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Calculus 1.Area Problem A1A1 A2A2 A3A3 A4A4 A = A 1 + A 2 + A 3 + A 4 A3A3 A4A4 A5A5 A 10 …… A = lim A n = πr 2 n -> ∞ A x y 0 y=x 2 x y 0 x y 0 Volume.

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Presentation on theme: "Calculus 1.Area Problem A1A1 A2A2 A3A3 A4A4 A = A 1 + A 2 + A 3 + A 4 A3A3 A4A4 A5A5 A 10 …… A = lim A n = πr 2 n -> ∞ A x y 0 y=x 2 x y 0 x y 0 Volume."— Presentation transcript:

1 Calculus 1.Area Problem A1A1 A2A2 A3A3 A4A4 A = A 1 + A 2 + A 3 + A 4 A3A3 A4A4 A5A5 A 10 …… A = lim A n = πr 2 n -> ∞ A x y 0 y=x 2 x y 0 x y 0 Volume Problem Integral Calculus

2 Calculus 2.Tangent Problem Differential Calculus x y 0 y=x 2 P k tangent line x y 0 y=x 2 P Q k PQ secant line k = lim k PQ QPQP Tangent to a curve at a point P. Slope? Calculus: deals with limits Integral calculus ↔ Differential calculus inverse problems

3 Main objects in Calculus: FUNCTIONS A function f is a rule that assigns to each element x in a set A (domain of f) exactly one element f(x) in a set B (range of f). x – independent variable, f(x) – dependent variable. domain Arange B = all possible values x1x1 x2x2 x3x3 f(x 1 ) f(x 2 ) f(x 3 ) r A(r)=πr 2 Example: black boxx f(x) inputoutput rule, machine function y x 0 y=x 2 1 1 Graph of a function: { (x,f(x)) | x  A}

4 Some properties: 1.Piecewise defined 2.Symmetric 3.Increasing/Decreasing 4.Periodic y x 0 y=x 2 1 1 y=1-x even: f(-x) = f(x), e.g. f(x) = x 2, symmetric w.r.t. y-axis odd: f(-x) = - f(x), e.g. f(x) = x 3, symmetric about the origin decreasing on I: if f(x 1 )>f(x 2 ) for any x 1 <x 2 in I increasing on I: if f(x 1 )<f(x 2 ) for any x 1 <x 2 in I y x 0 y=x 2 x2x2 x1x1 f(x 1 ) f(x 2 ) increasing on [0,∞) decreasing on (∞,0] with period T, if f(x+T) = f(x), e.g. cos(x+2π) = cos(x)

5 Some basic functions: Linear: f(x) = kx+b, graph is a line with slope k and y-intersect b, grow at constant rate Polynomial: f(x) = a n x n +a n-1 x n-1 +…+a 1 x+a 0, n ≥ 0 integer, coefficients a i – constants, i=0..n if a n ≠ 0 then n is called degree of polynomial, domain (−∞,∞) Power: f(x) = x a, a – constant. n > 0 integer if a = n  polynomial (i.e. ); if a = 1/n  root (i.e. ); if a = -n  reciprocal (i.e. ) y x 0 y=1/x hyperbola Algebraic: algebraic operations on polynomials (i.e. +, −, , ,  ) (e.g. ) Rational: f(x) = P(x)/Q(x) – ratio of two polynomials, domain: such x that Q(x) ≠ 0 (e.g., domain x ≠  2)

6 Trigonometric: f(x) = {sin(x), cos(x), tan(x), cot(x), sec(x), csc(x)}, x in radians y x 0 y=sin(x) π/2 1 π-π/2 y x 0 y=cos(x) π/2 1 π-π/2 domain: (−∞,∞) range: [-1,1] period: 2π (waves) zeros: πn for sin(x) π/2+πn for cos(x) y x 0 y=tan(x) π/2 1 π-π/2 domain: cos(x) ≠ 0 range: (−∞,∞) period: π zeros: πn for sin(x)

7 Exponential: f(x) = a x, constant a > 0 – base, x - exponent Special base e = 2.71828… Logarithmic: f(x) = log a x. Inverse of exponential: log a x = y  a y = x. lnx:= log e` x, ln e=1 Hyperbolic: certain combination of e x and e -x :

8 y x 0 y=e x /2 1/2 y=-e -x /2 y=sinhx y x 0 y=e x /2 1/2 y=coshx 1 y=e -x /2 domain: (−∞,∞) range: (−∞,∞) for sinh(x) [1,∞) for cosh(x) Application of cosh: shape of hanging wire = catenary (catena=chain in Latin) y=c+acosh(x/a) y x 0 y=tanhx 1 1 asymptote y=1 asymptote y=-1

9 Trigonometric vs. Hyperbolic  = twice the area of this region 0 (cos ,sin  ) r=1 circle 0 (cosh ,sinh  ) hyperbola Inverse hyperbolic: 0 y=sinh -1 x 0 y=cosh -1 x 1 0 y=tanh -1 x 1

10 Show that Proof. where Then Solving this quadratic equation with respect to z: But while z should be positive. Therefore

11 Limit of a function: A function f(x) is continuous at a point a if i.e. f(a) defined exists and A function f(x) is continuous on an interval if it is continuous at every point of this interval. Itermediate Value Theorem: f – continuous on [a,b]. i.e. continuous function takes on every intermediate value between the function values f(a) and f(b). ab f(a) f(b) N c

12 (tangent = touching in Latin) Tangent line has the same direction as the curve at the point of contact. y=f(x) P(a,f(a)) Q(a+h,f(a+h)) f(a+h)-f(a) h   Tangent line to the curve y=f(x) at the point P(a,f(a)) is the line through P with slope if this limit exists. The tangent line to the curve y=f(x) at the point (a,f(a)) is the line through (a,f(a)) whose slope is equal to f(a) = the derivative of f at a, i.e. If this limit exists then the function f is differentiable at a point a. Example.

13 Derivative as a function (let the point a vary): Given any x for which this limit exists  assign to x the number f(x). Example. Other notations: Th. If f is differentiable at a, then it is continuous at a. (Proof: see section 2.9) Example: f(x)=|x|, a=0 not differentiable at a, but continuous a Example: discontinuity not continuous at a  not differentiable at a a Example: vertical tangent f is continuous at a and not differentiable: a

14 Constant c: Power: Polynomial: derivative of polynomial = sum of all corresponding derivatives: Derivatives of some basic functions. Rules. Constant Multiple Rule: Sum Rule: Difference Rule: Exponential:  rate of change of any exponential function is proportional to the function itself. e is defined s.t. Product Rule: Quotient Rule: Trigonometric:

15 Hyperbolic: Chain Rule: Implicit differentiation… for inverse functions Logarithmic: Inverse hyperbolic:

16 Show that Proof. Let us differentiate both sides with respect to x and find y’: But since Therefore,


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