Day 3 Review of Basic Calculus

Transcription

1 Day 3 Review of Basic Calculus Sivaram sivaambi@stanford.edu Institute of Computational and Mathematical Engineering Stanford University September 21, 2011

2 Differential Calculus Product rule Quotient rule Chain rule Exponential and logarithmic function Trigonometric and Inverse trigonometric function L Hospital rule Maximum and Minimum of a function Rolle s theorem Mean Value theorem Implicit differentiation Taylor Series Well-known Taylor series

3 Integral Calculus and Multi-Variable calculus Integral Calculus Riemann Integral Some well-known summations Fundamental theorem of calculus Integrating rational functions, exponential, logarithmic, trigonometric and inverse trigonometric functions Tricks for performing integration Leibniz s rule Arc length Multi-Variable Calculus Partial derivatives Integral to compute surface areas and volumes Change of coordinates Cylindrical polar coordinates Spherical coordinates exp( x 2 ) = π

4 Derivative The left derivative of a function f at a point a is given by f (a) f (a + h) f (a) = lim. h 0 h The right derivative of a function f at a point a is given by f (a) + f (a + h) f (a) = lim. h 0 + h The function f is said to be differentiable at the point a if we have f (a) = f (a) +. The derivative at the point a is then denoted by f (a) (or) df a. When the left and right derivatives match, the derivative can be directly obtained as f f (a + h) f (a) (a) = lim. h 0 h dz

5 Differentiation The geometrical interpretation of the the derivative at the point a is the slope of the tangent at the point a. Figure: Geometrical interpretation of the the derivative at the point x 0

6 Properties of derivatives The derivative of a constant is zero i.e. if f (x) = c, x S, then f (x) = 0, x S. f f (x + h) f (x) c c (x) = lim = lim h 0 h h 0 h = lim h 0 0 = 0. (f (x) + g(x)) = f (x) + g (x) wherever f (x) and g (x) are defined. d(x n ) = nx n 1 where n R The derivative of a polynomial of degree n is a polynomial of degree at most n 1.

7 Properties of derivatives Product rule: If f and g are two functions whose derivative exists at x = a, then (f (x)g(x)) x=a = f (a)g(a) + f (a)g (a) Quotient rule: If f and g are two functions whose derivative exists at a and we have g(a) 0, then ( f (x) g(x) ) x=a = g(a)f (a) f (a)g (a) (g(a)) 2 Chain rule: If f and g are two functions whose derivative exists at a and g(a) respectively, then df (g(x)) = df (y) dy dg(x) y=g(a) x=a

8 Exponential and Logarithmic functions = exp(x) d(exp(x)) d(log(x)) = 1 x d(exp(f (x))) = f (x) exp(f (x)) d(log(f (x))) = f (x) f (x) = ax log(a) dax = x x (1 + log(x)) x

9 Trigonometric and Inverse trigonometric functions = cos(x) d(sin(x)) = sin(x) d(cos(x)) = sec 2 (x) d(tan(x)) = sec(x) tan(x) d(sec(x)) = cosec(x) cot(x) d(cosec(x)) = cosec 2 (x) d(cot(x))

10 C 1 (S) space of functions Just as we defined the C 0 (S) space of functions we can define another space of functions, viz, C 1 (S) space of functions. A function is said to be differentiable over a set S if the function is differentiable at every point in the set S. For example, f (x) = x is differentiable over the entire real line, R. The space of all differentiable functions over a set S, whose derivative is continuous over S, is denoted by C 1 (S). For instance, we have f (x) = x { C 1 (R). 0 x 0 f (x) = exp ( ) 1 x > 0 C 1 (R). x { 0 x 0 whereas f (x) = / C x x 0 1 (R) but is in C 0 (R). { 0 x = 0 Does f (x) = x 2 sin( 1 x ) x 0 C 1 (R)? It is not hard to see that C 0 (S) C 1 (S).

11 L Hopital s Rule Suppose f and g are differentiable on ]a, b[ and g (x) 0 for x ]a, b[. Suppose f (x) g (x) A as x a If f (x) 0 and g(x) 0 as x 0, or if g(x) as x a, then Example: f (x) g(x) A as x a 1 cos x d/(1 cos x) sin x lim x 0 x = lim 2 x 0 d/(x 2 ) = lim x 0 2x = 1 2 x sin x d/(x sin x) 1 cos x lim x 0 x = lim 3 x 0 d/(x 3 ) = lim x 0 3x = lim x 0 e x 1 x x 2 /2 x 3 = lim x 0 e x 1 x 3x 2 = lim x 0 e x 1 6x = 1 6

12 Higher derivatives and C k (S) space of functions The k th left derivative of a function f at a point a is given by f k (a) f k 1 (a + h) f k 1 (a) = lim. h 0 h The k th right derivative of a function f at a point a is given by f k (a) + f k 1 (a + h) f k 1 (a) = lim. h 0 + h The function f is said to have k th derivative at the point a if we have f k (a) = f k (a) +. The k th derivative at the point a is denoted by f k (a) (or) d k f dz k a. When the left and right k th derivatives match, the k th derivative can be directly obtained as f k f k 1 (a + h) f k 1 (a) (a) = lim. h 0 h The space of all k-differentiable functions over a set S, whose k th derivative is continuous over S, is denoted by C k (S).

13 Maximum and Minimum of a continuous function Theorem: If f is a real valued continuous function on a closed and bounded set X, then the function f (x) is bounded i.e. there exists m, M R such that we have m = inf x X (f (x)) R and M = sup x X (f (x)) R. Further, the bounds are hit i.e. we have p, q X such that f (p) = m and f (q) = M. Example: f (x) = x 3 for x X = [ 1, 1]. Clearly, we have f (x is continuous on X. We get m = 1, M = 1, p = 1 and q = 1. f (x) = sin(x) for x X = [ π 4, 7π 4 ]. Clearly, we have f (x is continuous on X. We get m = 1, M = 1, p = π 2 and q = 3π 2. The closed and bounded sets on the real line are called compact sets. Hence, the theorem can be restated as A continuous function attains it minimum and maximum on a compact set

14 Maximum and Minimum of a continuous function We look at some of the counterexamples of the previous theorem when the set X is not compact. f (x) = 1 x where x X = (0, ). f (x) is continuous on (0, ). We have m = inf x X (f (x)) = 0 R. However, we have M = sup x X (f (x)) = / R. Also, there exists no p, q X such that f (p) = 0 and f (q) =. This is so since (0, ) is neither closed nor bounded and hence not compact. f (x) = 1 x+1 where x X = [0, ). f (x) is continuous on [0, ). We have M = sup x X (f (x)) = 1 R and m = inf x X (f (x)) = 0 R. Taking q = 0 X, we get f (q) = 1 = M. However, there exists no p X such that f (p) = m = 0. This is so since [0, ) is not bounded and hence not compact. f (x) = 1 x where x X = (0, 1]. f (x) is continuous on (0, 1]. We have m = inf x X (f (x)) = 1 R. However, we have M = sup x X (f (x)) = / R. Choosing p = 1 X, we have f (p) = 1 = m. But, there exists no q X such that f (q) =. This is so since (0, 1] is not closed and hence not compact.

15 Rolle s theorem Rolle s Theorem: Rolle s theorem states that a differentiable function which attains equal values at two distinct points must have a point somewhere between them where the first derivative is zero i.e. there exists a point such that the slope of the tangent is zero. The technical version is as follows. If f : [a, b] is continuous, f is differentiable on (a, b) and we have f (a) = f (b), then there exists c (a, b) such that f (c) = 0.

16 Rolle s theorem Example: f (x) = 1 x 2, x [ 1, 1]. The function satisfies the criteria for Rolle s theorem with f ( 1) = f (1) = 0. Hence, there exists a point c ( 1, 1) such that f (c) = 0. In this case, we can easily see that c = 0 is a desired point.

17 Rolle s theorem Example: If the differentiability fails at even one interior point, then the conclusion of Rolle s theorem may not hold. Consider f (x) = x, x [ 1, 1]. The function doesn t have derivative at one point namely at x = 0. Hence, even though we have f ( 1) = f (1) = 1, we do not have a point c ( 1, 1) such that f (c) = 0.

18 Mean Value theorem This is nothing but tilted version of Rolle s theorem. Tilt your head and apply Rolle s theorem to get this result. Theorem: If f : [a, b] is continuous, f is differentiable on (a, b), then there exists c (a, b) such that f (c) = f (b) f (a) b a.

19 Implicit differentiation Implicit function of the form f (x, y) = 0. For example, x 2 + y 2 = 2 is an implicit function relating y in terms of x. We could rewrite this as f (x, y) = x 2 + y 2 2 = 0. Another example is tan(x + y) = x 2 y which could be rewritten as f (x, y) = tan(x + y) x 2 y = 0. We are interested in finding the derivative dy at a specific point say (x 0, y 0 ). f dy y + f x = 0. f y f x - treat x as a contant and differentiate w.r.t y. - treat y as a contant and differentiate w.r.t x. f (x, y) = x 2 + y 2 2 = 0 = 2y dy + 2x = 0 = dy = x y

20 Taylor s Series Suppose f is a real valued function on [a, b], n is a positive integer, f (n 1) is continuous on [a, b] and f (n) (t) exists for every t ]a, b[. Let α, β be distinct points in [a, b], and define P(t) = n 1 k=0 f (k) (α) (t α) k k! Then there exists a point x between α and beta such that f (β) = P(β) + f (n) (β α)n n! For n = 1 this is nothing but the mean value theorem. In general the theorem shows that f can be approximated by a polynomial of degree n 1 and the above equation allows us to estimate the error, if we know a prioiri, bounds on f (n) (x).

21 Taylor Series The most common way of stating the Taylor s theorem is where, f (x + h) = f (x) + f (x)h + f (x)h 2 2! + + f (n) (x)h n n! R n = f (n+1) (c) (n + 1)! hn+1 for some x c x + h + R n (h) The function R n is called the remainder of order n or the error term. If R n 0 as n, we say that the Taylor series generated by f converges and write f (x + h) = k=0 f (k) (x) h k k!

22 Some well-known taylor series exp(x) = 1 + x + x 2 2! + x 3 3! + x 4 4! + x 5 sin(x) = x x 3 3! + x 5 5! x 7 7! + x 9 cos(x) = 1 x 2 2! + x 4 4! x 6 6! + x 8 For x < 1, log(1 + x) = x x x 3 3 x ! + 9! + 8! +

23 Integral A motivation of an integral of a function is to find the area enclosed by the function From the figure, the area S is given by S = b a f (x). The integral is the limit of the sum of the area of the rectangles. Figure: Area under a curve Figure: Area under a curve

24 Some well-known summations Some summations which will be of use while performing integration from first principles n = n(n+1) n 2 = n(n+1)(2n+1) n 3 = 6 ( n(n+1) 2 1 k + 2 k + 3 k + + n k = nk+1 k+1 + c kn k + c k 1 n k c r + r r n = r n+1 1 r 1 ) 2

25 Elementary Integrals 1 0 x - Partition the interval [0, 1] into n intervals where x k = k n and k goes from 0 to n. Approximating the integral from below using rectangles we get 1 0 x = lim n 1 n k=0 1 x k n = lim n n 1 k=0 k n 2 = lim n(n 1) n 2n 2 = exp(x) - Partition the interval [0, 1] into n intervals where 0 x k = k n and k goes from 0 to n. Approximating the integral from below using rectangles we get exp(x) = lim exp(x) = lim n n k=0 n 1 exp(x k ) 1 n = lim exp(1) 1 n(exp(1/n) 1) 1 0 n 1 n k=0 exp(k/n) n = (exp(1) 1) lim n exp(x) = (exp(1) 1) 1 n(exp(1/n) 1)

26 Properties of Integral b a (f 1(x) + f 2 (x)) = b a f 1(x) + b a f 2(x). b a cf (x) = c b a f (x). f 1 (x) f 2 (x), then b a f 1(x) b a f 2(x). b a f (x) = c a f (x) + b c f (x). If f (x) M, then b a f (x) = M(b a).

27 Fundamental Theorem of Calculus Part 1: If f is continuous on a closed interval [a, b], then ( d x ) f (t)dt = f (x) for all x [a, b]. a Figure: Geometric Argument Part 2: If f (x) is differentiable and f (x) is integrable over [a, b], then where [c, d] [a, b]. d c df = f (d) f (c)

28 Standard Integrals Indefinite Integral Reversed Derivative ( ) x n = x n+1 n+1 + C, n 1 d x n+1 n+1 = x n x 1 d log(x) = log(x) + C = 1 x ( ) sin(kx) = cos(kx) d k + C cos(kx) k = sin(kx) ( ) cos(kx) = sin(kx) d k + C sin(kx) k = cos(kx) ( ) tan(kx) = log(sec(kx)) d k + C log(sec(kx)) k = tan(kx) ( ) sec(kx) = log(sec(kx)+tan(kx)) d k + C log(sec(kx)+tan(kx)) k = sec(kx) ( ) csc(kx) = log(tan(kx/2)) d k + C log(tan(kx/2)) k = csc(kx) ( ) cot(kx) = log(sin(kx)) d k + C log(sin(kx)) k = cot(kx) ( ) exp(kx) = exp(kx) d k + C exp(kx) k = exp(kx) log(x) = x log x x + C (x log x x) = log(x) d

29 Integration tricks Methodof substitution: I = 1 + x. Let x = tan(θ). Then = 2 sec2 (θ)dθ. Hence, we get I = I = sec 2 (θ)dθ 1 + tan 2 (θ) = sec 2 (θ)dθ sec 2 (θ) = dθ = θ = tan 1 (x) + C.. Let x = sin(θ). Then = cos(θ)dθ. Hence, 1 x 2 I = cos(θ)dθ 1 sin 2 (θ) = cos(θ)dθ cos(θ) = dθ = θ = sin 1 (x) + C. Method of partial fractions: I = 1 x = 1 ( x ) x 1 x x I = (x 1) = 3 (x 1) = 3 1 (x 1) + 2 = 1 ( ) x log + C. x 1 1 (x 1) 3 = 1 x 1 1 2(x 1) 2 + C.

30 Integration by parts: Integration tricks Relies on product rule for derivatives i.e. u dv = I = d(uv) d(uv) = u dv + v du. v du = u(x)v(x) v du. log(x). Choose u = log(x) and dv = 1 = v(x) = x. I = I = I = log(x) = x log(x) x 1 x = x log(x) = x log(x) x. x exp(x). Choose u = x, dv = exp(x) = v(x) = exp(x). x d exp(x) = x exp(x) exp(x) 1 = x exp(x) exp(x).

31 ( ) d b(t) b(t) f (x, t) = dt a(t) a(t) Leibniz s rule f (x, t) t +f (b(t), t) db(t) f (a(t), t) da(t) dt dt Example: f (x, t) = x 2 + t 3, a(t) = 1, b(t) = t, then we have b(t) a(t) Hence, We have b(t) a(t) t 1 f (x, t) = t ( ) d b(t) f (x, t) dt a(t) f (x, t) 1 ( x (x 2 + t 3 3 ) = 3 + xt3 = 3t 2, db(t) dt = d(t4 2 3 t3 1 2 ) dt = 1 and da(t) dt ) x=t x=1 = t t = 4t 3 2t 2. = 0. Hence, t f (x, t) + f (b(t), t) db(t) f (a(t), t) da(t) = t dt dt 3t 2 + t 2 + t 3 = 3t 2 t 3t 2 + t 2 + t 3 = 4t 3 2t 2.

32 Computing Arc length If S is the arc-length, then S Figure: Arclength allparts ( x)2 + ( y) 2. Taking the limit, xfinal xfinal ( ) 2 dy the arc-length is S = ()2 + (dy) 2 = 1 +. x start x start Example: Length of the curve y = 2 3 x 3/2 from 0 to 1. We get dy = x. Hence, S = 1 + x = (1 + x)3/2 0 3 = 2 ( 2 )

33 Computing Areas Say we want to compute the area enclosed by the ellipse x 2 a 2 + y 2 b 2 = 1. Small strip parallel to the Y -axis located at x of thickness x. The length of the strip is 2b 1 x 2 a. 2 The area is then well approximated by A 2b a In the limit, we get A = 2b 1 x 2 a 2. x = a sin(θ) gives us A = π/2 π/2 a 2b cos(θ)a cos(θ)dθ = 2ab π/2 π/2 1 x 2 a 2 x. cos 2 (θ)dθ = πab.

34 Partial Derivatives Extend the idea of a derivative to high dimensions For instance, if f (x, y) is a function of two variables, then f x = lim f (x + h, y) f (x, y) h 0 h f y = lim h 0 The cross-derivatives x the nice functions. ( f y f (x, y + h) f (x, y) h ) ( ) f, are equal for most of y x

35 Partial Derivatives Example: Suppose f : R 3 R and f (x, y, z) = exp(xy) sin(x) + x 2 y 4 cos 2 (z). f x = exp(xy) cos(x) + y exp(xy) sin(x) + 2xy 4 cos 2 (z) f y = x exp(xy) sin(x) + 4x 2 y 3 cos 2 (z) f z = x 2 y 4 sin(2z) ( ) f = 2xy 4 sin(2z) = z x x ) z ( f y ( ) f z ) = 4x 2 y 3 sin(2z) = ( f y z

36 Integration z = f (x, y) be the surface f (x k, y k ) A k gives the volume of the cuboid which approximates the volume element. Take the Riemann sum and take the limit to get the volume. V = y x 0 f (x,y) dzdy dy dz is the volume element in cartesian coordinates.

37 (x, y, z) (ρ, φ, z) Cylindrical Polar coordinates The radial distance ρ is the Euclidean distance from the Z-axis to the point P. The azimuthal angle φ is the angle made by the projection of P to the XY -plane with the X -axis. The height z is the signed distance of the point P from the XY -plane.

38 Cylindrical Polar coordinates Note that we have x = ρ cos(φ), y = ρ sin(φ), z = z. Expressing it the other-way around we get ρ = x 2 + y 2, φ = tan 1 ( y x ) and z = z. Note that x ρ = cos(φ) and ρ x = x x 2 +y = cos(φ). 2 Note that x ρ 1 ρ x Volume element in (ρ, φ, z) is given by J dρ dφ dz where x x x ρ φ z cos(φ) ρ sin(φ) 0 J = det y ρ z ρ y φ z φ y z z z = det sin(φ) ρ cos(φ) = ρ Example: Volume of a cylinder of radius r and height h is given by V = h z=0 2π r φ=0 ρ=0 ρdρdφdz = r 2 2 2π h = πr 2 h

39 (x, y, z) (r, θ, φ) Spherical coordinates The radial distance r is the Euclidean distance from the origin to the point P. The azimuthal angle φ is the angle made by the projection of P to the XY -plane with the X -axis. The inclination/polar angle is the angle between the Z axis and the line segment OP.

40 Spherical coordinates x = r sin(θ) cos(φ), y = r sin(θ) sin(φ), z = r cos(θ). r = x 2 + y 2 + z 2, φ = tan ( ) 1 y x and θ = z. x 2 +y 2 +z 2 Volume element in (r, θ, φ) is given by J dr dθ dφ where x x x r θ φ y y y J = det r θ φ = r 2 sin(θ) z r z θ z φ Example: Volume of a sphere of radius R is given by V = π θ=0 2π R φ=0 r=0 r 2 sin(θ)drdφdθ = R3 3 2π 2 = 4 3 πr3

41 Differential Calculus Integral Calculus Multi-Variable Calculus Let I = I 2 = I 2 = y= x= exp( x 2 ) = exp( x 2 ) y= x= exp( x 2 ) = π exp( y 2 )dy. exp( y 2 )dy. exp( (x 2 + y 2 ))dy. x = ρ cos(φ) and y = ρ sin(φ), we get dy = Jdρdφ where J = ρ. I 2 = φ=2π ρ= φ=0 ρ=0 ( ρ= exp( ρ 2 )ρdρdφ I 2 = 1 2 d ( exp( ρ 2 ) )) ( ) φ=2π dφ ρ=0 φ=0 I 2 = π = π Hence, we get exp( x 2 ) = π