《高中數(shù)學(xué) 第三章 導(dǎo)數(shù)應(yīng)用 3.2導(dǎo)數(shù)在實(shí)際問(wèn)題中的應(yīng)用 3.2.1 實(shí)際問(wèn)題中導(dǎo)數(shù)的意義課件 北師大版選修22》由會(huì)員分享，可在線閱讀，更多相關(guān)《高中數(shù)學(xué) 第三章 導(dǎo)數(shù)應(yīng)用 3.2導(dǎo)數(shù)在實(shí)際問(wèn)題中的應(yīng)用 3.2.1 實(shí)際問(wèn)題中導(dǎo)數(shù)的意義課件 北師大版選修22（12頁(yè)珍藏版）》請(qǐng)?jiān)谘b配圖網(wǎng)上搜索。

1、2 2導(dǎo)數(shù)在實(shí)際問(wèn)題中的應(yīng)導(dǎo)數(shù)在實(shí)際問(wèn)題中的應(yīng)用用2 2.1 1實(shí)際問(wèn)題中導(dǎo)數(shù)的意義MUBIAODAOHANG目標(biāo)導(dǎo)航DIANLI TOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHI SHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂

2、演練ZHISHISHULI知識(shí)梳理1.理解平均變化率與導(dǎo)數(shù)的關(guān)系.2.理解導(dǎo)數(shù)的實(shí)際意義.3.體會(huì)導(dǎo)數(shù)的意義在實(shí)際生活中的應(yīng)用.MUBIAODAOHANG目標(biāo)導(dǎo)航DIANLI TOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHI SHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOU

3、XI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理生活中的變化率問(wèn)題(1)在物理學(xué)中,通常稱力在單位時(shí)間內(nèi)做的功為功率,它的單位是瓦特.(2)在氣象學(xué)中,通常把在單位時(shí)間(如1時(shí)、1天等)內(nèi)的降雨量稱作降雨強(qiáng)度,它是反映一次降雨大小的一個(gè)重要指標(biāo).(3)在經(jīng)濟(jì)學(xué)中,通常把生產(chǎn)成本y關(guān)于產(chǎn)量x的函數(shù)y=f(x)的導(dǎo)函數(shù)稱為邊際成本,邊際成本f(x0)指的是當(dāng)產(chǎn)量為x0時(shí),生產(chǎn)成本的增加速度,也就是當(dāng)產(chǎn)量為x0時(shí),每增加一個(gè)單位的產(chǎn)量,需要增加f(x0)個(gè)單位的成本.MUBIAODAOHANG目標(biāo)導(dǎo)航DIANLI TOUXI典例透析SUITANGYANLIAN隨堂演練Z

4、HISHI SHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理MUBIAODAOHANG目標(biāo)導(dǎo)航DIANLI TOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHI SHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLI

5、TOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理MUBIAODAOHANG目標(biāo)導(dǎo)航DIANLI TOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHI SHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂

6、演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理【變式訓(xùn)練】 假設(shè)某國(guó)家在20年間通貨膨脹率為5%,物價(jià)p(單位:元)與時(shí)間t(單位:年)有如下函數(shù)關(guān)系:p(t)=p0(1+5%)t,其中p0為t=0時(shí)的物價(jià).假定某種商品的p0=1,則在第10個(gè)年頭,這種商品價(jià)格的上漲速度大約是多少?(精確到0.01

7、元/年)解:因?yàn)閜0=1,所以p(t)=(1+5%)t=1.05t.根據(jù)基本初等函數(shù)的導(dǎo)數(shù)公式,得p(t)=(1.05t)=1.05tln 1.05,所以p(10)=1.0510ln 1.050.08(元/年).故在第10個(gè)年頭,這種商品價(jià)格的上漲速度約為0.08元/年.MUBIAODAOHANG目標(biāo)導(dǎo)航DIANLI TOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHI SHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISH

8、ULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理1 2 3 4 51如果質(zhì)點(diǎn)A按規(guī)律s(t)=3t2運(yùn)動(dòng),那么在t=3時(shí)的瞬時(shí)速度為()A.6B.18C.54 D.81解析:瞬時(shí)速度v(t)=s(t)=(3t2)=6t,v(3)=63=18.答案:BMUBIAODAOHANG目標(biāo)導(dǎo)航DIANLI TOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHI SHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SU

9、ITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理1 2 3 4 5MUBIAODAOHANG目標(biāo)導(dǎo)航DIANLI TOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHI SHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練

10、ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理1 2 3 4 53做直線運(yùn)動(dòng)的某物體,其位移s與時(shí)間t的關(guān)系是s(t)=3t-t2,則該物體的初速度是.解析:s(t)=3-2t,s(0)=3,即該物體的初速度是3.答案:3MUBIAODAOHANG目標(biāo)導(dǎo)航DIANLI TOUXI典例透析SUITANGY

11、ANLIAN隨堂演練ZHISHI SHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理1 2 3 4 54.豎直向上彈射一個(gè)小球,小球的初速度為100 m/s,試求小球何時(shí)的瞬時(shí)速度為0 m/s?(g9.8 m/s2,

12、結(jié)果精確到0.1 s)MUBIAODAOHANG目標(biāo)導(dǎo)航DIANLI TOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHI SHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理1 2 3 4 55某企業(yè)每

13、天的產(chǎn)品均能售出,售價(jià)為490元/噸,其每天的成本C與每天的產(chǎn)量q之間的函數(shù)關(guān)系為C(q)=2 000+450q+0.02q2.(1)寫出收入函數(shù);(2)寫出利潤(rùn)函數(shù);(3)求利潤(rùn)函數(shù)的導(dǎo)數(shù),并說(shuō)明其經(jīng)濟(jì)意義.解:設(shè)收入函數(shù)為R(q),利潤(rùn)函數(shù)為L(zhǎng)(q).(1)收入函數(shù)為R(q)=490q.(2)利潤(rùn)函數(shù)為L(zhǎng)(q)=R(q)-C(q)=490q-(2 000+450q+0.02q2)=-2 000+40q-0.02q2.(3)利潤(rùn)函數(shù)的導(dǎo)數(shù)為L(zhǎng)(q)=(-2 000+40q-0.02q2)=40-0.04q.利潤(rùn)函數(shù)的導(dǎo)數(shù)稱為邊際利潤(rùn),其經(jīng)濟(jì)意義為當(dāng)產(chǎn)量達(dá)到q時(shí),再增加單位產(chǎn)量后利潤(rùn)的改變量.