MATH101 April 2010
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Question 02 (c)

FullSolution Problem. Justify your answer and show all your work. Simplification of the answer is not required.
Find the length of the curve
 $\displaystyle y=x^{2}{\frac {\ln x}{8}},\quad {\text{for }}1\leq x\leq 2.$

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint

The formula for the arc length of this curve on the interval from $1\leq x\leq 2$ is
$\int _{1}^{2}{\sqrt {1+\left({\frac {dy}{dx}}\right)^{2}}}\,dx.$

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 If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work.
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Solution

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Using the formula in the hint, we see that we require the derivative of the curve in question. Notice that
 ${\frac {dy}{dx}}=2x{\frac {1}{8x}}$
and so we compute the arc length to be
 ${\begin{aligned}\int _{1}^{2}{\sqrt {1+\left({\frac {dy}{dx}}\right)^{2}}}\,dx&=\int _{1}^{2}{\sqrt {1+\left(2x{\frac {1}{8x}}\right)^{2}}}\,dx\\&=\int _{1}^{2}{\sqrt {1+4x^{2}{\frac {1}{2}}+{\frac {1}{64x^{2}}}}}\,dx\\&=\int _{1}^{2}{\sqrt {4x^{2}+{\frac {1}{2}}+{\frac {1}{64x^{2}}}}}\,dx\\&=\int _{1}^{2}{\sqrt {(2x+{\frac {1}{8x}})^{2}}}\,dx\\&=\int _{1}^{2}(2x+{\frac {1}{8x}})\,dx\\&=\left.(x^{2}+{\frac {1}{8}}\ln x)\right_{1}^{2}\\&=(2)^{2}+{\frac {1}{8}}\ln(2)(1)^{2}{\frac {1}{8}}\ln(1)\\&=3+{\frac {1}{8}}\ln(2).\end{aligned}}$

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MER QGH flag, MER QGQ flag, MER QGS flag, MER RT flag, MER Tag Arc length, Pages using DynamicPageList3 parser function, Pages using DynamicPageList3 parser tag

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