Math1011_Week08_3_910

Math1011_Week08_3_910 - Math1011 Section 3.10 and use this...

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Math1011 Section 3.10 10/21/2009 3 Find the linearization of ( ) at -8 and use this to find the value of L(-7.9) fx x a = = Find the linearization of ( ) sin( ) at 0 and use this to find the value of L(0.1) = =
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Math1011 Section 3.10 10/21/2009 3 3 Find the linearization of ( ) at a=-8 and use this to find an estimate for 7.9 fx x = 2/3 2/3 (1/3) 1/3 -2/3 11 1 1 3 12 34 3x 3(-8) 8 12 12 ( ) f(-8)=(-8) 2 f'(x)= x f'(-8)= () () ' ( ) , 8 (8 ) ' ) ( 8 ) () 2 ( 8 ) 2 Lx f a f a x a a f =⇒ =+ = =− + + =− + + + 1 = = = 4 1 12 12 3 4 1 12 3 (7 .9 ) ) 1 .99 167 L −= −− Find the linearization of ( ) sin( ) at 0 and use this to find the value of L(0.1) = = ( ) sin (0) sin0 0 '( ) cos '(0) cos0 1 ( ) , 0 ( ) (0) '(0)( 0) () 0 1 ( 0 ) (0.1) 0.1 : sin0.1 0.099833 Note = + + = = = = = = (Doc #011w.310.01t)
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Math1011 Section 3.10 2 A bicycle rider writes down her speed at the various times in the table below. If at 3:00 she is at kilometer 125, where would you expect her to be at 3:30? At 4:00? 30 Velocity Km/hr 25 20 15 10 12:00 12:30 1:00 1:30 2:00 2:30 3:00 Time Would you expect the 3:30 or the 4:00 estimate to be better? Why?
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Math1011 Section 3.10 2 A bicycle rider writes down her speed at the various times in the table below. If at 3:00 she is at kilometer 125, where would you expect her to be at 3:30? At 4:00? 30 Velocity Km/hr 25 20 15 10 12:00 12:30 1:00 1:30 2:00 2:30 3:00 Time We can use linearization to estimate the distance at 3:30: () () ' () ( ) ( ) (3) '(3)( 3) () 1 2 5 2 0 ( 3 ) at 3:30: (3.5) 125 20(3.5 3) 135 at 4:00: (4) 125 20(4 3) 145 Lx f a f a x a f x Lkm =+ = = Would you expect the 3:30 or the 4:00 estimate to be better? Why? Linear Approximation uses the fact that curves tend to lie close to the tangent line for points close to the point of tangency. Linearization Approximations are thus usually more accurate for numbers closer to the original point. We could expect our estimate for 3:30 would be more accurate than our estimate for 4:00. (Doc #011w.310.02t)
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Math1011 Section 3.10 3 Use differentials to estimate the amount of paint needed to apply a coat of paint 0.05cm thick to a hemispherical dome with diameter 50 m.
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Math1011 Section 3.10 3
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This note was uploaded on 11/22/2009 for the course CHEM 2070 at Cornell.

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Math1011_Week08_3_910 - Math1011 Section 3.10 and use this...

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