Help me in these problems please thanks

asked Feb 10 in Calculus Answers by Goodboy123

Your answer

Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:

To avoid this verification in future, please log in or register.

1 Answer

11. If the variation in 2 is k corresponding to variation h in x near x=3 then


Squaring: 1+x±h=4±4k+k². When x=3, h=±4k+k².

When k=1, h=5 and -3, and x=(3-3,3+5)=(0,8) correspond to (1,3) which are (2-1,2+1).

When k=0.0002, h=±0.0008+0.00000004, that is to 4 decimal places h=±0.0008.

So x=(3-0.0008,3+0.0008)=(2.9992,3.0008).

12. Let x be the length of each wire then x² is the area 100±0.06 sq in. If the variation for x is h, then:


x+h=10√(1±0.0006)=10(1±0.0006)^½=10(1±0.0003) approx=10±0.003.

Whenever x=10, h=±0.003. So the variation in the length of each piece of wire is less than 0.003 in.

Another way to solve this: x²+2hx+h²=100±0.06, so, putting x=10, 20h=0.06 if h is small. So h=0.003 in. approx.

answered Feb 10 by Rod Top Rated User (510,360 points)

Related questions

1 answer
asked Feb 10 in Calculus Answers by Goodboy123 | 23 views
1 answer
asked Feb 10 in Calculus Answers by Joseph | 15 views
0 answers
asked Apr 2, 2013 in Calculus Answers by awi (140 points) | 225 views
0 answers
asked Jun 7, 2012 in Calculus Answers by anonymous | 74 views
Welcome to, where students, teachers and math enthusiasts can ask and answer any math question. Get help and answers to any math problem including algebra, trigonometry, geometry, calculus, trigonometry, fractions, solving expression, simplifying expressions and more. Get answers to math questions. Help is always 100% free!
80,492 questions
84,419 answers
67,804 users