3.9 – inverse trigonometric functions
DESCRIPTION
3.9 – Inverse Trigonometric Functions. 3.9 – Inverse Trigonometric Functions. 3.9 – Inverse Trigonometric Functions. 3.9 – Inverse Trigonometric Functions. 3.11 – Linearization and Differentials. 3.11 – Linearization and Differentials. 3.11 – Linearization and Differentials. - PowerPoint PPT PresentationTRANSCRIPT
3.9 – Inverse Trigonometric Functions
3.9 – Inverse Trigonometric Functions
3.9 – Inverse Trigonometric Functions
𝑦=𝑠𝑖𝑛− 1𝑥2
𝐷𝑥 𝑠𝑖𝑛− 1𝑢=1
√1−𝑢2𝑑𝑢𝑑𝑥
𝑢=𝑥2
𝑦 ′= 1
√1− (𝑥2 )2
𝑦 ′= 2 𝑥√1−𝑥4
𝑦=𝑡𝑎𝑛− 1(3 𝑥)
𝐷𝑥𝑡𝑎𝑛− 1𝑢=1
1+𝑢2𝑑𝑢𝑑𝑥
𝑢=3 𝑥
𝑦 ′= 31+9𝑥2
𝑦 ′= 11+ (3𝑥 )2
𝑦=𝑠𝑒𝑐−1(𝑒¿¿ 2𝑥)¿
𝐷𝑥 𝑠𝑒𝑐− 1𝑢=1
|𝑢|√𝑢2−1𝑑𝑢𝑑𝑥
𝑢=𝑒2𝑥
𝑦 ′= 1
|𝑒2𝑥|√ (𝑒2𝑥 )2−1
𝑦 ′= 2𝑒2𝑥
𝑒2 𝑥√𝑒4 𝑥−1
𝑦 ′= 2√𝑒4 𝑥−1
𝐸𝑥𝑎𝑚𝑝𝑙𝑒𝑠 :
2 𝑥 3 𝑒2𝑥2
3.9 – Inverse Trigonometric Functions
𝑦=(𝑠𝑖𝑛−1 (5 𝑥2 ) )3
2 𝑥
1
√1− (5 𝑥2 )2
𝑦=𝑥2𝑡𝑎𝑛− 1(𝑥4)
𝑦 ′=𝑥2
𝑦 ′=3( 𝑠𝑖𝑛−1 (5 𝑥2 ))2
𝑦 ′=30𝑥 (𝑠𝑖𝑛− 1 (5 𝑥2 ))2
√1−25 𝑥4
10 𝑥
11+(𝑥4 )24 𝑥
3+¿ 𝑡𝑎𝑛− 1(𝑥4)
𝑦 ′= 4 𝑥5
1+𝑥8+2𝑥 𝑡𝑎𝑛−1(𝑥4)
𝐸𝑥𝑎𝑚𝑝𝑙𝑒𝑠 :
𝑦=𝑥2+2𝑑𝑦𝑑𝑥 =2𝑥
𝑎𝑡 𝑥=1 𝑦=3
𝑎𝑡 𝑥=1 𝑑𝑦𝑑𝑥 =2
𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛𝑜𝑓 h𝑡 𝑒 𝑡𝑎𝑛𝑔𝑒𝑛𝑡 𝑙𝑖𝑛𝑒𝑎𝑡 𝑥=1𝑦−3=2(𝑥−1)𝑦=2 𝑥+1
1 1
1
3.11 – Linearization and Differentials
3.11 – Linearization and Differentials
𝑎
𝑑𝑦
∆ 𝑥 𝑜𝑟 𝑑 𝑥
∆ 𝑦
𝑓 (𝑥)𝐿(𝑥)
𝑓 (𝑎)
𝑓 (𝑎 )+𝑑𝑦
𝑥−𝑎
3.11 – Linearization and Differentials
𝑎
𝑑𝑦
∆ 𝑥 𝑜𝑟 𝑑 𝑥
∆ 𝑦
𝑓 (𝑥)𝐿(𝑥)
𝑓 (𝑎)
𝑓 (𝑎 )+𝑑𝑦
𝑥−𝑎
∆ 𝑦∆𝑥 =𝑚𝑠𝑒𝑐
𝑑𝑦𝑑𝑥 = 𝑓 ′ (𝑥 )=𝑚𝑡𝑎𝑛
𝑑𝑦= 𝑓 ′ (𝑥 )𝑑𝑥
𝐴𝑝𝑝𝑟𝑜𝑥𝑖𝑚𝑎𝑡𝑖𝑜𝑛 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛:𝐿(𝑥)
𝐿 (𝑥 )=¿𝑓 (𝑎 )+𝑑𝑦𝐿 (𝑥 )= 𝑓 (𝑎 )+ 𝑓 ′ (𝑎)𝑑𝑥
𝐿 (𝑥 )= 𝑓 (𝑎 )+ 𝑓 ′ (𝑎)(𝑥−𝑎)
𝐿 (𝑥 )≈ 𝑓 (𝑥 )
𝐿𝑖𝑛𝑒𝑎𝑟𝑖𝑧𝑎𝑡𝑖𝑜𝑛𝑜𝑓 𝑓 (𝑥 )
3.11 – Linearization and Differentials
3.11 – Linearization and Differentials
𝐹𝑖𝑛𝑑 h𝑡 𝑒𝑙𝑖𝑛𝑒𝑎𝑟𝑖𝑧𝑎𝑡𝑖𝑜𝑛 𝑓𝑜𝑟 𝑓 (𝑥 )=√𝑥2+16𝑛𝑒𝑎𝑟 𝑥=−3
2 𝑥
𝑓 (𝑥 )=√𝑥2+16¿ (𝑥2+16 )12
𝑓 ′ (𝑥 )= 𝑥√𝑥2+16
𝐿 (𝑥 )= 𝑓 (𝑎 )+ 𝑓 ′ (𝑎)(𝑥−𝑎)
𝐿 (𝑥 )= 𝑓 (−3 )+ 𝑓 ′ (−3 )(𝑥− (−3 ))
𝐿 (𝑥 )=√(−3 )2+16+−3
√ (−3 )2+16(𝑥+3)
𝐿 (𝑥 )=5− 35(𝑥+3 ) 𝐿𝑖𝑛𝑒𝑎𝑟𝑖𝑧𝑎𝑡𝑖𝑜𝑛𝑜𝑓 𝑓 (𝑥 )
𝐸𝑥𝑎𝑚𝑝𝑙𝑒 :
3.11 – Linearization and Differentials𝐹𝑖𝑛𝑑 h𝑡 𝑒𝑙𝑖𝑛𝑒𝑎𝑟 𝑎𝑝𝑝𝑟𝑜𝑥𝑖𝑚𝑎𝑡𝑖𝑜𝑛 𝑓𝑜𝑟 𝑓 (𝑥 )=√𝑥2+16𝑎𝑡 𝑥=−2.95 ,−3.05𝑎𝑛𝑑−4.𝐿 (𝑥 )=5− 35
(𝑥+3 )
𝐿 (−2.95 )=5− 35(−2.95+3 )
𝐿 (−2.95 )=4.97𝑓 (−2.95 )=4.97016
𝐿 (−3.05 )=5− 35(−3.05+3 )
𝐿 (−3.05 )=5.03𝑓 (−3.05 )=5.03016
𝐿 (−4 )=5− 35(−4+3 )
𝐿 (−4 )=5.6
𝑓 (−4 )=5.65685
3.11 – Linearization and Differentials
𝐹𝑖𝑛𝑑 h𝑡 𝑒𝑙𝑖𝑛𝑒𝑎𝑟𝑖𝑧𝑎𝑡𝑖𝑜𝑛 𝑓𝑜𝑟 𝑓 (𝑥 )= tan𝑥𝑛𝑒𝑎𝑟 𝑥=𝜋4
𝑓 (𝑥 )=tan 𝑥𝑓 ′ (𝑥 )=𝑠𝑒𝑐2𝑥𝐿 (𝑥 )= 𝑓 (𝑎 )+ 𝑓 ′ (𝑎)(𝑥−𝑎)
𝐿 (𝑥 )= 𝑓 ( 𝜋4 )+ 𝑓 ′( 𝜋4 )(𝑥− 𝜋4 )𝐿 (𝑥 )= tan 𝜋4
+1
𝑐𝑜𝑠2 𝜋4
𝐿 (𝑥 )=1+2(𝑥− 𝜋4 ) 𝐿𝑖𝑛𝑒𝑎𝑟𝑖𝑧𝑎𝑡𝑖𝑜𝑛𝑜𝑓 𝑓 (𝑥 )
𝐸𝑥𝑎𝑚𝑝𝑙𝑒 :
(𝑥− 𝜋4 )𝐿 (𝑥 )=1
+𝑠𝑒𝑐2 𝜋4
(𝑥− 𝜋4 )
3.11 – Linearization and Differentials
𝐹𝑖𝑛𝑑 h𝑡 𝑒𝑙𝑖𝑛𝑒𝑎𝑟 𝑎𝑝𝑝𝑟𝑜𝑥𝑖𝑚𝑎𝑡𝑖𝑜𝑛 𝑓𝑜𝑟 𝑓 (𝑥 )= tan 𝑥𝑎𝑡 𝑥=2𝜋9 ,
5𝜋18 𝑎𝑛𝑑 𝜋3 .
𝐿( 2𝜋9 )=1+2( 2𝜋9 − 𝜋4 )𝐿( 2𝜋9 )=0.82547𝑓 (2𝜋9 )=0.8391
𝐿 (𝑥 )=1+2(𝑥− 𝜋4 )𝐿( 5𝜋18 )=1+2 (5𝜋18 − 𝜋4 )𝐿( 5𝜋18 )=1.17453𝑓 ( 5𝜋18 )=1.19175
𝐿( 𝜋3 )=1+2(𝜋3 − 𝜋4 )
𝐿( 𝜋3 )=1.5236
𝑓 ( 𝜋3 )=1.73205
3.11 – Linearization and Differentials
𝑈𝑠𝑒𝑙𝑖𝑛𝑒𝑎𝑟𝑖𝑧𝑎𝑡𝑖𝑜𝑛𝑡𝑜𝑎𝑝𝑝𝑟𝑜𝑥𝑖𝑚𝑎𝑡𝑒 h𝑡 𝑒𝑣𝑎𝑙𝑢𝑒𝑜𝑓 √99.4𝑓 (𝑥 )=√𝑥𝑓 ′ (𝑥 )=1
2𝑥− 12
h𝐶 𝑜𝑜𝑠𝑒𝑎𝑣𝑎𝑙𝑢𝑒𝑜𝑓 𝑥 𝑛𝑒𝑎𝑟 99.4 .𝑥=100
𝐿 (𝑥 )=√100
𝑎𝑝𝑝𝑟𝑜𝑥𝑖𝑚𝑎𝑡𝑒𝑣𝑎𝑙𝑢𝑒𝑜𝑓 √99.4
𝐸𝑥𝑎𝑚𝑝𝑙𝑒 :
+12√100
¿ 𝑥12
¿12√𝑥
𝐿 (𝑥 )= 𝑓 (𝑎 )+ 𝑓 ′ (𝑎)(𝑥−𝑎)
𝐿 (𝑥 )= 𝑓 (100 )+ 𝑓 ′ (100 )(99.4−100)
(99.4−100)
𝐿 (𝑥 )=10+ 120(− .6 )
𝐿 (𝑥 )=10−0.03=9.97
𝑓 (99.4 )=√99.4=9.96995𝑎𝑐𝑡𝑢𝑎𝑙𝑣𝑎𝑙𝑢𝑒𝑜𝑓 √99.4
3.11 – Linearization and Differentials𝐷𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑙𝑠 𝑑𝑦
𝑑𝑥 = 𝑓 ′ (𝑥 ) 𝑑𝑦= 𝑓 ′ (𝑥 )𝑑𝑥
𝐸𝑥𝑎𝑚𝑝𝑙𝑒𝑠 :𝑦=𝑒3𝑥𝑑𝑦=¿𝑒3𝑥3𝑑𝑥𝑑𝑦=3𝑒3𝑥𝑑𝑥
𝑦=cos 𝑥2𝑑𝑦=¿−sin 𝑥22 𝑥𝑑𝑥𝑑𝑦=−2𝑥𝑠𝑖𝑛𝑥2𝑑𝑥
𝑥 𝑦 2−4 𝑥32− 𝑦=0
𝑥2 𝑦𝑑𝑦−6 𝑥12 𝑑𝑥−𝑑𝑦=0
2 𝑥𝑦𝑑𝑦−𝑑𝑦=6 𝑥12 𝑑𝑥
𝑑𝑦 (2𝑥𝑦−1 )=6 𝑥12𝑑𝑥
𝑑𝑦=6 𝑥
12
2 𝑥𝑦−1 𝑑𝑥
𝑑𝑦= 6√𝑥2 𝑥𝑦−1 𝑑𝑥
𝑎+𝑑𝑥𝑎
𝑑𝑦
∆ 𝑥 𝑜𝑟 𝑑𝑥
∆ 𝑦
𝑓 (𝑥)𝐿(𝑥)
𝑓 (𝑎)
𝑓 (𝑎+𝑑𝑥)
𝑓 (𝑎 )+𝑑𝑦
𝑥−𝑎
3.11 – Linearization and Differentials𝐴𝑝𝑝𝑟𝑜𝑥𝑖𝑚𝑎𝑡𝑖𝑜𝑛𝐸𝑟𝑟𝑜𝑟 :|∆ 𝑦−𝑑𝑦|
𝑨𝒑𝒑𝒓𝒐𝒙𝒊𝒎𝒂𝒕𝒊𝒐𝒏 𝑬𝒓𝒓𝒐𝒓 :|∆ 𝒚−𝒅𝒚 |
3.11 – Linearization and Differentials𝐴𝑝𝑝𝑟𝑜𝑥𝑖𝑚𝑎𝑡𝑖𝑜𝑛𝐸𝑟𝑟𝑜𝑟 :|∆ 𝑦−𝑑𝑦|𝐸𝑥𝑎𝑚𝑝𝑙𝑒 :
∆ 𝑦= 𝑓 (1+0.1 )− 𝑓 (1)
𝑑𝑦=0.8
𝑓 (𝑥 )=2𝑥2+4 𝑥−3 𝑓𝑜𝑟 𝑥=1𝑎𝑛𝑑𝑑𝑥=0.1
𝑑𝑦=4 𝑥𝑑𝑥+4 𝑑𝑥
∆ 𝑦= 𝑓 (1.1 )− 𝑓 (1)∆ 𝑦=3.82−3∆ 𝑦=0.82
𝑓 (𝑥 )=𝑦=2𝑥2+4 𝑥−3
𝑑𝑦= (4 𝑥+4 )𝑑𝑥𝑑𝑦= (4 (1 )+4 )(0.1)
𝐴𝑝𝑝𝑟𝑜𝑥𝑖𝑚𝑎𝑡𝑖𝑜𝑛𝐸𝑟𝑟𝑜𝑟 :0.2
|∆ 𝑦−𝑑𝑦||0.82−0.8|
3.11 – Linearization and Differentials