MATH
1205Test3F06key

1205Test3F06key - Name ID AN S WE R KE Y Mat h 1 205 Class...

• Test Prep
• 4

This preview shows pages 1–3. Sign up to view the full content.

Name ANSWER KEY Math 1205 Test 3 November 29, 2006 ID # Class Time or CRN READ THE DIRECTIONS. YOU MUST SHOW ALL WORK ON THIS TEST AND USE METHODS LEARNED IN CLASS OR FROM THE WORKSHEETS TO RECEIVE FULL CREDIT. CALCULATORS ARE PERMITTED. Pledge: I have neither given nor received help on this test. Signed 1. Use linearization to estimate 99.9 . f ( x ) = x ; a = 100 ; f (100) = 100 = 10 ! f ( x ) = 1 2 x ; ! f ( x ) = 1 2 100 = 1 20 L ( x ) = f (100) + ! f (100)( x " 100) L ( x ) = 10 + 0.05( x " 100) 99.9 # L (99.9) = 10 + 0.05(99.9 " 100) = 10 + 0.05( " 0.1) = 10 " .005 = 9.995 2 . The radius of a circular disk is given as 24 cm with a maximum error in its measurement of 0.2cm. a. Use differentials to estimate the maximum error in the calculated area of the disk. r = 24 cm A = ! r 2 " r = 0.2 cm " A # dA = 2 ! rdr dA r = 24 cm " r = 0.2 cm = 2 ! (24)(0.2) = 9.6 ! cm 2 # 30.15928947 cm 2 b. Use the part a answer to estimate the percentage error. % Error = ! A A (100) " dA A (100) % Error " 9.6 # # (24) 2 (100) = 9.6 # 576 # (100) = 0.01 6 6(100)= 1. 6 6% 3. Find the third degree Taylor Polynomial centered at x = ! 2 for f ( x ) = sin( x ) .

This preview has intentionally blurred sections. Sign up to view the full version.

f ( x ) = sin( x ) f ( ! 2 ) = sin( ! 2 ) = 1 " f ( x ) = cos( x ) " f ( ! 2 ) = cos( ! 2 ) = 0 "" f ( x ) = # sin( x ) "" f ( !
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern