提供各類精美PPT模板下載
當(dāng)前位置:首頁(yè) > Word文檔 >

人教版高中生物必修3第二章第四節(jié)《免疫調(diào)節(jié)》說(shuō)課稿

  • 直線的兩點(diǎn)式方程教學(xué)設(shè)計(jì)人教A版高中數(shù)學(xué)選擇性必修第一冊(cè)

    直線的兩點(diǎn)式方程教學(xué)設(shè)計(jì)人教A版高中數(shù)學(xué)選擇性必修第一冊(cè)

    解析:①過(guò)原點(diǎn)時(shí),直線方程為y=-34x.②直線不過(guò)原點(diǎn)時(shí),可設(shè)其方程為xa+ya=1,∴4a+-3a=1,∴a=1.∴直線方程為x+y-1=0.所以這樣的直線有2條,選B.答案:B4.若點(diǎn)P(3,m)在過(guò)點(diǎn)A(2,-1),B(-3,4)的直線上,則m= . 解析:由兩點(diǎn)式方程得,過(guò)A,B兩點(diǎn)的直線方程為(y"-(-" 1")" )/(4"-(-" 1")" )=(x"-" 2)/("-" 3"-" 2),即x+y-1=0.又點(diǎn)P(3,m)在直線AB上,所以3+m-1=0,得m=-2.答案:-2 5.直線ax+by=1(ab≠0)與兩坐標(biāo)軸圍成的三角形的面積是 . 解析:直線在兩坐標(biāo)軸上的截距分別為1/a 與 1/b,所以直線與坐標(biāo)軸圍成的三角形面積為1/(2"|" ab"|" ).答案:1/(2"|" ab"|" )6.已知三角形的三個(gè)頂點(diǎn)A(0,4),B(-2,6),C(-8,0).(1)求三角形三邊所在直線的方程;(2)求AC邊上的垂直平分線的方程.解析(1)直線AB的方程為y-46-4=x-0-2-0,整理得x+y-4=0;直線BC的方程為y-06-0=x+8-2+8,整理得x-y+8=0;由截距式可知,直線AC的方程為x-8+y4=1,整理得x-2y+8=0.(2)線段AC的中點(diǎn)為D(-4,2),直線AC的斜率為12,則AC邊上的垂直平分線的斜率為-2,所以AC邊的垂直平分線的方程為y-2=-2(x+4),整理得2x+y+6=0.

  • 空間向量基本定理教學(xué)設(shè)計(jì)人教A版高中數(shù)學(xué)選擇性必修第一冊(cè)

    空間向量基本定理教學(xué)設(shè)計(jì)人教A版高中數(shù)學(xué)選擇性必修第一冊(cè)

    反思感悟用基底表示空間向量的解題策略1.空間中,任一向量都可以用一個(gè)基底表示,且只要基底確定,則表示形式是唯一的.2.用基底表示空間向量時(shí),一般要結(jié)合圖形,運(yùn)用向量加法、減法的平行四邊形法則、三角形法則,以及數(shù)乘向量的運(yùn)算法則,逐步向基向量過(guò)渡,直至全部用基向量表示.3.在空間幾何體中選擇基底時(shí),通常選取公共起點(diǎn)最集中的向量或關(guān)系最明確的向量作為基底,例如,在正方體、長(zhǎng)方體、平行六面體、四面體中,一般選用從同一頂點(diǎn)出發(fā)的三條棱所對(duì)應(yīng)的向量作為基底.例2.在棱長(zhǎng)為2的正方體ABCD-A1B1C1D1中,E,F分別是DD1,BD的中點(diǎn),點(diǎn)G在棱CD上,且CG=1/3 CD(1)證明:EF⊥B1C;(2)求EF與C1G所成角的余弦值.思路分析選擇一個(gè)空間基底,將(EF) ?,(B_1 C) ?,(C_1 G) ?用基向量表示.(1)證明(EF) ?·(B_1 C) ?=0即可;(2)求(EF) ?與(C_1 G) ?夾角的余弦值即可.(1)證明:設(shè)(DA) ?=i,(DC) ?=j,(DD_1 ) ?=k,則{i,j,k}構(gòu)成空間的一個(gè)正交基底.

  • 點(diǎn)到直線的距離公式教學(xué)設(shè)計(jì)人教A版高中數(shù)學(xué)選擇性必修第一冊(cè)

    點(diǎn)到直線的距離公式教學(xué)設(shè)計(jì)人教A版高中數(shù)學(xué)選擇性必修第一冊(cè)

    4.已知△ABC三個(gè)頂點(diǎn)坐標(biāo)A(-1,3),B(-3,0),C(1,2),求△ABC的面積S.【解析】由直線方程的兩點(diǎn)式得直線BC的方程為 = ,即x-2y+3=0,由兩點(diǎn)間距離公式得|BC|= ,點(diǎn)A到BC的距離為d,即為BC邊上的高,d= ,所以S= |BC|·d= ×2 × =4,即△ABC的面積為4.5.已知直線l經(jīng)過(guò)點(diǎn)P(0,2),且A(1,1),B(-3,1)兩點(diǎn)到直線l的距離相等,求直線l的方程.解:(方法一)∵點(diǎn)A(1,1)與B(-3,1)到y(tǒng)軸的距離不相等,∴直線l的斜率存在,設(shè)為k.又直線l在y軸上的截距為2,則直線l的方程為y=kx+2,即kx-y+2=0.由點(diǎn)A(1,1)與B(-3,1)到直線l的距離相等,∴直線l的方程是y=2或x-y+2=0.得("|" k"-" 1+2"|" )/√(k^2+1)=("|-" 3k"-" 1+2"|" )/√(k^2+1),解得k=0或k=1.(方法二)當(dāng)直線l過(guò)線段AB的中點(diǎn)時(shí),A,B兩點(diǎn)到直線l的距離相等.∵AB的中點(diǎn)是(-1,1),又直線l過(guò)點(diǎn)P(0,2),∴直線l的方程是x-y+2=0.當(dāng)直線l∥AB時(shí),A,B兩點(diǎn)到直線l的距離相等.∵直線AB的斜率為0,∴直線l的斜率為0,∴直線l的方程為y=2.綜上所述,滿足條件的直線l的方程是x-y+2=0或y=2.

  • 傾斜角與斜率教學(xué)設(shè)計(jì)人教A版高中數(shù)學(xué)選擇性必修第一冊(cè)

    傾斜角與斜率教學(xué)設(shè)計(jì)人教A版高中數(shù)學(xué)選擇性必修第一冊(cè)

    (2)l的傾斜角為90°,即l平行于y軸,所以m+1=2m,得m=1.延伸探究1 本例條件不變,試求直線l的傾斜角為銳角時(shí)實(shí)數(shù)m的取值范圍.解:由題意知(m"-" 1"-" 1)/(m+1"-" 2m)>0,解得1<m<2.延伸探究2 若將本例中的“N(2m,1)”改為“N(3m,2m)”,其他條件不變,結(jié)果如何?解:(1)由題意知(m"-" 1"-" 2m)/(m+1"-" 3m)=1,解得m=2.(2)由題意知m+1=3m,解得m=1/2.直線斜率的計(jì)算方法(1)判斷兩點(diǎn)的橫坐標(biāo)是否相等,若相等,則直線的斜率不存在.(2)若兩點(diǎn)的橫坐標(biāo)不相等,則可以用斜率公式k=(y_2 "-" y_1)/(x_2 "-" x_1 )(其中x1≠x2)進(jìn)行計(jì)算.金題典例 光線從點(diǎn)A(2,1)射到y(tǒng)軸上的點(diǎn)Q,經(jīng)y軸反射后過(guò)點(diǎn)B(4,3),試求點(diǎn)Q的坐標(biāo)及入射光線的斜率.解:(方法1)設(shè)Q(0,y),則由題意得kQA=-kQB.∵kQA=(1"-" y)/2,kQB=(3"-" y)/4,∴(1"-" y)/2=-(3"-" y)/4.解得y=5/3,即點(diǎn)Q的坐標(biāo)為 0,5/3 ,∴k入=kQA=(1"-" y)/2=-1/3.(方法2)設(shè)Q(0,y),如圖,點(diǎn)B(4,3)關(guān)于y軸的對(duì)稱點(diǎn)為B'(-4,3), kAB'=(1"-" 3)/(2+4)=-1/3,由題意得,A、Q、B'三點(diǎn)共線.從而入射光線的斜率為kAQ=kAB'=-1/3.所以,有(1"-" y)/2=(1"-" 3)/(2+4),解得y=5/3,點(diǎn)Q的坐標(biāo)為(0,5/3).

  • 兩條平行線間的距離教學(xué)設(shè)計(jì)人教A版高中數(shù)學(xué)選擇性必修第一冊(cè)

    兩條平行線間的距離教學(xué)設(shè)計(jì)人教A版高中數(shù)學(xué)選擇性必修第一冊(cè)

    一、情境導(dǎo)學(xué)前面我們已經(jīng)得到了兩點(diǎn)間的距離公式,點(diǎn)到直線的距離公式,關(guān)于平面上的距離問(wèn)題,兩條直線間的距離也是值得研究的。思考1:立定跳遠(yuǎn)測(cè)量的什么距離?A.兩平行線的距離 B.點(diǎn)到直線的距離 C. 點(diǎn)到點(diǎn)的距離二、探究新知思考2:已知兩條平行直線l_1,l_2的方程,如何求l_1 〖與l〗_2間的距離?根據(jù)兩條平行直線間距離的含義,在直線l_1上取任一點(diǎn)P(x_0,y_0 ),,點(diǎn)P(x_0,y_0 )到直線l_2的距離就是直線l_1與直線l_2間的距離,這樣求兩條平行線間的距離就轉(zhuǎn)化為求點(diǎn)到直線的距離。兩條平行直線間的距離1. 定義:夾在兩平行線間的__________的長(zhǎng).公垂線段2. 圖示: 3. 求法:轉(zhuǎn)化為點(diǎn)到直線的距離.1.原點(diǎn)到直線x+2y-5=0的距離是( )A.2 B.3 C.2 D.5D [d=|-5|12+22=5.選D.]

  • 圓的標(biāo)準(zhǔn)方程教學(xué)設(shè)計(jì)人教A版高中數(shù)學(xué)選擇性必修第一冊(cè)

    圓的標(biāo)準(zhǔn)方程教學(xué)設(shè)計(jì)人教A版高中數(shù)學(xué)選擇性必修第一冊(cè)

    (1)幾何法它是利用圖形的幾何性質(zhì),如圓的性質(zhì)等,直接求出圓的圓心和半徑,代入圓的標(biāo)準(zhǔn)方程,從而得到圓的標(biāo)準(zhǔn)方程.(2)待定系數(shù)法由三個(gè)獨(dú)立條件得到三個(gè)方程,解方程組以得到圓的標(biāo)準(zhǔn)方程中三個(gè)參數(shù),從而確定圓的標(biāo)準(zhǔn)方程.它是求圓的方程最常用的方法,一般步驟是:①設(shè)——設(shè)所求圓的方程為(x-a)2+(y-b)2=r2;②列——由已知條件,建立關(guān)于a,b,r的方程組;③解——解方程組,求出a,b,r;④代——將a,b,r代入所設(shè)方程,得所求圓的方程.跟蹤訓(xùn)練1.已知△ABC的三個(gè)頂點(diǎn)坐標(biāo)分別為A(0,5),B(1,-2),C(-3,-4),求該三角形的外接圓的方程.[解] 法一:設(shè)所求圓的標(biāo)準(zhǔn)方程為(x-a)2+(y-b)2=r2.因?yàn)锳(0,5),B(1,-2),C(-3,-4)都在圓上,所以它們的坐標(biāo)都滿足圓的標(biāo)準(zhǔn)方程,于是有?0-a?2+?5-b?2=r2,?1-a?2+?-2-b?2=r2,?-3-a?2+?-4-b?2=r2.解得a=-3,b=1,r=5.故所求圓的標(biāo)準(zhǔn)方程是(x+3)2+(y-1)2=25.

  • 圓的一般方程教學(xué)設(shè)計(jì)人教A版高中數(shù)學(xué)選擇性必修第一冊(cè)

    圓的一般方程教學(xué)設(shè)計(jì)人教A版高中數(shù)學(xué)選擇性必修第一冊(cè)

    情境導(dǎo)學(xué)前面我們已討論了圓的標(biāo)準(zhǔn)方程為(x-a)2+(y-b)2=r2,現(xiàn)將其展開可得:x2+y2-2ax-2bx+a2+b2-r2=0.可見(jiàn),任何一個(gè)圓的方程都可以變形x2+y2+Dx+Ey+F=0的形式.請(qǐng)大家思考一下,形如x2+y2+Dx+Ey+F=0的方程表示的曲線是不是圓?下面我們來(lái)探討這一方面的問(wèn)題.探究新知例如,對(duì)于方程x^2+y^2-2x-4y+6=0,對(duì)其進(jìn)行配方,得〖(x-1)〗^2+(〖y-2)〗^2=-1,因?yàn)槿我庖稽c(diǎn)的坐標(biāo) (x,y) 都不滿足這個(gè)方程,所以這個(gè)方程不表示任何圖形,所以形如x2+y2+Dx+Ey+F=0的方程不一定能通過(guò)恒等變換為圓的標(biāo)準(zhǔn)方程,這表明形如x2+y2+Dx+Ey+F=0的方程不一定是圓的方程.一、圓的一般方程(1)當(dāng)D2+E2-4F>0時(shí),方程x2+y2+Dx+Ey+F=0表示以(-D/2,-E/2)為圓心,1/2 √(D^2+E^2 "-" 4F)為半徑的圓,將方程x2+y2+Dx+Ey+F=0,配方可得〖(x+D/2)〗^2+(〖y+E/2)〗^2=(D^2+E^2-4F)/4(2)當(dāng)D2+E2-4F=0時(shí),方程x2+y2+Dx+Ey+F=0,表示一個(gè)點(diǎn)(-D/2,-E/2)(3)當(dāng)D2+E2-4F0);

  • 圓與圓的位置關(guān)系教學(xué)設(shè)計(jì)人教A版高中數(shù)學(xué)選擇性必修第一冊(cè)

    圓與圓的位置關(guān)系教學(xué)設(shè)計(jì)人教A版高中數(shù)學(xué)選擇性必修第一冊(cè)

    1.兩圓x2+y2-1=0和x2+y2-4x+2y-4=0的位置關(guān)系是( )A.內(nèi)切 B.相交 C.外切 D.外離解析:圓x2+y2-1=0表示以O(shè)1(0,0)點(diǎn)為圓心,以R1=1為半徑的圓.圓x2+y2-4x+2y-4=0表示以O(shè)2(2,-1)點(diǎn)為圓心,以R2=3為半徑的圓.∵|O1O2|=√5,∴R2-R1<|O1O2|<R2+R1,∴圓x2+y2-1=0和圓x2+y2-4x+2y-4=0相交.答案:B2.圓C1:x2+y2-12x-2y-13=0和圓C2:x2+y2+12x+16y-25=0的公共弦所在的直線方程是 . 解析:兩圓的方程相減得公共弦所在的直線方程為4x+3y-2=0.答案:4x+3y-2=03.半徑為6的圓與x軸相切,且與圓x2+(y-3)2=1內(nèi)切,則此圓的方程為( )A.(x-4)2+(y-6)2=16 B.(x±4)2+(y-6)2=16C.(x-4)2+(y-6)2=36 D.(x±4)2+(y-6)2=36解析:設(shè)所求圓心坐標(biāo)為(a,b),則|b|=6.由題意,得a2+(b-3)2=(6-1)2=25.若b=6,則a=±4;若b=-6,則a無(wú)解.故所求圓方程為(x±4)2+(y-6)2=36.答案:D4.若圓C1:x2+y2=4與圓C2:x2+y2-2ax+a2-1=0內(nèi)切,則a等于 . 解析:圓C1的圓心C1(0,0),半徑r1=2.圓C2可化為(x-a)2+y2=1,即圓心C2(a,0),半徑r2=1,若兩圓內(nèi)切,需|C1C2|=√(a^2+0^2 )=2-1=1.解得a=±1. 答案:±1 5. 已知兩個(gè)圓C1:x2+y2=4,C2:x2+y2-2x-4y+4=0,直線l:x+2y=0,求經(jīng)過(guò)C1和C2的交點(diǎn)且和l相切的圓的方程.解:設(shè)所求圓的方程為x2+y2+4-2x-4y+λ(x2+y2-4)=0,即(1+λ)x2+(1+λ)y2-2x-4y+4(1-λ)=0.所以圓心為 1/(1+λ),2/(1+λ) ,半徑為1/2 √((("-" 2)/(1+λ)) ^2+(("-" 4)/(1+λ)) ^2 "-" 16((1"-" λ)/(1+λ))),即|1/(1+λ)+4/(1+λ)|/√5=1/2 √((4+16"-" 16"(" 1"-" λ^2 ")" )/("(" 1+λ")" ^2 )).解得λ=±1,舍去λ=-1,圓x2+y2=4顯然不符合題意,故所求圓的方程為x2+y2-x-2y=0.

  • 直線的點(diǎn)斜式方程教學(xué)設(shè)計(jì)人教A版高中數(shù)學(xué)選擇性必修第一冊(cè)

    直線的點(diǎn)斜式方程教學(xué)設(shè)計(jì)人教A版高中數(shù)學(xué)選擇性必修第一冊(cè)

    【答案】B [由直線方程知直線斜率為3,令x=0可得在y軸上的截距為y=-3.故選B.]3.已知直線l1過(guò)點(diǎn)P(2,1)且與直線l2:y=x+1垂直,則l1的點(diǎn)斜式方程為________.【答案】y-1=-(x-2) [直線l2的斜率k2=1,故l1的斜率為-1,所以l1的點(diǎn)斜式方程為y-1=-(x-2).]4.已知兩條直線y=ax-2和y=(2-a)x+1互相平行,則a=________. 【答案】1 [由題意得a=2-a,解得a=1.]5.無(wú)論k取何值,直線y-2=k(x+1)所過(guò)的定點(diǎn)是 . 【答案】(-1,2)6.直線l經(jīng)過(guò)點(diǎn)P(3,4),它的傾斜角是直線y=3x+3的傾斜角的2倍,求直線l的點(diǎn)斜式方程.【答案】直線y=3x+3的斜率k=3,則其傾斜角α=60°,所以直線l的傾斜角為120°.以直線l的斜率為k′=tan 120°=-3.所以直線l的點(diǎn)斜式方程為y-4=-3(x-3).

  • 直線與圓的位置關(guān)系教學(xué)設(shè)計(jì)人教A版高中數(shù)學(xué)選擇性必修第一冊(cè)

    直線與圓的位置關(guān)系教學(xué)設(shè)計(jì)人教A版高中數(shù)學(xué)選擇性必修第一冊(cè)

    切線方程的求法1.求過(guò)圓上一點(diǎn)P(x0,y0)的圓的切線方程:先求切點(diǎn)與圓心連線的斜率k,則由垂直關(guān)系,切線斜率為-1/k,由點(diǎn)斜式方程可求得切線方程.若k=0或斜率不存在,則由圖形可直接得切線方程為y=b或x=a.2.求過(guò)圓外一點(diǎn)P(x0,y0)的圓的切線時(shí),常用幾何方法求解設(shè)切線方程為y-y0=k(x-x0),即kx-y-kx0+y0=0,由圓心到直線的距離等于半徑,可求得k,進(jìn)而切線方程即可求出.但要注意,此時(shí)的切線有兩條,若求出的k值只有一個(gè)時(shí),則另一條切線的斜率一定不存在,可通過(guò)數(shù)形結(jié)合求出.例3 求直線l:3x+y-6=0被圓C:x2+y2-2y-4=0截得的弦長(zhǎng).思路分析:解法一求出直線與圓的交點(diǎn)坐標(biāo),解法二利用弦長(zhǎng)公式,解法三利用幾何法作出直角三角形,三種解法都可求得弦長(zhǎng).解法一由{■(3x+y"-" 6=0"," @x^2+y^2 "-" 2y"-" 4=0"," )┤得交點(diǎn)A(1,3),B(2,0),故弦AB的長(zhǎng)為|AB|=√("(" 2"-" 1")" ^2+"(" 0"-" 3")" ^2 )=√10.解法二由{■(3x+y"-" 6=0"," @x^2+y^2 "-" 2y"-" 4=0"," )┤消去y,得x2-3x+2=0.設(shè)兩交點(diǎn)A,B的坐標(biāo)分別為A(x1,y1),B(x2,y2),則由根與系數(shù)的關(guān)系,得x1+x2=3,x1·x2=2.∴|AB|=√("(" x_2 "-" x_1 ")" ^2+"(" y_2 "-" y_1 ")" ^2 )=√(10"[(" x_1+x_2 ")" ^2 "-" 4x_1 x_2 "]" ┴" " )=√(10×"(" 3^2 "-" 4×2")" )=√10,即弦AB的長(zhǎng)為√10.解法三圓C:x2+y2-2y-4=0可化為x2+(y-1)2=5,其圓心坐標(biāo)(0,1),半徑r=√5,點(diǎn)(0,1)到直線l的距離為d=("|" 3×0+1"-" 6"|" )/√(3^2+1^2 )=√10/2,所以半弦長(zhǎng)為("|" AB"|" )/2=√(r^2 "-" d^2 )=√("(" √5 ")" ^2 "-" (√10/2) ^2 )=√10/2,所以弦長(zhǎng)|AB|=√10.

  • 直線的一般式方程教學(xué)設(shè)計(jì)人教A版高中數(shù)學(xué)選擇性必修第一冊(cè)

    直線的一般式方程教學(xué)設(shè)計(jì)人教A版高中數(shù)學(xué)選擇性必修第一冊(cè)

    解析:當(dāng)a0時(shí),直線ax-by=1在x軸上的截距1/a0,在y軸上的截距-1/a>0.只有B滿足.故選B.答案:B 3.過(guò)點(diǎn)(1,0)且與直線x-2y-2=0平行的直線方程是( ) A.x-2y-1=0 B.x-2y+1=0C.2x+y=2=0 D.x+2y-1=0答案A 解析:設(shè)所求直線方程為x-2y+c=0,把點(diǎn)(1,0)代入可求得c=-1.所以所求直線方程為x-2y-1=0.故選A.4.已知兩條直線y=ax-2和3x-(a+2)y+1=0互相平行,則a=________.答案:1或-3 解析:依題意得:a(a+2)=3×1,解得a=1或a=-3.5.若方程(m2-3m+2)x+(m-2)y-2m+5=0表示直線.(1)求實(shí)數(shù)m的范圍;(2)若該直線的斜率k=1,求實(shí)數(shù)m的值.解析: (1)由m2-3m+2=0,m-2=0,解得m=2,若方程表示直線,則m2-3m+2與m-2不能同時(shí)為0,故m≠2.(2)由-?m2-3m+2?m-2=1,解得m=0.

  • 新人教版高中英語(yǔ)必修3Unit 3 Diverse Cultures-Discovering Useful Structure教學(xué)設(shè)計(jì)

    新人教版高中英語(yǔ)必修3Unit 3 Diverse Cultures-Discovering Useful Structure教學(xué)設(shè)計(jì)

    Step 4 PracticeRead the conversation. Find out which words have been left out.Justin: Linlin, I’m going to Guizhou Province next month. I’m super excited! Any recommendations for places to visit?Linlin: Wow, cool! Guizhou is a province with a lot of cultural diversity. Places to visit...well, definitely the Huangguoshu Waterfall first.Justin: What’s special about the waterfall?Linlin: Well, have you ever heard of the Chinese novel Journey to the West ?Justin: Yes, I have. Why ?Linlin: In the back of the waterfall, you will find a cave, which is the home of the Monkey King.Justin: Really? Cool! I’ll definitely check it out.Linlin:And I strongly recommend the ethnic minority villages. You’ll find Chinese culture is much more diverse than you thought.Justin:Sounds great, thanks.Answers:Justin: Linlin, I’m going to Guizhou Province next month. I’m super excited! Do you have any recommendations for places to visit?Linlin: Wow, that’s cool! Guizhou is a province with a lot of cultural diversity. What are some places to visit in Guizhou ? Well, definitely the Huangguoshu Waterfall is the first place to visit in Guizhou Province.Justin: What’s special about the waterfall?Linlin: Well, have you ever heard of the Chinese novel Journey to the West ?Justin: Yes, I have heard of the Chinese novel Journey to the West . Why do you ask if I have heard of the Chinese novel Journey to the West?Linlin: In the back of the waterfall, you will find a cave, which is the home of the Monkey King from Journey to the West.Justin: That’s really true? It’s Cool! I’ll definitely check it out.Linlin:And I strongly recommend the ethnic minority villages on your trip to Guizhou Province. You’ll find Chinese culture is much more diverse than you thought it was.Justin:This all sounds great, thanks.

  • 新人教版高中英語(yǔ)必修3Unit 3 Diverse Cultures-Reading for Writing教學(xué)設(shè)計(jì)

    新人教版高中英語(yǔ)必修3Unit 3 Diverse Cultures-Reading for Writing教學(xué)設(shè)計(jì)

    The topic of this part is “Describe a place with distinctive cultural identity”.This section focuses on Chinese culture by introducing Chinatown, whose purpose is to show the relationship between the Chinese culture and American culture. The Chinese culture in Chinatown is an important part of American culture. Chinatown is an important window of spreading Chinese culture and the spirit homeland of oversea Chinese, where foreigners can experience Chinese culture by themselves.Concretely, the title is “Welcome to Chinatown!”, from which we can know that the article aims at introducing Chinatown. The author used the “Introduction--Body Paragraph--Conclusion” to describe the people, language, architecture, business, famous food and drinks and people’s activities, which can be a centre for Chinese culture and shows its unique charm.1. Read quickly to get main idea; read carefully to get the detailed information.2. Learn the characteristics of writing and language.3. Learn to introduce your own town according to the text.4. Learn to correct others’ writing.1. Learn the characteristics of writing and language.2. Learn to introduce your own town according to the text.Step 1 Lead in ---Small talkIn the reading part, we mentioned the Chinatown of San Francisco. How much do you know about Chinatown of San Francisco ?Chinatown is a main living place for Chinese immigrants, where you can see many Chinese-style buildings, costumes, operas, restaurants, music and even hear Chinese.Step 2 Before reading ---Predict the contentWhat is the writer’s purpose of writing this text ? How do you know ?From the title(Welcome to Chinatown) and some key words from the text(tourist, visit, visitors, experience), we can know the purpose of the text is to introduce Chinatown and show the relationship between Chinese culture and American culture.

  • 新人教版高中英語(yǔ)必修3Unit 3 Diverse Cultures-Listening &Speaking&Talking教學(xué)設(shè)計(jì)

    新人教版高中英語(yǔ)必修3Unit 3 Diverse Cultures-Listening &Speaking&Talking教學(xué)設(shè)計(jì)

    1. In Picture 1 and Picture 2, where do you think they are from? How do you know?From their wearings, we can know they are from ethnic minority of China--- Miao and Dong.Picture 1, they are playing their traditional instrument lusheng in their traditional costumes.Picture 2. the girls are Miao because they wear their traditional costumes and silver accessory.2. In Picture 3, can you find which village it is? What time is it in the picture?It is Dong village. It is at night. Step 2 While-listeningJustin met a new friend while traveling in Guizhou. Listen to their conversation and complete the summaries below.Part 1Justin and Wu Yue watched some Miao people play the lusheng. The instrument has a history of over 3,000 years and it is even mentioned in the oldest collection of Chinese poetry. Then they watched the lusheng dance. Justin wanted to buy some hand-made silver/traditional accessories as souvenirs. He was told that the price will depend on the percentage of silver. Part 2They will go to a pretty Dong minority village called Zhaoxing. they will see the drum towers and the wind and rain bridges. They may also see a performance of the Grand Song of the Dong people.Step 3 Post-listening---TalkingWork in groups. Imagine Justin is telling some friends about his trip to Guizhou. One of you is Justin and the rest of you are his friends. Ask Justin questions about his trip and experience. The following expressions may help you.

  • 新人教版高中英語(yǔ)必修3Unit 3 Diverse Cultures-Reading and Thinking教學(xué)設(shè)計(jì)

    新人教版高中英語(yǔ)必修3Unit 3 Diverse Cultures-Reading and Thinking教學(xué)設(shè)計(jì)

    Discuss these questions in groups.Q1: Have you ever been to a place that has a diverse culture ? What do you think about the culture diversity ?One culturally diverse place that I have been to is Harbin, the capital city of Heilongjiang Province. I went there last year with my family to see the Ice and Snow Festival, and I was amazed at how the culture as different to most other Chinese cities. There is a big Russian influence there, with beautiful Russian architecture and lots of interesting restaurants. I learnt that Harbin is called “the Oriental Moscow” and that many Russians settled there to help build the railway over 100 years ago.Q2: What are the benefits and challenges of cultural diversity ?The benefits: People are able to experience a wide variety of cultures, making their lives more interesting, and it can deepen the feelings for our national culture, it is also helpful for us to learn about other outstanding culture, which helps improve the ability to respect others. The challenges: People may have trouble communicating or understanding each other, and it may lead to disappearance of some civilizations and even make some people think “The western moon is rounder than his own.”Step 7 Post reading---RetellComplete the passage according to the text.Today, I arrived back in San Francisco, and it feels good (1) _____(be) back in the city again. The city succeeded in (2)_________ (rebuild) itself after the earthquake that (3)________ (occur) in 1906, and I stayed in the Mission District, enjoying some delicious noodles mixed with cultures. In the afternoon, I headed to a local museum (4)____ showed the historical changes in California. During the gold rush, many Chinese arrived, and some opened up shops and restaurants in Chinatown to earn a (5)_____ (live). Many others worked on (6)______ (farm), joined the gold rush, or went to build the railway that connected California to the east. The museum showed us (7)____ America was built by immigrants from (8)________ (difference) countries and cultures. In the evening, I went to Chinatown, and ate in a Cantonese restaurant that served food on (9)________(beauty) china plates. Tomorrow evening, I’m going to (10)__ jazz bar in the Richmond District. 答案:1. to be 2. rebuilding 3. occurred 4. that 5.living6. farms 7.how 8. different 9. beautiful 10. a

  • 人教版高中政治必修4矛盾是事物發(fā)展的源泉和動(dòng)力精品教案

    人教版高中政治必修4矛盾是事物發(fā)展的源泉和動(dòng)力精品教案

    一、教材分析第一目,矛盾的統(tǒng)一性和斗爭(zhēng)性。世界上的一切事物都包含著兩個(gè)方面——矛盾的定義——矛盾的兩個(gè)基本屬性——矛盾的同一性——矛盾的斗爭(zhēng)性——同一性和斗爭(zhēng)性的辯證關(guān)系。這一目的重點(diǎn)是讓學(xué)生理解世界上的一切事物都包含著矛盾,沒(méi)有矛盾就沒(méi)有世界。第二目,矛盾的普遍性和特殊性。這一目邏輯順序是:事事有矛盾,時(shí)時(shí)有矛盾——承認(rèn)矛盾的普遍性是堅(jiān)持唯物主義的前途——矛盾的特殊性及其三層涵義——矛盾的普遍性和特殊性的辯證關(guān)系——矛盾普遍性和特殊性關(guān)系的原理是矛盾問(wèn)題的精髓。最后得出結(jié)論:馬克思主義普遍原理與中國(guó)具體實(shí)際相結(jié)合體現(xiàn)了矛盾普遍性和特殊性的具體的歷史的統(tǒng)一。學(xué)習(xí)了唯物辯證法的矛盾觀,就要學(xué)會(huì)理論聯(lián)系實(shí)際,學(xué)會(huì)在生活、學(xué)習(xí)和工作中進(jìn)一步運(yùn)用所學(xué)的知識(shí),處理好生活中的實(shí)際問(wèn)題

  • 人教版高中政治必修4唯物主義和唯心主義精品教案

    人教版高中政治必修4唯物主義和唯心主義精品教案

    一、教材分析《唯物主義和唯心主義》是人教版高中思想政治必修模塊4《生活與哲學(xué)》第一單元第二課第二框題內(nèi)容。這一框主要是通過(guò)對(duì)哲學(xué)存在和發(fā)展的具體形態(tài)的介紹,讓學(xué)生從中感受什么是哲學(xué)。圍繞著這個(gè)問(wèn)題,教材設(shè)計(jì)了兩目:第一目主要是通過(guò)對(duì)歷史上各種不同的唯物主義哲學(xué)的介紹,從中概括出唯物主義的三種基本形態(tài);第二目主要是通過(guò)對(duì)歷史上各種不同的唯心主義哲學(xué)的介紹,從中概括出唯心主義的兩種基本形態(tài)。二、教學(xué)目標(biāo)(一)知識(shí)目標(biāo)什么是唯物主義,什么是唯心主義 ;理解哲學(xué)基本問(wèn)題第一方面的內(nèi)容是劃分唯物主義和唯心主義的唯一標(biāo)準(zhǔn);如何區(qū)分唯物主義的三種基本形態(tài)和唯心主義的兩種基本形態(tài)。(二)能力目標(biāo)初步具有自覺(jué)運(yùn)用唯物主義理論知識(shí),分析和把握社會(huì)生活現(xiàn)象的 能力。(三)情感、態(tài)度與價(jià)值觀目標(biāo)在實(shí)踐中堅(jiān)持辨證唯物主義觀點(diǎn),自覺(jué)反對(duì)和批判唯心主義。三、教學(xué)重點(diǎn)難點(diǎn)1、唯物主義和唯心主義的根本觀點(diǎn)(重點(diǎn))

  • 人教版高中歷史必修3現(xiàn)代中國(guó)教育的發(fā)展教案

    人教版高中歷史必修3現(xiàn)代中國(guó)教育的發(fā)展教案

    2、確立教育優(yōu)先發(fā)展地位,提出“科教興國(guó)”戰(zhàn)略:①提出“三個(gè)面向”指導(dǎo)方針;(即教育要面向現(xiàn)代化,面向世界,面向未來(lái))1983年,當(dāng)我們國(guó)家的改革開放處在起步階段時(shí),鄧小平同志以歷史的眼光,從戰(zhàn)略的高度,為北京景山學(xué)校題詞:“教育要面向現(xiàn)代化,面向世界,面向未來(lái)?!倍嗄陙?lái),這“三個(gè)面向”的題詞所蘊(yùn)含的深刻的教育理念,已經(jīng)成為中國(guó)教育改革與發(fā)展的指針,“三個(gè)面向”的思想,已經(jīng)深入人心;成為我們教育改革的旗幟和靈魂。②改革教育制度,基礎(chǔ)、中等和高等教育全面發(fā)展;基礎(chǔ)教育——普及九年義務(wù)教育,制定《義務(wù)教育法》(2006年)中等教育——實(shí)行普通教育與職業(yè)教育并舉;高等教育——增設(shè)邊緣學(xué)科,建立學(xué)位制,擴(kuò)大自主權(quán)③實(shí)施發(fā)展高等教育的“211工程”計(jì)劃;211工程"就是面向21世紀(jì),重點(diǎn)建設(shè)100所左右的高等學(xué)校和一批重點(diǎn)學(xué)科點(diǎn)。

  • 人教版高中歷史必修3西方人文主義思想的起源教案2篇

    人教版高中歷史必修3西方人文主義思想的起源教案2篇

    在當(dāng)時(shí)雅典的公民大會(huì)和陪審法庭上,人們常常要發(fā)表意見(jiàn),要和自己的對(duì)手辯論,雅典法庭規(guī)定每個(gè)公民須替自己辯護(hù),不許旁人代辯。所以出現(xiàn)了這樣一批專門教授人辯論、演說(shuō)、修辭的技巧和參政知識(shí)的職業(yè)教師。①政治因素:雅典奴隸制民主政治發(fā)展到頂峰,成為希臘政治和文化中心。參與政治生活成為每個(gè)公民生活的重要內(nèi)容②古希臘工商業(yè)發(fā)展,奴隸制經(jīng)濟(jì)繁榮(在廣大奴隸的勞動(dòng)基礎(chǔ)上,古希臘的經(jīng)濟(jì)迅速發(fā)展起來(lái),為哲學(xué)的成長(zhǎng)提供了物質(zhì)條件)——根本原因③人的地位的提高(民主政治制度和每個(gè)公民參與政治意識(shí)的加強(qiáng),使人的中心地位日益突出)最后教師強(qiáng)調(diào):提示并強(qiáng)調(diào)學(xué)生學(xué)習(xí)時(shí)要注意理解“一定的文化是一定社會(huì)的政治和經(jīng)濟(jì)在觀念形態(tài)上的反映”。3、代表人物:普羅泰格拉4、研究領(lǐng)域:人和人類社會(huì)關(guān)注人與人之間的關(guān)系、社會(huì)組織、風(fēng)俗習(xí)慣和倫理規(guī)范

  • 人教版高中語(yǔ)文必修1《心音共鳴:寫觸動(dòng)心靈的人和事》教案3篇

    人教版高中語(yǔ)文必修1《心音共鳴:寫觸動(dòng)心靈的人和事》教案3篇

    《普通高中語(yǔ)文課程標(biāo)準(zhǔn)》關(guān)于“表達(dá)與交流”方面學(xué)生應(yīng)達(dá)到的目標(biāo)有如下的表述:“學(xué)會(huì)多角度地觀察生活,豐富生活經(jīng)歷和情感體驗(yàn),對(duì)自然、社會(huì)和人生有自己的感受和思考”,“進(jìn)一步提高記敘述、說(shuō)明、描寫、議論、抒情等基本表達(dá)能力”。觀察、感受、思考是寫好作文的必要的積累與條件,而用最恰當(dāng)?shù)恼Z(yǔ)言與形式傳達(dá)自己的所得則屬于“技巧”方面的范疇。教材“表達(dá)與交流”的編選采用的“話題探討—寫法借鑒—寫作練習(xí)”的體例,其優(yōu)點(diǎn)是就某一話題訓(xùn)練某一方面的寫作能力,能使教與學(xué)具有較強(qiáng)的操作性,目標(biāo)更具體,也就是“既講‘寫什么’,又講‘怎么寫’”,能克服“純技術(shù)性訓(xùn)練”;不足在于容易造成教與學(xué)上的“只見(jiàn)樹木、不見(jiàn)森林”現(xiàn)象。要讓學(xué)生確實(shí)形成能力,舉一反三,老師的備課量非常之大,好在現(xiàn)在網(wǎng)絡(luò)發(fā)達(dá),必修1和必修2還配了教案(不知為什么必修3和必修4沒(méi)有),總算應(yīng)對(duì)過(guò)來(lái),因此,我在此所講的教學(xué)設(shè)計(jì)之類的,有許多不是我個(gè)人的,是別人的成果,特此聲明。

上一頁(yè)123...383940414243444546474849下一頁(yè)
提供各類高質(zhì)量Word文檔下載,PPT模板下載,PPT背景圖片下載,免費(fèi)ppt模板下載,ppt特效動(dòng)畫,PPT模板免費(fèi)下載,專注素材下載!

PPT全稱是PowerPoint,LFPPT為你提供免費(fèi)PPT模板下載資源。讓你10秒輕松搞定幻燈片制作,打造?顏值的豐富演示文稿素材模版合集。