# Lectures_MATH_233.pdf - Definition(Definition of Derivative...

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This preview shows page 1 out of 551 pages. Unformatted text preview: Definition (Definition of Derivative) Let f : R ! R be a real valued function f (x). Then the derivative f 0 (x) is given by the following formula if the limit exists: f (x + h) h!0 h f 0 (x) = lim f (x) Definition (Definition of Derivative) Let f : R ! R be a real valued function f (x). Then the derivative f 0 (x) is given by the following formula if the limit exists: f (x + h) h!0 h f 0 (x) = lim Example f (x) = x 2 f (x) Definition (Definition of Derivative) Let f : R ! R be a real valued function f (x). Then the derivative f 0 (x) is given by the following formula if the limit exists: f (x + h) h!0 h f (x) f 0 (x) = lim Example f (x) = x 2 f (x + h) h!0 h lim f (x) = lim h!0 (x + h)2 h x2 = x 2 + 2hx + h2 h x2 Definition (Definition of Derivative) Let f : R ! R be a real valued function f (x). Then the derivative f 0 (x) is given by the following formula if the limit exists: f (x + h) h!0 h f (x) f 0 (x) = lim Example f (x) = x 2 f (x + h) h!0 h lim f (x) = lim h!0 = lim h!0 (x + h)2 h x2 = x 2 + 2hx + h2 h 2hx + h2 = 2x + h = 2x h x2 Powers, Constant and Sum Rules Let c 2 R and f (x), g (x) be real valued functions. Then: (x c )0 = cx c 1 . (cf (x))0 = cf 0 (x). (f (x) + g (x))0 = f 0 (x) + g 0 (x). Powers, Constant and Sum Rules Let c 2 R and f (x), g (x) be real valued functions. Then: (x c )0 = cx c 1 . (cf (x))0 = cf 0 (x). (f (x) + g (x))0 = f 0 (x) + g 0 (x). Example If F (x) = 10x 2 + 7x, then F 0 (x) Powers, Constant and Sum Rules Let c 2 R and f (x), g (x) be real valued functions. Then: (x c )0 = cx c 1 . (cf (x))0 = cf 0 (x). (f (x) + g (x))0 = f 0 (x) + g 0 (x). Example If F (x) = 10x 2 + 7x, then F 0 (x) = 10(2x) + 7 = 20x + 7. Exponential Rule For f (x) = e x , f 0 (x) = e x Exponential Rule For f (x) = e x , f 0 (x) = e x Derivatives of sin(x) and cos(x) sin0 (x) = cos(x) cos0 (x) = sin(x) Exponential Rule For f (x) = e x , f 0 (x) = e x Derivatives of sin(x) and cos(x) sin0 (x) = cos(x) cos0 (x) = sin(x) Example If f (x) = 5e x 3 sin(x), find f 0 (x). Exponential Rule f 0 (x) = e x For f (x) = e x , Derivatives of sin(x) and cos(x) sin0 (x) = cos(x) cos0 (x) = sin(x) Example If f (x) = 5e x 3 sin(x), find f 0 (x). f 0 (x) = (5e x 3 sin(x))0 Exponential Rule f 0 (x) = e x For f (x) = e x , Derivatives of sin(x) and cos(x) sin0 (x) = cos(x) cos0 (x) = sin(x) Example If f (x) = 5e x 3 sin(x), find f 0 (x). f 0 (x) = (5e x 3 sin(x))0 = (5e x )0 (3 sin(x))0 = 5e x 3 cos(x) Product and quotient rules (f (x)g (x))0 = f 0 (x)g (x) + f (x)g 0 (x) 0 f (x) f 0 (x)g (x) f (x)g 0 (x) = g (x) (g (x))2 Product and quotient rules (f (x)g (x))0 = f 0 (x)g (x) + f (x)g 0 (x) 0 f (x) f 0 (x)g (x) f (x)g 0 (x) = g (x) (g (x))2 Example Compute f 0 (x) for f (x) = ex 1+x 2 . Product and quotient rules (f (x)g (x))0 = f 0 (x)g (x) + f (x)g 0 (x) 0 f (x) f 0 (x)g (x) f (x)g 0 (x) = g (x) (g (x))2 Example Compute f 0 (x) for f (x) = ex 1+x 2 . Solution. f 0 (x) = d d (1 + x 2 ) dx (e x ) e x dx (1 + x 2 ) (1 + x 2 )2 Product and quotient rules (f (x)g (x))0 = f 0 (x)g (x) + f (x)g 0 (x) 0 f (x) f 0 (x)g (x) f (x)g 0 (x) = g (x) (g (x))2 Example Compute f 0 (x) for f (x) = ex 1+x 2 . Solution. f 0 (x) = = d d (1 + x 2 ) dx (e x ) e x dx (1 + x 2 ) (1 + x 2 )2 (1 + x 2 )e x e x (2x) (1 + x 2 )2 Product and quotient rules (f (x)g (x))0 = f 0 (x)g (x) + f (x)g 0 (x) 0 f (x) f 0 (x)g (x) f (x)g 0 (x) = g (x) (g (x))2 Example Compute f 0 (x) for f (x) = ex 1+x 2 . Solution. f 0 (x) = = d d (1 + x 2 ) dx (e x ) e x dx (1 + x 2 ) (1 + x 2 )2 (1 + x 2 )e x e x (2x) e x (1 x)2 = . 2 2 (1 + x ) (1 + x 2 )2 Product and quotient rules (f (x)g (x))0 = f 0 (x)g (x) + f (x)g 0 (x) 0 f (x) f 0 (x)g (x) f (x)g 0 (x) = g (x) (g (x))2 Product and quotient rules (f (x)g (x))0 = f 0 (x)g (x) + f (x)g 0 (x) 0 f (x) f 0 (x)g (x) f (x)g 0 (x) = g (x) (g (x))2 Example Compute f 0 (x) for f (x) = tan(x) = sin(x) . cos(x) Product and quotient rules (f (x)g (x))0 = f 0 (x)g (x) + f (x)g 0 (x) 0 f (x) f 0 (x)g (x) f (x)g 0 (x) = g (x) (g (x))2 Example Compute f 0 (x) for f (x) = tan(x) = Solution (using the quotient rule): ✓ ◆0 sin(x) tan0 (x) = cos(x) sin(x) . cos(x) Product and quotient rules (f (x)g (x))0 = f 0 (x)g (x) + f (x)g 0 (x) 0 f (x) f 0 (x)g (x) f (x)g 0 (x) = g (x) (g (x))2 Example Compute f 0 (x) for f (x) = tan(x) = sin(x) . cos(x) Solution (using the quotient rule): ✓ ◆0 sin(x) cos(x) cos(x) sin(x)( sin(x) tan0 (x) = = cos(x) cos2 (x) Product and quotient rules (f (x)g (x))0 = f 0 (x)g (x) + f (x)g 0 (x) 0 f (x) f 0 (x)g (x) f (x)g 0 (x) = g (x) (g (x))2 Example Compute f 0 (x) for f (x) = tan(x) = sin(x) . cos(x) Solution (using the quotient rule): ✓ ◆0 sin(x) cos(x) cos(x) sin(x)( sin(x) tan0 (x) = = cos(x) cos2 (x) = cos2 (x) + sin2 (x) cos2 (x) Product and quotient rules (f (x)g (x))0 = f 0 (x)g (x) + f (x)g 0 (x) 0 f (x) f 0 (x)g (x) f (x)g 0 (x) = g (x) (g (x))2 Example Compute f 0 (x) for f (x) = tan(x) = sin(x) . cos(x) Solution (using the quotient rule): ✓ ◆0 sin(x) cos(x) cos(x) sin(x)( sin(x) tan0 (x) = = cos(x) cos2 (x) = cos2 (x) + sin2 (x) 1 = 2 cos (x) cos2 (x) Product and quotient rules (f (x)g (x))0 = f 0 (x)g (x) + f (x)g 0 (x) 0 f (x) f 0 (x)g (x) f (x)g 0 (x) = g (x) (g (x))2 Example Compute f 0 (x) for f (x) = tan(x) = sin(x) . cos(x) Solution (using the quotient rule): ✓ ◆0 sin(x) cos(x) cos(x) sin(x)( sin(x) tan0 (x) = = cos(x) cos2 (x) = cos2 (x) + sin2 (x) 1 = = sec2 (x). 2 cos (x) cos2 (x) Chain Rule (f g )0 (x) = f 0 (g (x)) · g 0 (x) Chain Rule (f g )0 (x) = f 0 (g (x)) · g 0 (x) Example If f (x) = e 3x x 2 , find f 0 (x). Chain Rule (f g )0 (x) = f 0 (g (x)) · g 0 (x) Example If f (x) = e 3x x 2 , find f 0 (x). f 0 (x) = (e 3x x 2 )0 = (e 3x )0 (x 2 )0 Chain Rule (f g )0 (x) = f 0 (g (x)) · g 0 (x) Example If f (x) = e 3x x 2 , find f 0 (x). f 0 (x) = (e 3x x 2 )0 = (e 3x )0 (x 2 )0 = 3e 3x 2x Chain Rule (f g )0 (x) = f 0 (g (x)) · g 0 (x) Chain Rule g )0 (x) = f 0 (g (x)) · g 0 (x) (f Example Find F 0 (x) if F (x) = e x 2 +1 . Chain Rule g )0 (x) = f 0 (g (x)) · g 0 (x) (f Example Find F 0 (x) if F (x) = e x 2 +1 . Solution: Let g (x) = e x and f (x) = x 2 + 1. Chain Rule g )0 (x) = f 0 (g (x)) · g 0 (x) (f Example Find F 0 (x) if F (x) = e x 2 +1 . Solution: Let g (x) = e x and f (x) = x 2 + 1. Then F (x) = (g f )(x). Chain Rule g )0 (x) = f 0 (g (x)) · g 0 (x) (f Example Find F 0 (x) if F (x) = e x 2 +1 . Solution: Let g (x) = e x and f (x) = x 2 + 1. Then F (x) = (g So, f )(x). F 0 (x) = g 0 (f (x)) · f 0 (x) Chain Rule g )0 (x) = f 0 (g (x)) · g 0 (x) (f Example Find F 0 (x) if F (x) = e x 2 +1 . Solution: Let g (x) = e x and f (x) = x 2 + 1. Then F (x) = (g So, f )(x). F 0 (x) = g 0 (f (x)) · f 0 (x) = e x 2 +1 (2x) Chain Rule (f g )0 (x) = f 0 (g (x)) · g 0 (x) Chain Rule (f g )0 (x) = f 0 (g (x)) · g 0 (x) Example If f (x) = tan 1 (x), show f 0 (x) = 1 . 1 + x2 Chain Rule (f g )0 (x) = f 0 (g (x)) · g 0 (x) Example If f (x) = tan 1 (x), show f 0 (x) = 1 . 1 + x2 Solution: Draw the triangle with angle ✓ such that x = tan(✓). This side x, adjacent side 1 and hypotenuse p triangle has opposite 1 + x 2 , and so, sec2 [✓ = tan 1 (x)] = 1 + x 2 . Chain Rule (f g )0 (x) = f 0 (g (x)) · g 0 (x) Example If f (x) = tan 1 (x), show f 0 (x) = 1 . 1 + x2 Solution: Draw the triangle with angle ✓ such that x = tan(✓). This side x, adjacent side 1 and hypotenuse p triangle has opposite 1 + x 2 , and so, sec2 [✓ = tan 1 (x)] = 1 + x 2 . Be definition of inverse function, x = tan(tan 1 (x). Chain Rule (f g )0 (x) = f 0 (g (x)) · g 0 (x) Example If f (x) = tan 1 (x), show f 0 (x) = 1 . 1 + x2 Solution: Draw the triangle with angle ✓ such that x = tan(✓). This side x, adjacent side 1 and hypotenuse p triangle has opposite 1 + x 2 , and so, sec2 [✓ = tan 1 (x)] = 1 + x 2 . Be definition of inverse function, x = tan(tan 1 (x). Take the derivative of each side and the chain rule gives: 1 = tan0 (tan 1 (x)) · (tan 1 0 ) (x) Chain Rule (f g )0 (x) = f 0 (g (x)) · g 0 (x) Example If f (x) = tan 1 (x), show f 0 (x) = 1 . 1 + x2 Solution: Draw the triangle with angle ✓ such that x = tan(✓). This side x, adjacent side 1 and hypotenuse p triangle has opposite 1 + x 2 , and so, sec2 [✓ = tan 1 (x)] = 1 + x 2 . Be definition of inverse function, x = tan(tan 1 (x). Take the derivative of each side and the chain rule gives: 1 = tan0 (tan = sec2 (tan 1 (x)) · (tan 1 (x)) · (tan 1 0 ) (x) 1 0 ) (x) Chain Rule g )0 (x) = f 0 (g (x)) · g 0 (x) (f Example If f (x) = tan 1 (x), show f 0 (x) = 1 . 1 + x2 Solution: Draw the triangle with angle ✓ such that x = tan(✓). This side x, adjacent side 1 and hypotenuse p triangle has opposite 1 + x 2 , and so, sec2 [✓ = tan 1 (x)] = 1 + x 2 . Be definition of inverse function, x = tan(tan 1 (x). Take the derivative of each side and the chain rule gives: 1 = tan0 (tan = sec2 (tan 1 (x)) · (tan Solving we get, (tan 1 0 ) (x) = 1 (x)) · (tan 1 0 1 0 ) (x) ) (x) = (1 + x 2 ) · (tan 1 1+x 2 . 1 0 ) (x). Derivatives of classical functions (x c )0 = cx c 1 (e x )0 = e x sin0 (x) = cos(x) cos0 (x) = sin(x) 0 tan (x) = sec2 (x) ln0 (x) = x1 , where ln(x) is the natural log. log0a (x) = 1 x ln(a) , 1 d dx (sin (x)) d 1 dx (tan (x)) = = where loga (x) is the log in base a. p 1 1 x2 1 1+x 2 Definition (Antiderivatives) A function F is called an antiderivative of f on an interval I if Z 0 F (x) = f (x) for all x in I. We use f (x) dx to denote F (x). Definition (Antiderivatives) A function F is called an antiderivative of f on an interval I if Z 0 F (x) = f (x) for all x in I. We use f (x) dx to denote F (x). Example Let f (x) = x 2 . Then an antiderivative F (x) for x 2 is F (x) = x3 3 . Definition (Antiderivatives) A function F is called an antiderivative of f on an interval I if Z 0 F (x) = f (x) for all x in I. We use f (x) dx to denote F (x). Example Let f (x) = x 2 . Then an antiderivative F (x) for x 2 is F (x) = x3 3 . Theorem If F is an antiderivative of f on an interval I, then the most general antiderivative of f on I is F (x) + C where C is an arbitrary constant. Table of Anti-di↵erentiation Formulas Let F (x), G (x) be the antiderivative respectively for the functions f (x), g (x). Function c · f (x) f (x) + g (x) x n (n 6= 1) 1 x x e cos x Particular antiderivative c · F (x) F (x) + G (x) x n+1 n+1 ln |x| ex sin x Table of Anti-di↵erentiation Formulas Function sin x sec2 x sec x tan x p 1 1 x2 1 1+x 2 Particular antiderivative cos x tan x sec x sin 1 x tan 1 x Area under y = x 2 from 0 to 1. Example Use rectangles to estimate the area under the parabola y = x 2 from 0 to 1. Area estimate using right end points 1 R4 = · 4 Note area A < 15 32 ✓ ◆2 ✓ ◆2 ✓ ◆2 1 1 1 1 3 1 15 + · + · + · 12 = 4 4 2 4 4 4 32 = .46875 Area estimate using left end points L4 = 1 2 1 ·0 + · 4 4 ✓ ◆2 ✓ ◆2 ✓ ◆2 1 1 1 1 3 7 + · + · = = .21875 4 4 2 4 4 32 Note area A satisfies .21875 A .46875 Area definition using right end points Definition The area A of the region S that lies under the graph of the continuous function f is the limit of the sum of the areas of approximating rectangles: A = lim Rn n!1 = lim [f (x1 ) x + f (x2 ) x + · · · + f (xn ) x] n!1 Note in the above definition that if I = [a, b], then x= b a n , where n is the number of rectangles or divisions. Definition of a Definite Integral Definition If f is a continuous function defined for a x b, we divide the interval [a, b] into n subintervals of equal width x = (b a)/n. We let x0 (= a), x1 , x2 , . . . , xn (= b) be the endpoints of these subintervals and we let x1⇤ , x2⇤ , . . . , xn⇤ be any sample points in these subintervals, so xi⇤ lies in the i-th subinterval [xi 1,xi ]. Then the definite integral of f from a to b is Z b f (x) dx = lim a n!1 X i=1 f (xi⇤ ) x Figure: Here there are n subintervals with b = xn . Example (Trapezoidal Approximation) The best simple estimate of A with n subintervals is the ”trapezoidal approximation”: A= b a n f (a)/2 + f (b)/2 + n 1 X i=1 f (xi ) ! Properties of the Integral Z b Z [f (x) + g (x)] dx = a Z Z b c · f (x) dx = c a Z a f (x) dx + a Z b g (x) dx a b f (x) dx, where c is any constant a b [f (x) b g (x)] dx = Z b f (x) dx a Z b g (x) dx a Definition A function F defined on real valued functions is said to be linear if given two such functions, f , g , F (f + g ) = F (f ) + F (g ) For c 2 R, F (c · f ) = c · F (f ). Question Why do you think integration is said to be linear? Theorem (Fundamental Theorem of Calculus, Part 1) If f is continuous on [a, b], then the function g defined by Z x g (x) = f (t) dt a x b a is continuous on [a, b] and di↵erentiable on (a, b), and g 0 (x) = f (x). Theorem (Fundamental Theorem of Calculus, Part 2) If f is continuous on [a, b], then Z b f (x) dx = F (b) F (a), a where F (x) is any antiderivative of f (x), that is, a function such that F 0 (x) = f (x). Application of FTC Example Evaluate Z 6 3 1 dx. x Application of FTC Example Evaluate Z 6 3 1 dx. x Solution: An antiderivative for x1 is F (x) = ln x. So, by FTC, Z 6 1 dx = F (6) F (3) = ln 6 x 3 ln 3. Example Evaluate Z 3 1 e x dx. Example Evaluate Z 3 e x dx. 1 Solution: Note that an antiderivative for e x is F (x) = e x . So, by FTC, Z 3 e x dx = F (3) F (1) = e 3 1 e. Example Find area A under the cosine curve from 0 to b, where 0 b ⇡ 2. Example Find area A under the cosine curve from 0 to b, where 0 b ⇡ 2. Solution: Since an antiderivative of f (x) = cos(x) is F (x) = sin(x), we have Z b b A= cos(x) dx = sin(x) = sin(b) sin(0) = sin(b). 0 0 Theorem (Fundamental Theorem of Calculus) Suppose f is continuous on [a, b]. Z x If g (x) = f (t) dt, then g 0 (x) = f (x). Z a b f (x) dx = F (b) a F (a), where F (x) is any antiderivative of f (x), that is, F 0 (x) = f (x). Notation: Indefinite integral R f (x) dx = F (x) means F 0 (x) = f (x). R We use the notation f (x) dx to denote an antiderivative for f (x) and it is called an indefinite integral. A definite integral has the form: Z b f (x) dx = a Z b f (x) dx = F (b) a F (a) Table of Indefinite Integrals Z Z c · f (x) dx = c · f (x) dx Z [f (x) + g (x)] dx = Z Z Z f (x) dx + Z g (x) dx k dx = kx + C x n dx = Z Z x n+1 +C n+1 (n 6= 1 dx = ln |x| + C x e x dx = e x + C 1) The Substitution Rule The Substitution Rule is one of the main tools used in this class for finding antiderivatives. It comes from the Chain Rule: 0 [F (g (x))] = F 0 (g (x))g 0 (x). So, Z F 0 (g (x))g 0 (x) dx = F (g (x)). Substitution Rule: If u = g (x) is a di↵erentiable function whose range is an interval I and f is continuous on I, then Z Z f (g (x))g 0 (x) dx = f (u) du. Example Find Z x 3 cos(x 4 + 2) dx. Substitution Rule: If u = g (x) is a di↵erentiable function whose range is an interval I and f is continuous on I, then Z Z f (g (x))g 0 (x) dx = f (u) du. Example Find Z x 3 cos(x 4 + 2) dx. Solution: Make the substitution: u = x 4 + 2. Get du = 4x 3 dx. Z Z Z 1 1 3 4 x cos(x + 2) dx = cos u · du = cos u du 4 4 1 1 sin u + C = sin(x 4 + 2) + C . 4 4 Note at the final stage we return to the original variable x. = Example Evaluate Z p 2x + 1 dx. Example Evaluate Z p 2x + 1 dx. Solution: Let u = 2x + 1. Then du = 2 dx, so dx = du 2 . The Substitution Rule gives Z Z Z p p du 1 1 2x + 1 dx = u = u 2 du 2 2 3 3 1 u2 1 3 1 = · 3 + C = u 2 + C = (2x + 1) 2 + C . 2 2 3 3 Substitution Rule: If u = g (x) is a di↵erentiable function whose range is an interval I and f is continuous on I, then Z Z 0 f (g (x))g (x) dx = f (u) du. Example Calculate Z e 5x dx. Substitution Rule: If u = g (x) is a di↵erentiable function whose range is an interval I and f is continuous on I, then Z Z 0 f (g (x))g (x) dx = f (u) du. Example Calculate Z e 5x dx. Solution: If we let u = 5x, then du = 5 dx, so dx = Therefore Z e 5x 1 dx = 5 Z 1 5 du 1 1 e u du = e u + C = e 5x + C . 5 5 Substitution Rule: If u = g (x) is a di↵erentiable function whose range is an interval I and f is continuous on I, then Z Z 0 f (g (x))g (x) dx = f (u) du. Example Calculate R tan x dx. Substitution Rule: If u = g (x) is a di↵erentiable function whose range is an interval I and f is continuous on I, then Z Z 0 f (g (x))g (x) dx = f (u) du. Example Calculate Solution: R tan x dx. Z tan x dx = Z sin x dx. cos x This suggests substitution u = cos x, since then du = and so, sin x dx = du: Z Z Z sin x du tan x dx = dx = cos x u = ln |u| + C = ln | cos x| + C . sin x dx The Substitution Rule for Definite Integrals Theorem (Substitution Rule for Definite Integrals) If g 0 is continuous on [a, b] and f is continuous on the range of u = g (x), then Z b 0 f (g (x))g (x) dx = a Z g (b) f (u) du. g (a) Integration by parts Recall the method of substitution came from the chain rule. Integration by parts comes from the product rule: d [f (x)g (x)] = f (x)g 0 (x) + g (x)f 0 (x). dx So, f (x)g 0 (x) = (f (x)g (x))0 g (x)f 0 (x). Thus, after taking antiderivatives, we get Z Z 0 f (x)g (x) dx = f (x)g (x) g (x)f 0 (x) dx. Integration by parts Z f (x)g 0 (x) dx = f (x)g (x) Z f 0 (x)g (x) dx The above is called the formula for integration by parts. If we let u = f (x) and v = g (x), then du = f 0 (x) dx and dv = g 0 (x) dx. Z Z So the formula becomes: u dv = uv v du. Strategy for using integration by parts Recall the integration by parts formula: Z Z u dv = uv v du. To apply this formula we must choose dv so that we can integrate it! Frequently, we choose u so that the derivative of u is simpler than u. If both properties hold, then you have made the correct choice. R R Examples using strategy: u dv = uv v du Z xe x dx : Choose u = x and dv = ex dx Z Z Z Z t 2 e t dt : Choose u = t2 and dv = et dt ln x dx : Choose u = ln x and dv = dx x sin x dx : u = x and dv = sin x dx x 2 sin 2x dx : u = x2 and dv = sin 2x dx R u dv = uv Example Find R v du Z xe x dx. R u dv = uv Example Find R v du Z xe x dx. Solution: Let dv = ex dx. u=x Then du = dx Integrating by parts gives Z xe x dx = xe x Z v = ex . e x dx = xe x ex + C . Example Evaluate Z ln x dx. Example Evaluate Z ln x dx. Solution: Let Then u = ln x du = dv = dx. 1 dx x Integrating by parts, we get Z ln x dx = x ln x = x ln x Z v = x. Z dx = x ln x x dx x x + C. Definition Let n be a positive integer. Then the Cartesian product of n copies of the real number line R is: Rn = R ⇥ R ⇥ . . . ⇥ R = {(a1 , a2 , . . . , an ) | aj 2 R}, which is the set of all ordered n-tuples of real numbers. Example (a) R2 = R ⇥ R = {(a1 , a2 ) | ai 2 R} is the Euclidean plane. (b) R3 = R ⇥ R ⇥ R = {(a1 , a2 , a3 ) | ai 2 R} is Euclidian three-space. Notation and coordinates There are two standard notations for points in R3 , or more generally Rn . If P 2 R3 , then P = (a1 , a2 , a3 ) for some scalars a1 , a2 , a3 2 R. The book also denotes this point by writing P(a1 , a2 , a3 ). The scalar a1 is called the x-coordinate of P, a2 is called the y -coordinate of P and a3 is called the z-coordinate of P. Example The point P = (1, 0, 7) in R3 can also be written as P(1, 0, 7). Its z-coordinate is 7. The coordinate axes and finding coordinate points The coordinate axes and finding coordinate points Plotting the points ( 4, 3, 5) and (3, 2, 6) Example What surfaces in R3 are represented by the following equations? (a) z = 3 (b) y = 5 Example What surfaces in R3 are represented by the following equations? (a) z = 3 (b) y = 5 Solution (a) z = 3 is the set {(x, y , z) | z = 3}, which is the set of points in R3 whose z-coordinate is 3. This is the horizontal plane parallel to the xy -plane and 3 units above it as in Figure (a). (b) y = 5 is the set of all points in R3 whose y -coordinate is 5. This is the vertical plane parallel to the xz-plane and 5 units to the right of it as in Figure (b). Example Describe and sketch the surface in R3 represented by the equation y = x. Example Describe and sketch the surface in R3 represented by the equation y = x. Solution The equation represents the set of points in R3 whose x- and y -coordinates are equal: {(x, x, z) | x 2 R, z 2 R}. This is a vertical plane that intersects the xy -plane in the line y = x, z = 0. The portion of this plane that lies in the first octant is sketched below. Distance Formula in Three Dimensions The d...
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