1Q1: Mid Level Math Course. Niels Walet, Fall 2000

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1 Q: Mid Level Mth Course Niels Wlet, Fll 2 Lst chnged on November 3, 2

2 Contents Revision 4. Powers, logs, exponentils Powers The product of two powers Exponentil Function The Logrithmic Function Completing the Squre Method Prtil frctions Trigonometric functions Trigonometric formule Inverse Trig Functions Polr Coordintes Differentition 5 2. Assumed knowledge First principles definition Mening s slope of curve Differentil of sum Differentil of product Differentil of quotient Function of function (chin rule) Differentil of inverse function Mxim nd minim Higher Derivtives Other techniques Implicit Differentition Logrithmic differentition Differentition of prmetric equtions Prtil derivtives Multiple prtil derivtives Appliction of differentition: Clcultion of smll errors Integrtion Wht is integrtion? Inverse of differentition Are under curve Strtegy Integrtion by substitution Type Type Integrtion by Prts Integrls of the form I = ()/(x + b) Integrls of the form I = (px + q)/(x 2 + x + b) Integrtion of rtionl Functions Integrls with squre roots in denomintor

3 CONTENTS 2 4 Applictions of Integrtion Finding res Are between two curves Bsic Derivtion of Are Formul Volumes of Revolution Centroids (First moment of re) First moment of the re bout the y xis First Moment of the re bout the x xis Centroid of plne re Mening of the centroid Second Moment of Are Differentil Equtions introduction Some specil types of DE Seprble type liner type Homogeneous Type Bernouilli s Eqution Vectors in 2-spce nd 3-spce 5 6. solid geometry Direction cosines Vectors nd vector rithmetic Wht is vector? Grphicl representtion Equlity nd line of ction Negtive of vector mgnitude of vector Multipliction by sclr Unit vectors Vector Addition Tringle Lw Prllelogrm Lw Generl Addition Associtivity Closed sets of vectors: null vector Subtrction of vectors Zero or Null Vector Vectors: Component Form Components in 2 dimensions Vectors in 3 dimensions Sum nd Difference of vectors in Component Form Direction cosines Scling of Vector Vector products The sclr or dot product Component form of dot product Angle between two vectors Work The vector product triple products Component Form Vector Triple Product The stright Line Stndrd form of L The vector eqution of plne Crtesin Form Perpendiculr distnce from origin to plne π

4 CONTENTS Angle between two plnes Perpendiculr distnce from point to plne ppliction

5 Chpter Revision. Powers, logs, exponentils See Stroud, F.4.. Powers Here we summrise the properties of the powers...2 The product of two powers First of ll the product of two powers, x y = x+y, (.) e.g., = 3 8, nd 3 /2 3 3/2 = 3 2 (we see tht x nd y do not hve to be integers (whole numbers)). Question: Evlute 5 7/ 5 3/. The power of power If we tke the power of power, we multiply the exponents, ( x ) y = xy (.2) e.g., (2 3 ) 2 = 8 2 = 64 = = 2 6 = 64. This gin works for x, y not integers. Question: Evlute 2 /4 4 3/8. Reltion with roots If the exponent is /n we re tking the nth root of, /n = n, (.3) e.g., 2 2 = 2, 2 3 = 3 2. If x = n then x n =. This cn be shown by tking both sides to the power n, x n = ( /n ) n = =. The number n is often tken to be n integer, but it does not hve to be. (E.g., (3 /9.5 ) 9.5 = 3.) Zeroth power of number If we tke number to the power zero, we find = for ny >. (.4) This follows from x = x+ = x, nd therefore =. (Note tht there is slight problem with : x = for x >. One usully defines =.) 4

6 .. POWERS, LOGS, EXPONENTIALS exp(x) exp( x) y x Figure.: A plot of the exponentil exp(x) nd exp( x). Negtive powers nd frctions If we tke number to negtive power, we write the result s frction involving positive power, since x x = x+x = =. Therefore x = x. E.g., 2 = /2. Common error Remember tht nd not As n exmple, = 2 8 = 64, but = = Exponentil Function x = x (.5) x+y = x y CORRECT!!!, (.6) x+y x + y WRONG!!! (.7) The exponentil function is specil cse of power, where y = e x, with e = (Euler s number). One lso writes exp(x) insted of e x. As we cn see from Fig.., e x is never less thn for ny x. From the properties of powers we know tht e x = e. This function is lso shown in Fig.., nd is positive s well. x Differentil (derivtive w.r.t. x) of e x is e x, i.e., de x = ex. (This is the only function with the property tht the derivtive equls the function itself.) If y = e x then dy = ex (this is form of the chin rule, which will be discussed lter), e.g., if y = 3e 7x then dy = 3 7e7x = 2e 7x...4 The Logrithmic Function Reltion between Logs nd Exponentils The inverse f of function f is defined such tht if y = f(x), then x = f (y). The functions ln(x) nd exp(x) re the inverse functions of ech other. This mens tht if y = ln(x) then x = e y. The reverse is lso true, if x = e y then y = ln x. Clerly it follows tht, using these reltions, exp(ln x) = e ln x = e y = x, ln(exp y) = ln(e y ) = ln x = y.

7 .2. COMPLETING THE SQUARE 6 5 ln(x) /x y x Figure.2: A plot of the nturl logrithm ln(x) nd its derivtive /x. A grph of the logrithm is shown in Fig..2. If we swp the x nd y xes, we recognise the exponentil. Normlly we use logs to bse e (inverse of e x )- clled nturl logrithms, hence the nme ln(x), but we lso write Logs to other bses log(x) = ln(x). Just s y = ln x x = e y for the logrithm corresponding to bse e (i.e., the inverse of e x ) for other bses we hve y = log x x = y. Here we use the nottion tht if we men log to bse, sy, we write log (x), i.e., if y = log (x), x = y. Chnge of bse Using this we cn chnge from one bse to nother. Let y = log x, then x = y. Now let b = ln (= log ), so = e b. Therefore x = (e b ) y = e by, so by = ln x, y = ln x b = ln x ln. Hence log x = ln x ln. Question: Determine α such tht log (x) = α log 2 (x). Differentil of ln x If y = ln x then dy = x. (Remember tht the differentil of ln x is /x, not the integrl! This is common error!) Log of product Using the fct tht e x e x 2 = e x +x 2, i.e., the product of exponents is the exponent of the sum, we conclude tht the inverse reltion holds for logrithms. Thus, the logrithm of product is the sum of the logrithms,.2 Completing the Squre ln(y y 2 ) = ln(e x e x 2 ) = ln e x +x 2 = x + x 2. Completing the squre is simple ide tht is surprisingly useful. First definition: See Stroud, F6.7-2 A polynomil is sum of powers of vrible x (sy). The degree of this polynomil is its highest power.

8 .2. COMPLETING THE SQUARE 7 Let us look t few exmples: polynomil degree () x + Also clled liner, since if we plot (b) 4x the functions y = x +, y = 4x, etc. (c) x + b we get stright line (d) x 2 + 2x + 2 (e) 7x (lso known s qudrtic) (f) x 2 + bx + c 2 (g) x 3 /9 πx 3 cubic (h) 2x A polynomil of infinite degree is usully clled n infinite power series. Any polynomil of degree 2, i.e., qudrtic, cn lwys be rerrnged to hve the form (x + b) 2 + c, s the squre of liner term plus constnt. Bringing qudrtic polynomil to this form is clled completing the squre..2. Method Completing the squre is bringing qudrtic to the form (x + b) 2 + c. In generl, if two polynomils re equl, it mens tht the coefficient of ech power of the vrible re equl, since ech power vries t different rte with the vrible. So in order to complete the squre, we must equte coefficients of powers of x on both sides. We shll do this by exmple.. Complete the squre in x 2 + x + : Put x 2 + x + = (x + b) 2 + c = x 2 + 2bx + c + b 2. Now equte coefficients of x 2 on both sides. We find =, or =. Then compre the coefficients of x. We conclude = 2b. Using = we find b = /2. Now equte the constnt term, = b 2 + c = 4 + c. We conclude tht c = 3/4. Collecting ll the results we find x 2 + x + = ( x + ) Complete the squre in 2x 2 x. Solve 2x 2 x = (x + b) 2 + c. We compre coefficients of x 2 : 2 =, x: = 2b, therefore b = /4, const: = b 2 + c, therefore c = /8. Thus ( 2x 2 x = 2 x ) It is often useful to write the constnt s c = { d 2 d 2 (if c is positive) (if c is negtive)

9 .3. PARTIAL FRACTIONS 8.3 Prtil frctions Before deling with prtil frctions, we need to define rtionl function. See Stroud, F.8 A rtionl function is one with the form f(x) = P (x)/q(x) (Q(x) = ), where P (x) nd Q(x) re polynomils. Prtil frctions is method of simplifying rtionl function. For the present we shll only consider rtionl functions where the degree of the numertor is less thn tht of the denomintor (not equl). If this is not true then we cn convert it into this form see lter (integrtion section). First fctorise the denomintor Q(x) into mixture of liner nd qudrtic fctors. This cn lwys be done without using complex numbers (use liner fctors only if possible). E.g., x 3 2x 2 + x 2 = (x 3)(x 2 + x + 4). We cn now simplify the rtionl function using prtil frctions. We do this by mens of exmples s prt of the revision. Exmple.: Simplify 3x 2x 2 x using prtil frctions. We hve fctorised the denomintor). Now put 3x 2x 2 x = 3x (x )(2x + ). 3x (x )(2x + ) = A x + B 2x + (A, B constnts). Multiply both sides by (x )(2x + 2), the denomintor of the left-hnd side. We find 3x = A(2x + ) + B(x ). (.8) Now compre coefficients on both sides. First x: 3 = 2A + B, nd for the constnt term we find = A B. We solve these simultneous liner equtions, nd find A = 2 3, B = 5 3. So 3x (x )(2x + ) = 2 3(x ) + 5 3(2x + ). Alterntively we cn find A nd B by choosing vlues for x. If we choose x = then (.8) becomes 2 = 3A + B, nd therefore A = (2/3). If we choose x = (/2) then it becomes (5/2) = A (3/2)B, nd therefore B = 5/3, in greement with our previous results. Exmple.2: Simplify x+ x(x 2 4) using prtil frctions. x + x(x 2 4) = x + x(x 2)(x + 2) = A x + (left s n exercise, A = /4, B = 3/8, C = /8). B x 2 + C x + 2,

10 .4. TRIGONOMETRIC FUNCTIONS 9 Exmple.3: Simplify x 2 (x )(x 2) 3 using prtil frctions. x 2 (x )(x 2) 3 = A (x ) + B (x 2) + C (x 2) 2 + D (x 2) 3, where we hve one term for ech power of the fctor up to the mximum. Multiply by (x )(x 2) 3 nd equte coefficients. x 2 = A(x 2) 3 + B(x )(x 2) 2 + C(x )(x 2) + D(x ). Substitute x = 2: 4 = D so D = 4. x = : = A so A =. Equte the coefficients of x 3 : = A + B + +, so B =, nd the coefficients of constnt term: = 8A 4B + 2C D, nd thus C =. x 2 (x )(x 2) 3 = (x ) + (x 2) (x 2) 3. Exmple.4: Simplify x+5 x 3 using prtil frctions. First fctorise Q(x), x 3 = (x )(x 2 + x + ). We cnnot fctorise x 2 + x + into their fctors with rel coefficients. Write Multiply with x 3, x + 5 x 3 = A x + B + Cx x 2 + x +. x + 5 = A(x 2 + x + ) + B + Cx(x ), substitute x = : 3A = 6, or A = 2. Equte coefficients of x 2 : A + C =, C = 2. Equte coefficients of the constnt prt: A B = 5, B = 3..4 Trigonometric functions x + 5 x 3 = 2 x 3 + 2x x 2 + x +. Trigonometric functions re the sine (sin(x)), cosine (cos(x)), tngent (tn(x) = sin(x)/ cos(x)), cotngent (cot(x) = / tn(x)), secns (sec(x) = / cos(x)) nd cosecns (cosec(x) = / sin(x))..4. Trigonometric formule We shll ssume tht you re fmilir with the sine nd cosine of the sum of two ngles, We lso expect you to know tht sin(a + B) = sin A cos B + cos A sin B, (.9) cos(a + B) = cos A cos B sin A cos B. (.) cos 2 θ + sin 2 θ = (.) for ll θ. Substitute A = B in Eq. (.9), nd find sin 2A = sin A cos A + cos A sin A, nd thus sin 2A = 2 sin A cos A. (.2)

11 .4. TRIGONOMETRIC FUNCTIONS sin(x) cos(x).5 y x Figure.3: A plot of the sine nd cosine. 3 tn(x) cotn(x) y x Figure.4: A plot of the tngent nd cotngent. 6 sec(x) cosec(x) 3 y x Figure.5: A plot of the secns nd cosecns.

12 .4. TRIGONOMETRIC FUNCTIONS In (.) put B = A, cos 2A = cos A cos A sin A sin A, so cos 2A = cos 2 A sin 2 A. (.3) However from (.) we hve sin 2 A = cos 2 A so we cn rewrite (.3) s cos 2A = cos 2 A ( cos 2 A) = 2 cos 2 A. (.3b) Similrly (left s exercise) cos 2A = 2 sin 2 A. (.3c) Exmple.5: Evlute cos(75 ). cos(75 ) = cos( ) = cos 45 cos 3 sin 45 sin 3 = = 2 = Note: We shll use rdins more often thn degrees, 8 = π rdins, so A = A π 8 rdins. E.g., cos 45 = cos π 4, sin 3 = sin π 6. Usully, if there is no degree sign ( ) then the ngle is specified in rdins. Exmple.6: Hint: Show from the equtions bove tht tn 2A = Other formule sin 2A cos 2A. You will sometimes need other formule such s (there re four of these), nd tn 2A = 2 tn A tn 2 A. sin A + sin B = 2 sin( A + B 2 (there re three of these). One formul you my not hve seen before is To find R nd φ we use formul (.9)nd find ) cos( A B ) 2 2 sin A cos B = sin(a + B) + sin(a B) sin x + b cos x = R sin(x + φ). sin x + b cos x = R[sin x cos φ + cos x sin φ] = R cos φ sin x + R sin φ cos x.

13 .4. TRIGONOMETRIC FUNCTIONS 2 rcsin(x) rccos(x) π y π/2 π/2 π x Figure.6: A plot of the inverse of the sine nd cosine. We equte the coefficient of sin x nd cos x on both sides of the eqution, nd find Therefore nd thus We lso find so = R cos φ, b = R sin φ. 2 + b 2 = R 2 cos 2 φr 2 + sin 2 φ = R 2, R = 2 + b 2. b/ = R sin φ = tn φ, R cos φ nd (tn will be discussed lter.) Exmple.7: tn φ = b φ = tn (b/). Express 3 sin x + 2 cos x in the form R sin(x + φ). We find R cos φ = 3, R sin φ = 2, R 2 cos 2 φ = 9, R 2 sin 2 φ = 9, R 2 (cos 2 φ + sin 2 φ) = = 3. Therefore R 2 = 3, R = 3. Also (R sin φ)/(r cos φ) = 2/3, nd thus tn φ = 2/3, φ = tn (2/3) = rctn(2/3) =.588 rdins = Inverse Trig Functions See Stroud, 9.-7

14 .5. POLAR COORDINATES 3 π rctn(x) rccot(x) π/2 y π/2 π 6 4 x Figure.7: A plot of the inverse tngent nd cotngent. rcsin The two lterntive forms y = sin (x) or y = rcsin(x) indicte tht y is n ngle whose sine is x. Exmple.8: Find sin () nd sin (/2). y = sin () mens sin(y) =. Therefore y = 9 = π/2 rds. y = sin (/2) mens sin(y)) = /2, nd thus y = 3 = π/6 rds. Note: sin 3 = /2, nd sin 5 = /2, nd sin 39 = /2, etc., so sin (x) is multivlued function. We need extr informtion, e.g., from the engineering sitution or common sense to sy which ngle we re looking t. The eqution y = sin (x) mens the sme s x = sin y, (grph of y = sin x but with xis switched), note x. Note: sin (x) is not the sme s sin x = sin(x)! The nottion is very poor here but unfortuntely very widely used. rcsin x would be better but not too common! rccos Similrly y = cos x = rccos x mens cos y = x. Once gin, x. rctn y = tn x = rctn x mens tn y = x. Exmple.9: Find x given 2 cos x = sin x. Divide by cos x: 2 = tn x, or x = tn (2)..5 Polr Coordintes See Stroud, The position of ny point P in two-dimensionl spce cn be specified by giving its (x, y) coordintes. However we could lso sy where P is by giving the distnce from the origin, nd the direction we need to go.

15 .5. POLAR COORDINATES 4 y r θ x Figure.8: The mening of polr coordintes. x=- θ r y=- Figure.9: A sketch of (-,-) nd polr coordintes. These two quntities re the polr coordintes (r, θ) of P. From right ngled tringle we see tht r cos θ = x, nd r sin θ = y, so x 2 + y 2 = r 2 cos 2 θ + r 2 sin 2 θ = r 2, nd thus r = x 2 + y 2. (N.B. We lwys tke positive squre root here!) Also y x = r sin θ r cos θ = tn θ, Therefore θ = tn (y/x). In this cse we must lwys drw digrm. The reson is tht two different ngles cn hve the sme tngent. The only relevnt once for polr coordintes re tht tn θ = tn θ 2, when θ 2 = 8 + θ = π + θ. If P is in first or second qudrnt we use θ, nd if P is in third or fourth qudrnt we use θ 2. So lwys drw little sketch! Exmple.: Find the polr coordintes corresponding to x =, y =. r = = 2, nd tn θ = y/x = =. From the sketch we see tht θ = 225 = 5π 4

16 Chpter 2 Differentition See Stroud, F.9 2. Assumed knowledge 2.. First principles definition If y = f(x) nd x increses from x to x + δx then the chnge in y is give by δy = f(x + δ) f(x), see Fig. 2.. The differentil is defined s dy = lim δy δx δx = lim f(x + δx) f(x). δx δx 2..2 Mening s slope of curve The derivtive cn lso be interpreted s the slope of curve, see Fig If the slope t given point hs n ngle θ, we find tht tn θ is dy. In other words, the line y y = tn θ (x x ) is tngent to the curve t (x, y ) Differentil of sum The differentil of sum is the sum of differentils, d(u + v) = du + dv. y+δ y y x x+δ x Figure 2.: The definition of the differentil. 5

17 2.. ASSUMED KNOWLEDGE 6 y θ x Figure 2.2: The definition of the differentil Differentil of product There exists simple rule to clculte the differentil of product, See Stroud, F E.g., if y = x 2 sin x, d(uv) = u dv + v du. dy = x2 cos(x) + 2x sin(x) Differentil of quotient In the sme wy we cn find reltion for the differentil of quotient, See Stroud, F d( u v ) = v du u dv v 2. E.g., if y = sin x x, dy x cos(x) sin(x) = x 2 = cos(x) sin(x) x x Function of function (chin rule) See Stroud, F ,7.5-8 Often we tke function of function. In such cse, where y = f(g(x)) we put z = g(x), nd find dy = df dz dz. This rule is sometimes expressed in words s the differtive of the function, times the derivtive of its rgument. Exmple 2.: Find dy for y = cos(ln x). Put z = ln x so y = cos z, dy = dy dz, dz = sin z x) = sin(ln x x.

18 2.. ASSUMED KNOWLEDGE 7 minimum mximum Figure 2.3: The mening of minimum nd mximum. Exmple 2.2: Find dy for y = sin3 (2x ). Put z = sin(2x ) so y = z 3, dy = dy dz dz = 3z2 2 cos(2x ) = 6 sin 2 (2x ) cos(2x ) Differentil of inverse function See Stroud, When we wish to clculte the differentil of n inverse function, i.e, function g such tht g(f(x)) = x, we cn use our knowledge of the derivtive of f to find tht of g. Exmple 2.3: Find the derivtive of y = sin x. We clculte dy first. dy = sin y = cos y. dy Now cos y = ± sin 2 y, but the slope of the inverse sine is lwys positive. Thus 2..8 Mxim nd minim dy = ( ) = dy x 2. See Stroud, At mximum or minimum the slope is so tht dy =. To find which cse it is, we look t d2 y, which 2 cn esily be done from plot of the slope. Exmple 2.4: Find ll mxim nd minim of y = x(3 x) nd determine their chrcter. We find tht dy = x( ) + (3 x) = 3 2x. For mximoum or minimum the slope must be. This hppens for 3 2x =, i.e., x = 3 2. For tht vlue of x, d2 y = 2. So the point x = 3/2, 2 y = 9/4 is (nd the only) mximum.

19 2.2. OTHER TECHNIQUES Higher Derivtives See Stroud, F Higher derivtives re obtined by differentition 2 or more times, d2 y 2 Exmple 2.5: = d(dy/), d3 y = d(d2 y/ 2 ) 3. y = ln x, dy = /x, d2 y = 2 x, d3 y 2 = 2 3 x, etc. 3 Exmple 2.6: The eqution for simple hrmonic motion (SHM) is d2 x dt 2 = ω 2 x. Prove tht x = (A cos ωt) + B sin ωt stisfies this eqution. We must differentite twice, strt with first derivtive, dt tht QED. d 2 x / dt2 = ω 2 A cos ω ω 2 B sin ωt = ω 2 (A cos ωt + B sin ωt) = ω 2 x. = ( ω)a sin ωt + ωb cos ωt), nd find N.B.: SHM not studied here, but in next semester. The constnts A, B cn only be obtined with extr input. 2.2 Other techniques 2.2. Implicit Differentition See Stroud, The eqution of circle x 2 + y 2 = 2 is not in the form y = f(x), (lthough it cn be rerrnged to y = ± 2 x 2 ). In this cse it is esier to find dy directly without rerrnging. Differentite both sides of the eqution x 2 + y 2 = 2 with respect to x, ssuming y to be function of x. We find 2x + dy2 =. Now use d(y 2 ) dy = 2y. (Proof: Put z = y2 - need dz, dz = dz dy dy dy dy = 2y.) So 2x + 2y =, or dy = x y. N.B.: This method usully gives dy Exmple 2.7: in terms of both x nd y. Find dy for x2 + 4x + 3xy + y 3 = 6. Differentiting both sides with respect to x we find we thus conclude tht 2x y + 3x dy dy + 3y2 =, dy y) = (2x (3x + 3y 2 ).

20 2.2. OTHER TECHNIQUES Logrithmic differentition See Stroud, If function hs lrge number of fctors it my be esier to tke the logrithm before differentiting, using the ft tht the logrithm of product is the sum of logrithms. Exmple 2.8: Find dy for y = +x b x x c. ln y = ln( + x) + ln( b x) ln(x c) = 2 ln( + x) + ln(b x) ln(x c). 2 Differentite both sides with respect to x: So nd thus dy y = 2 d ln y = y dy. ( + x) + ( ) 2 (b x) (x c). dy = ( ) 2 ( + x) (b x) 2 + x b x (x c) x c Differentition of prmetric equtions See Stroud, Some equtions cn be written in prmetric form, i.e., x = x(t), y = y(t) with t prmeter. We cn then find its differentil in terms of the prmeter. We shll study this by mens of n exmple only. Exmple 2.9: Given circle of rdius 4, use the prmetric form to find dy nd d2 y 2 t (2 3, 2). The prmetric form is which clerly stisfies (2.). Now x 2 + y 2 = 6 (2.) x = 4 cos θ, y = 4 sin θ, dy = dy dy df df = dθ dθ = 4 cos θ = cot θ. 4 sin θ Note: result is in terms of θ. When y = 2 sin θ = 2, θ = π 6 cos θ = 3 dy 2 therefore = 3 2 = 3. Now do d2 y. 2 Note: d2 y d2 y 2 dθ / d2 x 2 dθ 2 2 d 2 y 2 = d dy = d ( cot θ) = d dθ ( cot θ) dθ = cosec2 θ /dθ = (cosec2 θ)/( 4 sin θ) = (/4) cosec 3 θ. (must be in first qudrnt), nd

21 PD.nb /3/ 2.3. PARTIAL DERIVATIVES 2.5 f x, y.5 2 Π Π y Π x 2 Π PD.nb /3/ Figure 2.4: The surfce z = f(x, y) = sin(x) sin(y). f x, y Π Π y x Π Π Figure 2.5: The surfce z = f(x, y) = sin( x 2 + y 2 ). Other exmples of prmetric curves re. Ellipses x 2 / 2 + y 2 /b 2 = : put x = cos θ nd y = b sin θ, 2. Prbol x 2 / 2 y 2 /b 2 = : put x = cosh θ nd y = b sinh θ. 3. Use of time t, e.g., for x = 2t +, y = gt 2 /2 + 3t. 2.3 Prtil derivtives In Figs. 2.4 nd 2.5 we show n exmple of functions of more thn one vrible. Clerly it is very esy to pick out the minim nd mxim, since we cn mke very visul representtion of such function s surfce by the identifiction of the height z with the output of the function. In more thn two dimensions, i.e., when we hve function tht tkes three or more rguments, nd returns one vlue, we cn t use the visul nlogy. So how do we del with tht? We need to generlise derivtives to more thn one dimension. Let us study the sitution in two dimensions, nd generlise to three nd more dimensions lter. We shll look t very smll prt of the surfce, s in Fig The chnge in the fuction due to tking smll steps in both vribles simultneously (the most generl one possible), is f(x + δx, y) f(x, y) f(x + δx, y + δy) f(x + δx, y) f(x + δx, y + δy) f(x, y) = δx + δy +..., (2.2) δx δy where, just s in one dimension, the thre dots denote terms of higher power in the smll numbers δx nd δy. The expression is not symmetric under the interchnge of x nd y, nd we need to do one more step. The

22 2.3. PARTIAL DERIVATIVES 2 δy δx Figure 2.6: A smll squre δx by δy on the surfce of 2D function. c b Figure 2.7: A cuboid b c. second term cn be trnsformed bck to refer to x rther thn x + δx by mking n error proportionl to δx. But tht corresponds to term δxδy which is much smller thn the two terms lredy there if δx nd δy re smll. Thus f(x + δx, y) f(x, y) f(x, y + δy) f(x, y) f(x + δx, y + δy) f(x, y) = δx + δy +..., (2.3) δx δy This show tht generl chnge in the function cn be referred bck to chnge in the individul vribles, keeping the other fixed. In the limit of δx nd δy going to zero this gives rise to the prtil derivtives, denoted by curly. In mthemticl nottion Exmple 2.: f(x + δx, y) f(x, y) δx f(x, y + δy) f(x, y) δy Given u(x, y) = x 3 + x 2 y + xy + y 3, find u x f x = d (f(x, y)) y fixed, (2.4) f y = d dy (f(x, y)) x fixed. (2.5) nd u y. u x = 3x2 + 2xy + y +, u y = + x 2 + x + 3y 2. where the terms re the prtil derivtives of ech of the four terms in the function. From f(x + δx, y + δy) f(x, y) f x f δx + y δy we obtin tht when both prtil derivtives re zero we hve n extremum (minimum or mximum or...), where the fuction is flt in first pproximtion. We thus need to solve simultneous set of equtions for such thing to occur. Exmple 2.: Clculte the minimum surfce re for cuboid of size b c, Fig. 2.7, for fixed volume V.

23 2.3. PARTIAL DERIVATIVES 22 The volume V is simply bc. The surfce is the re of the six rectngulr sides, S = 2b+2c+2bc. The only problem is the constrint of constnt volume. We cn use tht to eliminte one of the three vribles from the problem, let us use c: c = V/(b). Thus S = 2b + 2V b + 2V. (2.6) Now differentite this with respect to nd b, nd find S S b = 2b 2V 2, = 2 2V b 2. (2.7) These must both equl zero, nd we get the equtions 2b = 2V 2, 2 = 2V b 2. (2.8) Substitute the first eqution into the l.h.s. of the second eqution, nd find = V 4, (2.9) which cn be rewritten s ( V 3 ) =. Clerly the solution = is nonsensicl (since b must be infinite), nd we find nd the minimum surfce is found for cube. = b = c = V /3, 2.3. Multiple prtil derivtives Multiple prtil derivtives re defined strightofwrdly s the prtil derivtive of the prtil derivtive, Slightly more complicted re the mixed ones, 2 f x 2 = x 3 f x 3 = x 2 f x y ( f x ( 2 f x 2 = ( f x y ( f x 2 f y x = y Even though this looks complicted, it cn be shown tht the order of differentition ctully doesn t mtter! Exmple 2.2: Find ll first nd second derivtives of f(x, y) = x sin y + cos(x y). ), ). ), ).

24 2.4. APPLICATION OF DIFFERENTIATION: CALCULATION OF SMALL ERRORS 23 f x = sin y sin(x y), f y = x cos y + sin(x y), 2 f x 2 = cos(x y), 2 f y 2 = x sin y cos(x y), 2 f y x = cos y + cos(x y), 2 f x y = cos y + cos(x y), where the lst two terms hve been clculted in the order indicted in the denomintor, nd we see the equlity lluded to bove. 2.4 Appliction of differentition: Clcultion of smll errors We know tht if y = f(x) then dy = lim f(x + δx) f(x) δx δx. Provided tht δx is smll enough (but not infinitesimlly smll) dy δy δx, so δy dy δx. Exmple 2.3: We cn mesure the volume of sphere by mesuring its rdius r nd then use the formul, V = (4/3)πr 3. Suppose we mesure r = 6.3 ±.2 m. Find the pproximte error in V. If r = 6.3 m then V = 4 3 π6.33 = 47.4 m 3. The smll error δr =.2 m will cuse n error in V given by δv dv dr δr = 4πr2 δr = 4π =. m 3. Hence V = 47.4 ±. m 3. Exmple 2.4: We mesure the height h of tower t distnce d, by mesuring d nd the ngle α with the horizontl. We then use the formul tn α = (h/d). Find error in h due to n error δα in α ssuming d to be known exctly. We solve for h, h = d tn α, dh/dα = d sec 2 α. Therefore δh dh dα δα = d sec2 αδα.

25 2.4. APPLICATION OF DIFFERENTIATION: CALCULATION OF SMALL ERRORS 24 Exmple 2.5: Given the reltion between current, voltge nd resistnce, I = V/R, with V = 25 V, R = 5 Ω, find the chnge in the current I if V chnges by V, nd R by Ω. We use the rule for smll chnges for prtil derivtives, We find Using the numericl vlues, we find δi I I δv + V R δr. I = V R, I R = V R 2. δi = = 5 = 2 25 =.8 A

26 Chpter 3 Integrtion This chpter should contin prtilly things you know -essentilly the bsis of integrtion- nd quite few new things tht build on tht, extending your knowledge of integrls nd integrtion. 3. Wht is integrtion? There re two wys of thinking bout integrtion, nd they both hve their uses. Inverse of differentition 2. Are under curve The first is more useful for indefinite integrls (when there re no boundries), nd the second is only useful for definite ones (with upper nd lower boundries). We cn use the first wy to get results for the second interprettion, but not the other wy round. 3.. Inverse of differentition See Stroud, F.. If f(x) = dg(x), then f(x) = g(x) + c, Exmple 3.: Integrte 4x 3. 4x 3 = d(x4 ), so 4x 3 = x 4 + c This type of integrtion is clled n indefinite integrl. We lwys get constnt of integrtion c for n indefinite integrl. Note: The result of f(x) is nother function of x. Mny of the integrls in the formul book were obtined this wy, Some exmples: d sin x = cos x cos x = sin x + c d ln x = = ln x + c x x d(e x ) = e x e x = ex + c d(sin x) = = x 2 x 2 sin x + c 25

27 3.2. STRATEGY 26 y y f(x) b x δ x b x Figure 3.: The re under curve Are under curve See Stroud, F..22 Plot the curve y = x, s in Fig. 3.. The shded re under curve between x = nd b equl number A. We cn clculte this s A = b (This is clled definite integrl.) This is defined s the sum from x = to x = b of the re of ll the smll strips under the curve, in the limit tht they become vnishingly thin. If we know tht f(x) = g(x) + c, thn the resulting re is A = g(b) g(). Note: There is no constnt of integrtion in definite integrl. Note: The result is number not function. Exmple 3.2: x, Find re under curve y = 2e 3x between x = nd x =. The re is given by the integrl A = = 2 2e 3x e 3x = 2 ( e 3x /3 ) = 2(e 3 /3 e 3 /3) Finl Remrk: Some integrls cn never be done in terms of known functions. Exmple 3.3: e x 2, 2 /(x + cos x). For these numericl method will give results for definite integrl, e.g., computer version of summing the re of the strips under curve. 3.2 Strtegy Since there is no gurnteed method of doing integrls we proceed s follows. Drw up list of s mny s possible stndrd integrls tht cn be done. (This hs lredy been done for you nd is given in the formul book.)

28 3.3. INTEGRATION BY SUBSTITUTION When given new integrl you must try to rerrnge into one of the stndrd types. This my involve some or ll of the following () directly rerrngement (rther trivil); (b) substitution; (c) integrtion by prts; (d) specil methods for prticulr types. 3.3 Integrtion by substitution See Stroud, 5. (no exct mtch!) This is the integrl equivlent of the chin rule. If z = f(x) nd x = g(t) then the chin rule sys, dz We cn rerrnge this by multiplying by dt to get,. (This cn be proven from the rule for finite steps, dz = dz dt = dz dt. which cn be rerrnged s δz δz δt = δt δx δx, δz = δz deltx δx. In the limit tht δx goes to zero, s it must in the integrl, we find the required result). This is the bsic formul we need to convert n integrl with respect to new vrible z. It is true s substitution rule inside the integrl, not s generl equlity Type Replce some function of x by z. Exmple 3.4: Evlute I = 2 x sin(x 2 ). Substitute z = x 2 (try this), then dz = (dz/) = 2x. We cn only use this substitution if we cn identify 2x s prt of I. To tht end write I = (/2) 2 sin x2 2x. We cn now substitute for x 2 = z nd for 2x = dz, nd thus I = 2 sin z dz, where the limits still need to be filled in. Since I is now n integrl w.r.t. z, the limits must be strting nd finishing vlues of z. At the strt, where x =, z = x 2 =. At the finish x = 2, z = 4, so I = 2 4 sin z dz = 2 [ cos z]4 = ( cos 4 + ) Note: The integrnd, (i.e., the object being integrted) chnges from x sin(x 2 ) to (/2) sin z. Prt of this chnge is due to the chnge from to dz. Note: The integrtion limits chnge (for definite integrls only).

29 3.3. INTEGRATION BY SUBSTITUTION 28 Exmple 3.5: Clculte the indefinite integrl I = 2 + x 2. Use substitution, nd tke z = (x/), dz = (/), x = z. Finlly we must substitute bck using z = x/, I = 2 + x 2 = z 2 dz = 2 ( + z 2 ) dz = dz ( + z 2 ) = tn z + c. I = ( x ) tn + c. Severl stndrd integrls cn be generlised using this substitution (left s exercise). Exmple 3.6: Evlute I = 2 x. 2 Using the substitution x = z we find dz I = = z 2 sin z + c = sin (x/ + c) Thus 2 x 2 = sin (x/ + c) Type 2 Replce x by function of z. Sometimes, insted of putting e.g., z = x 2, we replce x directly by putting z = f(x), (3.) x = g(z). (3.2) This is relly sme s using () since we cn rerrnge this eqution, (i.e., solve for x) to get (3.2). However, we cn work directly from (3.2) by clculting /dz. We then use the formul = dz dz. (Remember tht we lso must chnge limits on definite integrl!)

30 3.4. INTEGRATION BY PARTS 29 Exmple 3.7: Evlute I = 4 + x. Put x = z 2, /dz = 2z, = 2zdz. The limits chnge, x = z =, x = 4 z = 2. We obtin I = = Integrtion by Prts 2 + z 2zdz = 2 2(z + ) 2 dz = (z + ) 2 2z ( + z) dz ( 2 2 (z + ) ) dz = [2z 2 ln(z + )] 2 = (4 2 ln 3) = 4 2 ln 3 =.828 This is the integrl equivlent to the differentil of product. Strt with Integrte both sides with respect to x, uv = d(uv) = u dv + v du. u dv + v du. Now use (dv/) = dv nd (du/) = du. Rerrnge the terms, nd find uv = udv + vdu. See Stroud, 5,2-3 This lst eqution is minly used in the form u dv = uv v du. Exmple 3.8: Evlute I = xe x. Put u = x nd e x = dv. u prt: u = x, therefore du/ = nd du =. v prt: e x = dv therefore dv/ = e x nd v = e x du = e x (constnt of integrtion not needed here). Thus I = uv v du = xe x e x = xe x (e x + c). Note tht the x prt of the originl integrnd (i.e., u) ws differentited, but the e x prt (i.e., dv/) ws integrted. We obtined new integrl which ws esier thn the old one becuse (du/) ws simpler thn u but v ws no hrder thn v. It is requirement tht the resulting integrl is no more complicted thn the originl! Exmple 3.9: Evlute I = x 2 sin x.

31 3.4. INTEGRATION BY PARTS 3 Put u = x 2, dv = sin x. du/ = 2x, therefore du = (2x). dv/ = sin x, therefore v = sin x = cos x. We thus obtin I = uv v du = x 2 ( cos x) ( cos x)2x = x 2 cos x + 2 x cos x. Now repet this procedure: Put u = x, cos x = dv. We find du/ =, nd therefore du =. Finlly dv/ = cos x nd thus v = sin x. [ I = x 2 cos x + 2 x sin x ] sin x = x 2 cos x + 2x sin x 2( cos x) + k. (We hve put the constnt of integrtion in t the end.) Exmple 3.: Evlute I = ln x. Even though this does not look like integrtion by prts, we cn use trick! Use the fct tht the derivtive of the logrithm is much more mngeble thn the logrithm itself, nd use function v with derivtive. Thus u = ln x, dv =, du = x, dv =, du = (/x), v = x. I = uv v du = x ln x x /x = x ln x Exmple 3.: Find I = π/2 e x cos x. = x ln x x + k. Here we cn integrte or differentite e x, nd differentite or integrte cos x, since the integrls nd derivtives of both functions re s simple s the originl function. We choose u = e x, therefore du/ = cos x nd v = sin x. Now integrte by prts gin. I = (uv) π/2 π/2 = (e x sin x) π/2 = e π/2 π/2 vdu π/2 sin xe x. sin x e x Note: Initilly we differentited u = e x, tking cos x s derivtive. We must use the sme procedure gin, nd not switch u nd v. I.e., we must put u = e x nd dv = sin x. Therefore

32 3.5. INTEGRALS OF THE FORM I = ()/(AX + B) DX 3 u = e x, du/ = e x, nd thus du = e x, dv = sin x. It follows tht dv/ = sin x, nd so v = cos x. Bring I to the left-hnd side, nd thus, finlly, I = e π/2 ( e x cos x) π/2 + = e π/2 [ ( )] = e π/2 I. π/2 π/2 2I = e π/2. cos xe x e x cos x I = 2 ( ) e π/ Integrls of the form I = ()/(x + b) The integrl I = ()/(x + b), cn be done by substitution, z = x + b, = dz/, I = z dz = (ln z + C). Thus I = ()/(x + b) = (ln(x + b) + C 3.6 Integrls of the form I = (px + q)/(x 2 + x + b) See Stroud, We now study the integrl I = (px + q)/(x 2 + x + b), i.e., liner over qudrtic, where the qudrtic does not fctorize. Step Clculte the differentil of the denomintor, Use this to rerrnge the numertor into form d (x2 + x + b) = 2x +. p (2x + ) + (q p/2), 2 i.e., s constnt times the derivtive of the denomintor plus nother constnt. We cn now split the integrl, I = p 2x + 2 x 2 + (q p/2) + x + b x 2. + x + b The first integrl on the r.h.s. cn be done using the substitution z = x 2 + x + b, 2x + x 2 + x + b = z dz = ln z = ln(x2 + x + b). Step 2 The second integrl is more complicted, nd we del with J = x 2 + x + b

33 3.7. INTEGRATION OF RATIONAL FUNCTIONS 32 seprtely. The technique used is bsed on completing the squre, x 2 + x + b = (x + c) 2 ± d 2, which fter the substitution z = x + c leds to stndrd integrl dz z 2 ± d 2. Depending on the sign we get either n inverse tngent or rtio of logrithms, z 2 + d 2 dz = d tn (z/d) + c, z 2 d 2 dz = 2d ( z d z + d ) dz = 2d ln ( ) z d + c. z + d Exmple 3.2: Evlute I = 4x x 2 + 2x + 3. Step Differentiting the denomintor gives 2x + 2. numertor s 4x = 2(2x + 2) 5. 2x + 2 I = x 2 + 2x x 2 + 2x + 3 = ln(x 2 + 2x + 3) 5 x 2 + 2x 3 Now complete the squre for the denomintor, nd find tht x 2 + 2x + 3 = (x + ) = (x + ) Tke prt into tow pieces, by rerrnging J = x 2 + 2x + 3 = (x + ) Substitute z = x +, dz = (dz/) =, dz J = z = (/ 2) tn (z/ 2) + c. Thus we find I = 2 ln(x 2 + 2x + 3) (5/ 2) tn ((x + )/ 2) + c. 3.7 Integrtion of rtionl Functions A rtionl function is function of the form f(x) = P (x)/q(x) where P nd Q re both polynomils. Integrtion of such functions re delt with ccording to the following procedure: Step If the degree of P is equl or greter tht of Q then rerrnge the numertor to get where L nd M re polynomils nd M hs lower degree thn Q, P (x) = L(x)Q(x) + M(x) (3.3)

34 3.7. INTEGRATION OF RATIONAL FUNCTIONS 33 Exmple 3.3: Bring f(x) = 2x3 + x 2 + x + x 3 x to the form (3.3). Put 2x 3 +x 2 +x+ = 2(x 3 x 2 +2)+3x 2 +x 3 This corresponds to L = 2 nd M = 3x 2 +x+3. Thus f(x) = LQ+M Q = L + M Q. We cn clerly integrte L directly (why?). Step 2 We now hve to integrte the new rtionl function M Q with by. fctorising Q in liner nd/or qudrtic fctors. 2. using the technique of prtil frctions. We now obtin integrls with one or more of the following types () x+ : integrtes to ln(x + ). (b) : integrtes to (x+) 2 x+ (c) px+q x 2 +x+b : integrtes see bove (Sec. 3.6) Exmple 3.4: Integrte (3x 2 + x + 3)/(x 3 x 2 + 2). This integrnd cn be rewritten s 3x 2 + x + 3 x 3 x = To find A, B, C, we need to solve where M hs lower degree thn Q. This is delt 3x 2 + x + 3 (x + )(x 2 2x + 2) = A x + + Bx + C x 2 2x + 2 3x 2 + x 3 = A(x 2 2x + 2) + (Bx + C)(x + ). We cn get one of the vlues for lmost free, using x = : 5A =, or A = /5. We solve for the rest by equting the coefficients of identicl powers of x, x 2 : 3 = A + B, therefore B = 6/5, constnt: 3 = 2A + C, so tht C = 3/5. We hve reexpressed the integrl s 3x 2 + x + 3 x 3 x = 5 x x 3 x 2 2x + 2. The first term (/(x + )) is esy to integrte nd gives ln(x + ). Let us therefore concentrte on the second term 6x 3 8(2x 2) + 5 x 2 2x + 2 = x 2 2x + 2 (x 2 2x + 2) = 8 x 2 2x x 2 2x + 2 = 8 ln(x 2 5 2x + 2) + x 2 2x + 2. Here we hve used the fct tht the differentil of the denomintor is 2x 2. The remining integrl is treted by completing the squre, which llows us to write x 2 2x + 2 = (x ) 2 +, 5 x 2 2x + 2 = 5 tn (x ).

35 3.8. INTEGRALS WITH SQUARE ROOTS IN DENOMINATOR 34 Using the two previous exmples we conclude tht 2x 3 + x 2 + x + (x 3 x 2 + 2) = 2x 5 ln(x + ) ln(x2 2x + 2) 3 5 tn (x ). 3.8 Integrls with squre roots in denomintor See Stroud, 6. (some overlp). We shll consider only one type + bx x 2. The coefficient of x 2 must be negtive, if it is positive we need different pproch which involves hyperbolic functions (not discussed here). The method is s follows. Complete the squre, + bx x 2 = d 2 (x + c) 2, with c = b/2 nd d 2 = + b 2 /4. 2. Substitute z = x c, which gives us the derivtive of the rcsin. Exmple 3.5: Clculte I = 3 + 4x x 2. Complete the squre: 3 + 4x x 2 = d 2 (x + c) 2. Equte the coefficients of ech power. x 2 : =, contins no unknowns. x: 4 = 2c (therefore c = 2). The constnt term gives 3 = d 2 c 2 = d 2 4, nd thus d 2 = 7, d = 7, nd I = 7 (x 2) 2. We substitute z = x 2, dz = dz/ =, which leds to I = dz = sin z ( ) x 2 + k = sin + k. 7 2 z (The integrl is stndrd integrl nd cn be found in the tbles, but is esily checked by using the chin rule nd d dy sin y =. ) y 2

36 Chpter 4 Applictions of Integrtion 4. Finding res See Stroud, 8.-8 We hve lredy discussed how n integrl corresponds to n re. Exmple 4.: Evlute the re A under y = x 2 from x = to x = 3. A = 3 x2 which is 27/3 /3 = 26/3, see Fig Are between two curves Exmple 4.2: Find the re A of the region bounded by y = e x nd y = x, for x rnging from to, see Fig y=x 2 5 y x [htb] Figure 4.: The surfce below x 2 between nd 3. 35

37 4.. FINDING AREAS x exp(x) y [htb] x Figure 4.2: The re between x nd e x for x between nd. y f(x) δ x b x Figure 4.3: Integrtion s the sum of re of smll strips. From the grph we see tht e x is bove x, so tht A = (re below y = e x ) (re below y = x) = = = = e x (e x + x) ( x) ) (e x x + x2 2 ( e + ) 2 = e Here we hve mde the optionl choice to combine the two integrnds before evlution of the integrl Bsic Derivtion of Are Formul See Stroud, 8.-8 To find re beneth the curve y = f(x) between x = nd x = b, we divide the re into strips s shown in Fig Let the thickness of strip t x be δx. The height t x is f(x), nd therefore the re of the strip

38 4.2. VOLUMES OF REVOLUTION 37 f(x) x Figure 4.4: A surfce of revolution. r δ x Figure 4.5: The volume of smll disc. is δa f(x)δx. Now sum up ll strips from to b. The res is A b f(x)δx. In the limit tht δx becomes infinitesiml (i.e., pproches zero), we replce δx by, the b by b nd so 4.2 Volumes of Revolution A = b f(x). (4.) See Stroud, 9.- If we tke re under the curve y = f(x) between x = nd x = b, s bove nd then rotte it round the x xis through 36 we sweep out volume clled volume of revolution V. This sitution is shown in Fig Clerly V hs n xis of symmetry, i.e., the x xis. Mny volumes tht occur in prctice hve such n xis. We cn use integrtion to find the volume. Agin divide the re into strips of width δx. Since the height is f(x), when we rotte the strip we get disc of rdius r = f(x), see Fig The re of this disc is πr 2 = πf(x) 2, nd the volume of the disc is δv = πr 2 δx. The totl volume is gin sum, b V = πr 2 δx = π Now tke limit where δx becomes infinitesiml, nd thus b f(x) 2 δx. V = π b This is the formul for the volume of solid of revolution. Exmple 4.3: f(x) 2. Find the volume formed when the curve y = /x, between x = nd x = 2 is rotted round the x xis, see Fig. 4.6

39 4.3. CENTROIDS (FIRST MOMENT OF AREA) y z x.75 2 Figure 4.6: The surfce of revolution for y = (/x), < x < 2. V = π 2 (/x) 2 = π ( /x) 2 = π( (/2) ( )) = π/2. Exmple 4.4: Find the volume formed when equilterl tringle with corners t O = (, ), A = (, 3), B = (2, ) is rotted round the x xis, see Fig Along OA the curve is y = 3x, long AB the curve is y = 2 3 3x. Thus V = π ( 3x) 2 + π 2 (2 3 3x) 2 = 3π ( x 3 /3 ) + 3π ( (2 x) 3 /3 ) 2 = π + π( + ) = 2π. 4.3 Centroids (First moment of re) See Stroud,

40 4.3. CENTROIDS (FIRST MOMENT OF AREA) 39 y - z -.5 x.5 2 Figure 4.7: 4.3. First moment of the re bout the y xis Agin consider curve y = f(x) from to b, divided into strips of thickness δx. The re of the strip is given by (δa f((x))δx). The totl re is given by the sum, A b δa = b f(x)δx b f(x). If the strip is very thin then ll of it is pproximtely t distnce x from y xis. If we now dd up NOT δa but insted δa times x, i.e., δa weighted by x, we get the first moment of the re bout the x xis, M x b xδa = b xf(x)δx b xf((x)). This is usully clled M x, even though it is the first moment round the y xis. Exmple 4.5: Find the first moment of re under y = + x + x 2 from x = to x = 2 bout the y xis. M x = = 2 2 x( + x 2 + x 3 ) (x + x 2 + x 3 ) = ( x 2 /2 + x 3 /3 + x 4 /4 ) 2 = 2 + 8/3 + 4 = 26/3.

41 4.3. CENTROIDS (FIRST MOMENT OF AREA) 4 δ x b Figure 4.8: Subdividing the strips of width δx in ones of height δy. Exmple 4.6: Find the first moment of the re under y = e x from x = to x = bout the y xis. M x = xe x. Integrte by prts: u = x, du/ =, du =, dv = e x. Therefore dv/ = e x, nd thus v = e x, M x = ( xe x) ( e x ) = e + e x = e + ( e x ) = e + ( e + ) = 2 e = First Moment of the re bout the x xis Now consider the sme strip of thickness δx. On this strip y goes from to f(x). Divide strip into segments of length δy s shown in Fig The re of such segment is δyδx. The totl re of strip is δa f(x) y= δyδx. In the limit tht δy becomes infinitesiml we get δa f(x) y= dyδx = (y) f(x) δx = f(x)δx, s before. Now insted of summing segments we cn weight ech of them by the vlue of y to get δm y = = ( f(x) yδyδx y= f(x) y dy)δx = ( y2 2 )f(x) δx = 2 f(x)2 δx

42 4.3. CENTROIDS (FIRST MOMENT OF AREA) 4 To find M y we hve to dd the contributions of ll strips M y = = = 2 b δm y b b 2 f(x)2 δx f(x) 2 This is the formul for the first moment of the re bout the x xis (This integrl is sme s tht for the volume of revolution except for the fctor 2 outside the integrl rther thn π). Exmple 4.7: Find M y for re under curve y = +x+x 2 from x = to x = 2 (sme re s in exmple xxxx()) Therefore f(x) = + x + x 2, f(x) 2 = ( + x + x 2 ) 2 = + 2x + 3x 2 + 2x 3 + x 4. 2 M y = ( + 2x + 3x 2 + 2x 3 + x 4 ) 2 = ) 2 (x + x 2 + x 3 + x x5 5 = ( ) Centroid of plne re = = 7 5 = 4.2. For ny plne shpe with re A, the centroid is point with coordintes (x C, y C ) given by x C = /AM x, y C = /AM y, where M x is first moment of re bout the y xis, nd M y is first moment of re bout the x xis Exmple 4.8: Find the centroid of the re under y = + x + x 2 from x = to x = 2 using the previous two exmples. We know tht M x = 26/3 nd M y = 7/5, nd we just need to determine A, A = 2 ( + x + x 2 ) Therefore = (x + x 2 /2 + x 3 /3) 2 = /3 = 2/3. x C = M x A = 3 2 y C = M y A = = 26 2 =.3, 7 5 = 23 = 2.3.

43 4.4. SECOND MOMENT OF AREA Mening of the centroid If we hve thin plte with constnt thickness then the centroid is the position of centre of mss (C of M). The C of M is the point t which ll mss cn be regrded s cting. Let mss per unit re be ρ: This will be constnt if the thickness is constnt (nd mteril is of uniform composition). The totl mss m = Aρ where A is re. Turning effect bout y xis of mss m t (x, y) would be mx = Aρx. A strip of thickness δx, height f(x) hs re f(x)δx. Mss would be ρf(x)δx. Totl turning effect is b xρf(x)δx b xf(x) = ρm x, therefore Aρx = ρm x, therefore x C = /AM x. 4.4 Second Moment of Are The first moment of re (bout the y xis) ws M x b xδa = b xf(x) b xf(x). Similrly second moment is sme but with x 2 insted of x, δx = δa = b x 2, b x 2 f(x) b x 2 f(x). Exmple 4.9: Find the second moment of re under y = + x + x 2 bout the y xis from x = to x = 2. δx = = = 2 2 ( x 3 x 2 ( + x 2 + x 3 ) (x 2 + x 3 + x 4 ) 3 + x4 4 + x5 5 ) 2 = = 5 = 96 5 = 3 5. Note: To find second moment bout x xis is more complicted: δy = b 3 f(x)3. This will not be done here. Note: Recll tht first moments re used in clculting centroids which re relted to centres of mss. Second moments re used in clculting moments of inerti of flt plnes.