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1 Maima 在微積分上之應用三角函數國立屏東教育大學應用數學系研究助理徐偉玲日期 :009//09 除另有說明外, 本文件採用創用 CC 姓名標示非商業性 5 台灣條款

2 7 Trigonometry 7 Derivatives of Trigonometric Functions Eample Find the derivative of y = tan () Solution: dy = tan d(tan ) = tan sec d() = 6 tan sec d Eample Evaluate cost lim π t / t / π Solution:(%i) limit(cos(t)/(t-%pi/),t,%pi/); 數, 範圍 ) cost // 方程式為 t π / 極限指令 :limit( 方程式, 極限變, 極限變數為 t, 範圍為 t 趨近於 π / This is a limit of the form 0/0 because lim cost = 0, t π / lim ( t t π / π ) = 0 By l Hospital s Rule (Section 5), cost lim = π / t π / t sin t lim π / t = sin( π ) = Eample A particle travels around a vertical circle of radius r 0 with constant angular velocity ω = d θ / dt, beginning with θ = 0 at time t = 0 If the sun is directly overhead, find the position, velocity, and acceleration of the shadow Let us center the circle at the origin in the (, y) plane (Figure 75) Then = r0 cosθ, y = r0 sinθ At time t, θ has the value θ = ωt So the motion of the particle is given by the parametric equations = r0 cos( ω t), y = r0 sin( ωt) The shadow is directly below the particle, and its position is given by the -component = r 0 cos( ωt)

3 The velocity and acceleration of the shadow are d v = = r 0 ω sin( ωt), dt dv a = = r 0 ω cos( ωt) dt Eample A light beam on a 00 ft tower rotates in a vertical circle at the rate of one revolution per second Find the speed of the spot of light moving along the ground at a point 000 ft from the base of the tower We start by drawing the picture in Figure 76 Solution: Assume the rotation is counterclockwise Let t be time and let and θ be as in the figure Then dθ = π radians/sec, dt We wish to find d dt dθ = 00 sec θ dt When = 000, = 00 tanθ ft d / dt when = 000 =00π sec θ sec θ = + tan θ =+ ( /00) = + 0 = 0 d Therefore = 000π ~ 6000 ft/sec dt Eample 5 Find sin t costdt Let u = sin t, du = costdt Solution:(%i) integrate(sin(t)^*cos(t),t); 圍 ) 函數為 sin t cost 積分指令 :integrate( 數式, 變數, 範, 變數為 t, 因為不為定積分所以不用打範圍 Then u sin t = C sin t costdt u du = + C = + Eample 6 Find the area under one arch of the curve y = cos

4 Solution:(%i) plotd(cos(),[,-*%pi,*%pi]); // 畫出 cos 的圖形, 軸範圍為 π ~ π (%i) integrate(cos(),,-%pi/,%pi/); 積分指令 :integrate( 數式, 變數, 範圍 ) 函數為 sin t cost, 變數為 t, 定積分範圍為 π / ~ π / From Figure 77 we see that one arch lies between the limits = π / and = π /, therefore the area is π / π / π / = π / cost dt sin t = ( ) = Eample 7 Evaluate sec d Solution:(%i) integrate(sec()^,); 積分指令 :integrate( 數式, 變數, 範圍 ) 函數為 sec, 變數為, 因為不為定積分所以不用打範圍

5 Using the identity sec = + tan, we have sec d ( + = tan )sec d tan = ( + tan ) d(tan ) = tan + + C Eample 8 Find cos d Solution:(%i) integrate(sqrt(-cos()),); 積分指令 :integrate( 數式, 變數, 範圍 ) 函數為 cos, 變數為, 因為不為定積分所以不用打範圍 Using the identity sin + cos =, we have cos = cos + cos + cos = cos + cos = sin sin = + cos + cos Case In an interval where sin 0, sin 0 cos ) + cos + cos cos d = d = d( + = + cos + C Case In an interval where sin 0, cos d = + cos + C 7 Inverse Trigonometric Functions Eample

6 Find arccos( / ) Solution:(%i) plotd(acos(y),[y,-,]); // 畫出 arccos 圖形,y 軸範圍為 -~ (%i) acos(sqrt()/); arccos 指令 :acos( 數值 ) // 此題為求 arccos( / ) 的值 From Table 7, cos( π / ) = / Since 0 π / π, arccos( / ) = π / Eample Find arcsin( ) Solution:(%i) plotd(asin(y),[y,-,]); // 畫出 arcsin 圖形,y 軸範圍為 -~

7 (%i) asin(-); arcsin 指令 :asin( 數值 ) // 此題為求 arcsin() 的值 From Table 7, sin( π / ) = But π / is not in the interval [ π /, π / ] Using sin( θ + n π ) = sinθ, we have sin( π / ) = sin(π / ) =, So arcsin( ) = π / Eample Find arctan( )

8 Solution:(%i) plotd(atan(y),[y,-,]); // 畫出 arctan 圖形,y 軸範圍為 -~ (%i) atan(-sqrt()); arctan 指令 :atan( 數值 ) // 此題為求 arctan( ) 的值 We must find a θ in the interval [ π /, π / ] such that tanθ = From Table 7, sin( π / ) = /, cos( π / ) = / Then sin( π / ) = /, cos( π / ) = / So tan( π / ) = / / =,

9 arctan( ) = π / Eample Find cos(arctan y ) Solution:(%i) cos(atan(y)); Let θ = arctan y Thus tan θ = y Using sin θ + cos θ = sinθ = cosθ y, we solve for cos θ sinθ = y cosθ, ( y cosθ ) + cos θ =, cos θ ( y + ) =, cos θ = y + Thus cosθ = ± y + By definition of arctan y, we know that π / θ π / In this interval, cosθ 0 Therefore cosθ = y + Eample 5 Show that arcsin y + arccos y = π / (Figure 76) Let θ = arcsin y

10 We have y = sinθ = cos( π / θ ) Also, when π / θ π /, we have π / θ π /, π π / θ 0 Thus π / θ = arccos y, arcsin y + arccos y = θ + ( π / θ ) = π / Eample 6 (a) Find the area of the region under the curve y = + for (b) Find the area of the region under the same curve for < < The regions are shown in Figure 77 Solution: (a) (%i) integrate(/(+^),,-,); 積分指令 :integrate( 數式, 變數, 範圍 ) 函數為 +, 變數為, 定積分範圍為 -~ π π π A = d = arctan = ( ) = + (b) (%i) integrate(/(+^),,minf,inf); 積分指令 :integrate( 數式, 變數, 範圍 ) 函數為 +, 變數為, 定積分範圍為 ~ A = 0 d = d d = lim a 0 b 0 d + lim a + b + d

11 = lim (arctan 0 arctan a) + lim(arctan b arctan 0) a b = lim arctan a + lim arctan b a b From the graph of is π /, so arctan we see that the first limit is π / and the second limit π π A = ( ) + = π Thus the region under y = /( + ) has eactly the same area as the unit circle, and half of this area is between = and = Eample 7 Find d Solution:(%i) integrate(/(*sqrt(^-)),,-,-sqrt()); 積分指令 :integrate( 數式, 變數, 範圍 ) 函數為, 變數為, 定積分範圍為 ~ The region is shown in Figure 78 Since is negative, = Thus d = d = arc sec = ( arc sec( ) arcsec( )) π π π = ( ) =

12 7 Integration by Parts Eample Evaluate sin d Our plan is to break sin d into a product of the form udv, evaluate the integrals dv and vdu, and then use integration by parts to get udv There are several choices we might make for u and dv, and not all of them lead to a solution of the problem Some guesswork is required Solution:(%i) integrate(*sin(),); 積分指令 :integrate( 數式, 變數, 範圍 ) 函數為 sin, 變數為, 因為不為定積分所以不用打範圍 First try: u =, dv = d = C Take sin dv d = + v = Net we find du and try to evaluate vdu du = cos d, vdu = cos d This integral looks harder than the one we started with, so we shall start over with another choice of u and dv Second try: u =, dv = sin d dv d = cos + = sin C We take v = cos This time we find du and easily evaluate vdu du = d, vdu cos d = sin + C = Finally we use the rule

13 uv udv = vdu, sin d = ( cos ) ( sin + C), or sin d = cos + sin + C Eample Evaluate arcsin d A choice of u and dv which works is u = arcsin, dv = d Solution:(%i) integrate(asin(),); 積分指令 :integrate( 數式, 變數, 範圍 ) 函數為 arcsin, 變數為, 因為不為定積分所以不用打範圍 We may take v = Then d du =, d vdu = C = + Finally, arcsin d = arcsin ( + C ), arcsin d = arcsin + + C Eample

14 Evaluate sin d This requires two integrations by parts Solution:(%i) integrate(^*sin(),); 積分指令 :integrate( 數式, 變數, 範圍 ) 函數為 sin, 變數為, 因為不為定積分所以不用打範圍 Step : u =, dv = sin d, dv = sin d = cos + du = d, C We take v = cos sin d = uv vdu = cos + cos d Step : Evaluate cos d u = =, dv cos d, du dv = cos d = sin + = d, C We take v = sin cos d uv v du = = sin sin d = sin + cos + C Combining the two steps, sin d = cos + sin + cos + C

15 Eample Evaluate sin θdθ Solution:(%i) integrate(sin()^,); 積分指令 :integrate( 數式, 變數, 範圍 ) 函數為 sin, 變數為, 因為不為定積分所以不用打範圍, 由於 θ 在 Maima 中部好表示, 於是我們用代替 θ 這個變數 Let u = sin θ, dv = sinθdθ Then du = cosθdθ, v = cosθ sin θdθ = θ θ sin cos cos θdθ + = sin θ cosθ cos θdθ + ( = sinθ cosθ sin θ ) dθ = sin θ cosθ + θ sin θdθ We solve this equation for sin θdθ, sin θdθ = sinθ cosθ + θ + C Eample 5 Evaluate π 0 sin d (Figure 7)

16 Solution:(%i) integrate(*sin(),,0,%pi); 圍 ) 函數為 sin, 變數為, 定積分範圍為 0~π 積分指令 :integrate( 數式, 變數, 範 (%i) plotd(*sin(),[,0,%pi]); // 畫出 sin 的圖形, 軸範圍為 0~π Take u =, dv = sin d as in Eample Then v = cos and π 0 sin d = cos π 0 π 0 cos d = cos + sin 0 π π 0 = ( π ( ) + 0 ) + (0 0) = π

17 75 Integrals of Powers of Trigonometric Functions Eample tan = C 0 tan d tan d = tan + (%i) integrate(tan()^,); 積分指令 :integrate( 數式, 變數, 範圍 ) 函數為 tan, 變數為, 因為不為定積分所以不用打範圍 Eample tan tan tan d = tan d = tan + + C (%i) integrate(tan()^,); 積分指令 :integrate( 數式, 變數, 範圍 ) 函數為 tan, 變數為, 因為不為定積分所以不用打範圍 Eample tan tan d = tan d (%i) integrate(tan()^,); 積分指令 :integrate( 數式, 變數, 範圍 ) 函數為 tan, 變數為, 因為不為定積分所以不用打範圍 Eample sin d = sin cos + d = sin cos + + C cos d = cos sin + d = cos sin + + C (%i) integrate(sin()^,); 積分指令 :integrate( 數式, 變數, 範圍 ) 函數為 sin, 變數為, 因為不為定積分所以不用打範圍

18 (%i) integrate(cos()^,); 積分指令 :integrate( 數式, 變數, 範圍 ) 函數為 cos, 變數為, 因為不為定積分所以不用打範圍 Eample 5 d = cos cos sin + cos d = cos sin + sin + C (%i) integrate(cos()^,); 積分指令 :integrate( 數式, 變數, 範圍 ) 函數為 cos, 變數為, 因為不為定積分所以不用打範圍 Eample 6 sec d = sec sin + sec d (%i) integrate(sec()^,); 積分指令 :integrate( 數式, 變數, 範圍 ) 函數為 sec, 變數為, 因為不為定積分所以不用打範圍 Eample 7 = sec d sec sin + sec d

19 = sec sin + tan + C (%i) integrate(sec()^,); 積分指令 :integrate( 數式, 變數, 範圍 ) 函數為 sec, 變數為, 因為不為定積分所以不用打範圍 Eample 8 sin cos d Solution:(%i) integrate((sin()^)*(cos()^),); 積分指令 :integrate( 數式, 變數, 範圍 ) 函數為 sin cos, 變數為, 因為不為定積分所以不用打範圍 Let u = sin, du = cos d sin cos d = u ( u ) du 5 7 = u u + C 5 7 = sin sin + C 7 Eample 9 cos sin d Solution:(%i) integrate(sqrt(cos())*(sin()^),); 積分指令 :integrate( 數式, 變數, 範圍 ) 函數為 cos sin, 變數為, 因為不為定積分所以不用打範圍 Let u = cos, du = sin d = cos sin d u ( u )( ) du

20 = u / + u 5 / du = u / + u 7 / + C 7 = (cos ) / + (cos ) 7 / + C 7 Eample 0 5 sin d Solution:(%i) integrate(sin()^5,); 積分指令 :integrate( 數式, 變數, 範圍 ) 函數為 sin 5, 變數為, 因為不為定積分所以不用打範圍 Let u = cos, du = sin d 5 sin ( u ) ( d = ) du 5 = ( u + u ) du = u + u u + C 5 5 = cos + cos cos + C 5 Eample sin cos d = sin ( sin ) d 6 8 = sin sin + sin d

21 (%i) integrate((sin()^)*(cos()^),); 積分指令 :integrate( 數式, 變數, 範圍 ) 函數為 sin cos, 變數為, 因為不為定積分所以不用打範圍 Eample When m is even use tan = sec tan sec d = (sec ) sec d 5 = sec sec + sec d (%i) integrate((tan()^)*(sec()),); 積分指令 :integrate( 數式, 變數, 範圍 ) 函數為 tan sec, 變數為, 因為不為定積分所以不用打範圍 Eample When m is odd use the new variable u = sec or u = csc cot csc d = cot csc (cot csc d) 5 u u = ( u ) u du = + C 5 csc = 5 5 csc + + C

22 76 Trigonometric Substitutions Eample Find / ( a + ) d Solution:(%i) integrate((a^+^)^(-/),); 積分指令 :integrate( 數式, 變數, 範圍 ) 函數為 ( a + ) /, 變數為, 因為不為定積分所以不用打範圍 Let θ = arctan( / a) Then from Figure 76, = a tanθ, d = asec θdθ, a + = a secθ So / ( a + ) d = ( asecθ ) asec θdθ = a (secθ ) dθ = a cosθdθ = a tanθ sinθ + C = + C = + C a secθ a a + Eample Find a d

23 Solution:(%i) integrate(sqrt(^-a^),); 積分指令 :integrate( 數式, 變數, 範圍 ) 函數為 a, 變數為, 因為不為定積分所以不用打範圍 Let θ = arcsec( / a) (Figure 76), so = asecθ, d = a tanθ secθdθ, a = a tanθ So a d = a tanθa tanθ secθdθ = a tan θ secθdθ = a (sec θ ) secθdθ = a sec θdθ a secθdθ = ( a sec θ sinθ + a secθdθ ) a secθdθ = a sec θ sinθ a secθdθ = a a secθdθ Eample a d

24 Solution:(%i) integrate(/(^*(sqrt(a^-^))),); 積分指令 :integrate( 數式, 變數, 範圍 ) 函數為, 變數為, 因為不為定積分所以不用打範圍, a 第二式是問 a 是否為 0 或不為零的數, 在此我們令 a 為不為零的數 Let θ = arcsin( / a) (Figure 76) Then = asinθ, d = a cosθdθ, a = a cosθ d = a cosθdθ = a a θa θ sin cos a sin = sec θdθ = cotθ + C a a dθ θ = a a + C Eample a d Solution:(%i) integrate((sqrt(^-a^))/,); 積分指令 :integrate( 數式, 變數, a 範圍 ) 函數為, 變數為, 因為不為定積分所以不用打範圍, 第二式是問 a 是否為 0 或不為零的數, 在此我們令 a 為不為零的數 Put θ = arcsec( / a) (Figure 765) Then

25 = asecθ, d = a tanθ secθdθ, a = a tanθ a a tanθ a secθ d = a tanθ secθdθ = a tan θdθ sec θdθ a dθ = a tanθ aθ + = a C = a aarcsec( / a) + C Eample 5 The basic integrals: (a) d = arcsin + C, d (b) d = arctan + C +, d (c) = arcsec + C, > can be evaluated very easily by a trigonometric substitution Solution: (a) (%i) integrate(/sqrt(-^),); 積分指令 :integrate( 數式, 變數, 範圍 ) 函數為, 變數為, 因為不為定積分所以不用打範圍 d Let θ = arcsin (Figure 766) Then = sinθ, d = cosθdθ, = cosθ

26 d = cosθdθ cosθ = dθ = θ + C, d = arcsin + C (b) (%i) (%i) integrate(/(+^),); 積分指令 :integrate( 數式, 變數, 範圍 ) 函數為, 變數為, 因為不為定積分所以不用打範圍 + d + Let θ = arctan (Figure 767 Then = tanθ, d = sec θdθ, + = secθ d + = sec θ sec θ dθ = dθ = θ + C, d = arctan + C + (c) (%i) integrate(/(*sqrt(^-)),); 積分指令 :integrate( 數式, 變數, 範圍 ) 函數為, 變數為, 因為不為定積分所以不用打範圍 d, > Let θ = arcsec (Figure 768) Then = secθ, d = tanθ secθdθ, = tanθ = dθ = dθ = θ + d tanθ secθ secθ tanθ C

27 d = arc sec + C, + > 77 Polar Coordinates Eample Plot the following points in polar coordinates (, π / ), (-, π / ), (, π / ), (, -π /), (-, -π / ) The solution is shown in Figure 77 Each point P has infinitely many different polar coordinate pairs We see in Figure 77 that the point P (, π / ) has all the coordinates (, π / + nπ ), n (,π / + nπ ), an integer Eample The graph of the equation (Figure 776(a)) The graph of the equation origin (Figure 776(b)) r = a is the circle of radius a centered at the origin θ = b is a straight line through the Eample The graph of the system of formulas r = θ, 0 θ is the spiral of Archimedes formed by moving a pencil along the line OX while the line is rotating with the pencil moving at the same speed as the point X The graph is shown in Figure 776(c)

28 Eample The parabola y = has the polar equation sinθ r sinθ = ( r cosθ ), or r = = tanθ secθ cos θ Eample 5 The curve y = / has the polar equation r sinθ =, or r = secθ cscθ r cosθ The graph is shown in Figure 778 Eample 6 The graph of the equation r = asinθ is the circle one of whose diameters is the line from the origin to a point a above the origin This can be seen from Figure 779, if we remember that a diameter and a point on the circle form a right triangle As θ increase, the point ( a sinθ, θ ) goes around this circle once for every π radians Eample 7 (a) The spiral r = θ has the parametric equations

29 = θ cos θ, y = θ sinθ (b) The circle r = asinθ has the parametric equations = a sinθ cosθ, y = asin θ 78 Slopes and Curve Sketching in Polar Coordinates Eample Sketch the curve r = + cosθ Solution:(%i) load(draw); 由於極坐標畫圖並沒有直接的指令, 於是我們需要讀取模組 draw, 此模組裡頭包含極坐標畫圖的指令 // 讀取模組 draw (%i) drawd( nticks=00,polar(+cos(theta),theta,0,*%pi) ); 極坐標畫圖指令 : draw(nticks= 點數,polar( 函數式, 角度, 最小的角度, 最大的角度 )) // 在此題畫極坐標表示的函數圖形為 + cosθ, 角度為 θ, 最小的角度是 0, 最大的角度是 π, 用 00 個點描繪曲線

30 Step : dr / dθ = sinθ Step : r = 0 when θ = π dr / dθ = 0 where θ = 0, π Step : See Figure 786 Step : θ r = + cosθ dr / dθ tan ψ Comments ma π / - - r decreasing π Min, cusp at 0 π / r increasing Step 5: We draw the curve in Figure 787 The curve is called a cardioid because of its heart shape Eample

31 Sketch the curve r = sin θ Solution:(%i) load(draw); 讀取模組 draw, 此模組裡頭包含極坐標畫圖的指令由於極坐標畫圖並沒有直接的指令, 於是我們需要 // 讀取模組 draw (%i) drawd(nticks=00,polar(sin(*theta),theta,0,*%pi)); 極坐標畫圖指令 : draw(nticks= 點數,polar( 函數式, 角度, 最小的角度, 最大的角度 )) // 在此題畫極坐標表示的函數圖形為 sin( θ ), 角度為 θ, 最小的角度是 0, 最大的角度是 π, 用 00 個點描繪曲線 Step : dr / dθ = cos θ Step : r = 0 at π π θ = 0,, π, dr / dθ = 0 at π π 5π 7π θ =,,, Step : See Figure 788

32 Step : We take values at intervals of π beginning at θ = 0 We can save some 8 time by observing that the values from π to π are the same as those from 0 to π θ r = sin θ dr / dθ tan ψ Comments 0 and π 0 0 Crosses origin π / 8 and 9π / 8 / / r increasing π / 8 and 0π / ma π /8 and π / 8 / -/ r decreasing π / 8 and π / Crosses origin 5π / 8 and π / 8 / / r increasing 6π /8 and π / min 7π /8 and 5π / 8 / -/ r decreasing Step 5: We plot the points and trace out the curve as θ increases from 0 to π Figure 789 shows the curve at various stages of development The graph looks like a four-leaf clover If r approaches as θ approaches 0 or π, the curve may have a horizontal asymptote which can be found by computing the limit of y At θ = π / or π / there may be vertical asymptotes The method is illustrated in the following eample Eample Sketch r = tan( θ ) Solution:(%i) load(draw); 由於極坐標畫圖並沒有直接的指令, 於是我們需要讀取模組 draw, 此模組裡頭包含極坐標畫圖的指令 // 讀取模組 draw (%i) drawd(nticks=50,polar(tan((/)*theta),theta,0,*%pi)); 極坐標畫圖指令 :

33 draw(nticks= 點數,polar( 函數式, 角度, 最小的角度, 最大的角度 )) // 在此題畫極坐標表示的函數圖形為 tan( θ ), 角度為 θ, 最小的角度是 0, 最大的角度是 π, 用 50 個點描繪曲線 Step : dr / dθ = sec ( θ ) y = r sin θ = sin θ sinθ / cos θ = sin θ (sin θ cos θ ) / cos θ = sin ( θ ) Step : r = 0 at θ = 0 r is undefined at dr / dθ is never 0 θ = π

34 Step : See Figure 780 Step : θ r or lim r lim y dr / dθ tan ψ Comments 0 0 / crosses origin π / r increasing θ π asymptote y = + θ π asymptote y = π / - - r decreasing Step 5: The curve crosses itself at the point = 0, y =, because this point has both polar coordinates ( r =, θ = π / ), ( r =, θ = π / ) Figure 78 shows the graph for various stages of development 79 Area in Polar Coordinates Eample Find the area of one loop of the four-leaf clover r = sin θ Solution:(%i) (/)*integrate(sin(*theta)^,theta,0,%pi/); 積分指令 : integrate( 數式, 變數, 範圍 ) 函數為 sin (θ ), 變數為 θ, 定積分範圍為 0~ π /, z 積分算出來的數值後再乘上 / From Figure 796, we see that one loop is traced out then θ goes from 0 to π / Therefore the area is

35 A = π / π sin (θ ) dθ = 0 0 sin φdφ π = sin φ φ ( sinφ cosφ φ) π d = + = 0 8 π 0 As one would epect, all four loops have the same area On the loop form θ = π / to θ = π, the value of r = sin θ is negative However, the area is again π A = sin (θ ) θ π d = π / 8 Eample Find the area of the region inside the circle r = sinθ (Figure 797) Solution:(%i) integrate((/)*(sin(theta)^),theta,0,%pi); 積分指令 :integrate( 數式, 變數, 範圍 ) 函數為 sin θ, 變數為 θ, 定積分範圍為 0~π The point ( r, θ ) goes around the circle once when 0 θ π with r positive, and again when π θ π with r negative The theorem says that we will get the correct area if we take either 0 and π, or π and π, as the limits of integration Thus π A = sin θdθ = ( sinθ cosθ + θ ) 0 π 0 = ( π 0) = π / Alternatively,

36 π A = sin θdθ = ( sinθ cosθ + θ ) π π π = (π π ) = π / Since the curve is a circle of radius, our answer π / agrees with the usual formula A = πr Integrating from 0 to π would count the area twice and give the wrong answer Eample Find the area of the region inside both the circles r = sinθ and r = cosθ Solution:(%i) integrate((/)*(sin(theta)^),theta,0,%pi/)+integrate((/)*(cos(theta)^),theta,%pi/,%pi/); 積分指令 :integrate( 數式, 變數, 範圍 ) 函數為 sin θ 和變數為 θ, 定積分範圍為 0 ~ π / 和 π / ~ π /, 將兩積分後數值相加 cos θ, (%i) factor(%); // 用 factor 指令將上式做整理,% 表示引用前式之結果 The first thing to do is draw the graphs of both curves The graphs are in Figure 798 We see that the two circles intersect at the origin and at θ = π / The region is divided into two parts, one bounded by bounded by A = π / 0 sin r = cosθ for π / θ π / Thus θdθ + π / π / cos θdθ r = sinθ for 0 θ π / and the other

37 = π / ( sinθ cosθ + θ ) + ( sinθ cosθ + θ ) 0 π / π / = ( π 0) + ( 0) + (0 8 ) π π π + ( ) = Length of A Curve in Polar Coordinates Eample Find the length of the spiral 70 r = θ from θ = π to θ = π, shown in Figure Solution:(%i) integrate(/*(sqrt(u)),u,(%pi)^+,6*(%pi)^+); 積分指令 : integrate( 數式, 變數, 範圍 ) 函數為 u π + ~ 6π +, 變數為 u, 定積分範圍為 π = π s r + ( dr / dθ ) dθ = π π π π θ + θ θ = θ + d θ dθ Let u = θ +, du = θdθ Then s = 6π + π + udu = u 6π + / π + / = ((6π + ) / ( π + ) )

38 Eample Find the length of the curve 70 dr / dθ = cosθ, so r = sinθ from θ = α to θ = β, shown in Figure s θ + cos θ dθ = dθ = β = β α sin α β α (%i) integrate(sqrt(sin(theta)^+cos(theta)^),theta,alpha,beta); 積分指令 : integrate( 數式, 變數, 範圍 ) 函數為 sin θ + cos θ, 變數為 θ, 定積分範圍為 α ~ β

Untitled-3

Untitled-3 SEC.. Separable Equations In each of problems 1 through 8 solve the given differential equation : ü 1. y ' x y x y, y 0 fl y - x 0 fl y - x 0 fl y - x3 3 c, y 0 ü. y ' x ^ y 1 + x 3 x y 1 + x 3, y 0 fl