∫ ππ(π₯π₯) ππππ ∫ ππ(π₯π₯) ± ππ(π₯π₯) ππππ = ∫ ππ(π₯π₯) ππππ ± ∫ ππ(π₯π₯) ππππ Integration by substitution: ∫ ππ(ππ(π₯π₯)) · ππ′(π₯π₯) ππππ = ∫ ππ(π‘π‘)ππππ Integration by parts: ∫ f ( x) ⋅ g ( x)dx = F ( x) ⋅ g ( x) − ∫ F ( x) ⋅ g´(x)dx Make a Good Start Course, SDU-S, DWN, 2020 , t = g (x) ππ ∫ππ ππ(π₯π₯) ππππ = [πΉπΉ(π₯π₯)]ππππ = πΉπΉ(ππ) − πΉπΉ(ππ) Definite integrals: ππ ππ ππ οΏ½ ππ(π₯π₯) ππππ = οΏ½ ππ(π₯π₯) ππππ + οΏ½ ππ(π₯π₯) ππππ ππ ππ ππ ππ ππ ∫ππ ππ · ππ(π₯π₯) ππππ = ππ · ∫ππ ππ(π₯π₯) ππππ ππ ππ ππ ∫ππ ππ(π₯π₯) ± ππ(π₯π₯) ππππ = ∫ππ ππ(π₯π₯) ππππ ± ∫ππ ππ(π₯π₯) ππππ ππ Integration by substitution: ππ(ππ) ππ(ππ) ∫ππ ππ(ππ(π₯π₯)) · ππ′(π₯π₯) ππππ = ∫ππ(ππ) ππ(π‘π‘) ππππ = [πΉπΉ(π‘π‘)]ππ(ππ) = πΉπΉοΏ½ππ(ππ)οΏ½ − πΉπΉοΏ½ππ(ππ)οΏ½ b ∫ f ( x) ⋅ g ( x)dx = [ F ( x) ⋅ g ( x)] Integration by parts: b a a − b ∫ F ( x) ⋅ g´( x)dx a Volume of solid of revolution b Vx= π ⋅ ∫ f ( x) 2 dx 10 a Differential Equations Equation Solution π¦π¦ ′ = β(π₯π₯) π¦π¦ = ∫ β(π₯π₯)dx π¦π¦ ′ = ππ − ππ · π¦π¦ π¦π¦ = + ππ · ππ −ππ·π₯π₯ π¦π¦ ′ = ππ · π¦π¦ π¦π¦ ′ = π¦π¦ · (ππ − ππ · π¦π¦) π¦π¦ ′ = ππ · π¦π¦ · (ππ − π¦π¦) π¦π¦ ′ + ππ(π₯π₯) · π¦π¦ = β(π₯π₯) Separation of variables: ( y´ = dy dx x : the independent variable) π¦π¦ = ππ · ππ ππ·π₯π₯ ππ ππ π¦π¦ = π¦π¦ = ππ ππ 1+ππ·ππ −ππ·π₯π₯ ππ 1+ππ·ππ −ππ·ππ·π₯π₯ π¦π¦ = ππ −πΊπΊ(π₯π₯) ∫ β(π₯π₯) · ππ πΊπΊ(π₯π₯) ππππ G ( x) : an antiderivative of g ( x) y´= h( x ) ⋅ g ( y ) ⇔ 1 ∫ g ( y) dy =∫ h( x) dx Make a Good Start Course, SDU-S, DWN, 2020 Geometry Equiangular Triangles ππ1 ππ1 ππ1 = = = ππ ππ ππ ππ Right-angled Triangle B c a 11 A b C Pythagoras cosine, sine & tangent in right-angled triangles ππ2 + ππ 2 = ππ 2 cos(π΄π΄) = sin(π΄π΄) = tan(π΄π΄) = ππ ππ ππ ππ ππ ππ Make a Good Start Course, SDU-S, DWN, 2020 Any triangle B c a A b C Sine rule ππ ππ ππ = = sin (π΄π΄) sin (π΅π΅) sin(πΆπΆ) Cosine rule ππ 2 = ππ2 + ππ 2 − 2 · ππ · ππ ππππππ(πΆπΆ) πΆπΆ = cos−1 οΏ½ Area of triangle ππ = Circle Area of circle ππ2 + ππ 2 − ππ 2 οΏ½ 2ππππ 1 · ππ · ππ · π π π π π π (πΆπΆ) 2 π΄π΄ = ππ · ππ 2 Circumference of circle ππ = 2 · ππ · ππ Geometry in 2 D Distance between 2 points π΄π΄(π₯π₯1 , π¦π¦1 ) og π΅π΅(π₯π₯2 , π¦π¦2 ) AB = οΏ½(π₯π₯1 − π₯π₯2 )2 + (π¦π¦1 − π¦π¦2 )2 The midpoint M between two points π΄π΄(π₯π₯1 , π¦π¦1 ) og π΅π΅(π₯π₯2 , π¦π¦2 ) π₯π₯1 +π₯π₯2 2 M =οΏ½ , π¦π¦1 +π¦π¦2 οΏ½ 2 Make a Good Start Course, SDU-S, DWN, 2020 12 Vectors in 2 D → Length of a vector ππβ |ππβ| = οΏ½ππ12 + ππ22 a ο£Ά a =  1 ο£·ο£· ο£ a2 ο£Έ Coordinates of a vector ππ1 ππ · ππ1 ππ οΏ½ππ οΏ½ = οΏ½ οΏ½ ππ · ππ2 2 Multiplication by a scalar ∧ → ∧ → → Unit vector a in the direction of a : a= → a → (the concept ”tværvektor” is not used outside DK) a Sum of vectors / difference of vectors ππ1 ππ ππ ± ππ1 ππβ ± πποΏ½β = οΏ½ππ οΏ½ ± οΏ½ 1 οΏ½ = οΏ½ 1 οΏ½ ππ2 ππ2 ± ππ2 2 Dot product πποΏ½β ππβ · πποΏ½β = ππ1 ππ1 + ππ2 ππ2 → → → 2 a⋅a= a v ππβ ππβ · πποΏ½β = |ππβ||πποΏ½β| · cos (π£π£) v is the angle between ππβ and πποΏ½β. Perpendicular vectors ππβ ππβ · πποΏ½β = 0 ⇔ ππβ ⊥ πποΏ½β πποΏ½β Projection of πποΏ½β on ππβ πποΏ½β πποΏ½βππ ππβ πποΏ½βππ = Length of projection vector ππβ · πποΏ½β ππβ |ππβ|2 |πποΏ½βππ | = Make a Good Start Course, SDU-S, DWN, 2020 οΏ½ππβ · πποΏ½βοΏ½ |ππβ| 13 Determinant of vectors detοΏ½ππβ, πποΏ½βοΏ½ = ππ1 ππ2 − ππ2 ππ1 = οΏ½ ππ1 ππ2 detοΏ½ππβ, πποΏ½βοΏ½ = |ππβ||πποΏ½β| · sin(π£π£) ππ1 οΏ½ ππ2 v is the angle from ππβ to πποΏ½β (positive anticlockwise and negative clockwise) Parallel vectors: detοΏ½ππβ, πποΏ½βοΏ½ = 0 ⇔ ππβ β₯ πποΏ½β Area of the parallelogram spanned by ππβ og πποΏ½β π΄π΄ = |det (ππβ, πποΏ½β)| πποΏ½β ππβ 14 Make a Good Start Course, SDU-S, DWN, 2020 Vectors in 3 D Most definitions and theorems concerning vectors in 2 D can be transferred to 3 D by adding a 3rd coordinate. Specific definitions and theorems for vectors in 3 D: Vektor product (cross product) ππβ × πποΏ½β πποΏ½β v ππ οΏ½ 2 ππ β 3 ππ ππβ × πποΏ½β = βοΏ½ 3 β ππ1 ππ οΏ½ 1 β ππ2 Lenght of a vector product ππ2 οΏ½ ππ3 β ππ3 β οΏ½ ππ1 β ππ1 οΏ½ ππ2 β οΏ½ππβ × πποΏ½βοΏ½ = |ππβ||πποΏ½β|sin (π£π£) ππβ v is the angle between ππβ and πποΏ½β . ππβ × πποΏ½β Area of the parallelogram spanned by ππβ and πποΏ½β π΄π΄ = οΏ½ππβ × πποΏ½βοΏ½ πποΏ½β v ππβ Geometry in 3 D Sphere Equation of a sphere with centre πΆπΆ(π₯π₯0 , π¦π¦0 , π§π§0 ) and radius r (π₯π₯ − π₯π₯0 )2 + (π¦π¦ − π¦π¦0 )2 + (π§π§ − π§π§0 )2 = ππ 2 Make a Good Start Course, SDU-S, DWN, 2020 15 Line in 3 D Parametric form of the line l through ππ0 (π₯π₯0 , π¦π¦0 , π§π§0 ) with ππ1 direction vector ππβ = οΏ½ππ2 οΏ½ ππ3 π₯π₯0 ππ1 π₯π₯ οΏ½π¦π¦οΏ½ = οΏ½π¦π¦0 οΏ½ + π‘π‘ οΏ½ππ2 οΏ½ π§π§0 ππ3 π§π§ Distance between the point P and the line l intersecting the point ππ0 and having the direction vector ππβ P ππβ ππ0 ππππππππ(ππ, ππ) = l |ππβ × οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½β ππ0 ππ| |ππβ| 16 Plane in 3 D πποΏ½β α P0 The equation of a plane α through ππ0 (π₯π₯0 , π¦π¦0 , π§π§0 ) where ππ πποΏ½β = οΏ½πποΏ½ is a vector perpendicular to the plane ππ ππ · (π₯π₯ − π₯π₯0 ) + ππ · (π¦π¦ − π¦π¦0 ) + ππ · (π§π§ − π§π§0 ) = 0 P α Distance between the point ππ(π₯π₯1 , π¦π¦1 , π§π§1 ) and the plane α given by ππ · π₯π₯ + ππ · π¦π¦ + ππ · π§π§ + ππ = 0 ππππππππ(ππ, πΌπΌ) = Make a Good Start Course, SDU-S, DWN, 2020 |ππ · π₯π₯1 + ππ · π¦π¦1 + ππ · π§π§1 + ππ| √ππ2 + ππ 2 + ππ 2
