Taylor series and quadratic approximation

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To approximate the square root of 36.03 using a local quadratic approximation, rewrite it as √(36 + 0.03), which can be expressed as 6√(1 + ε). The value of ε is determined by the small increment from 36, specifically ε = 0.03/36. Expanding this expression to the second order of ε allows for an accurate approximation. The discussion emphasizes the importance of understanding the Taylor series expansion for effective problem-solving in calculus.
ookt2c
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Homework Statement



use an appropriate local quadratic approximation to approximate the square root of 36.03

Homework Equations



not sure

The Attempt at a Solution



missed a day of class
 
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ookt2c said:

Homework Statement



use an appropriate local quadratic approximation to approximate the square root of 36.03

Homework Equations



not sure

The Attempt at a Solution



missed a day of class

You need to rwwrite
\sqrt{36+0.03}
in the form 6 \sqrt{1+\epsilon} and then expand this to order epsilon squared. I will let you figure out what the value of epsilon is.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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