Answer to Question #264742 in Calculus for JaytheCreator

Question #264742

Question 3

Determine the length of the curve ๐‘ฅ = ๐‘ฆ^2 /2 for 0 โ‰ค ๐‘ฅ โ‰ค 1/2 . Assume ๐‘ฆ positive.


1
Expert's answer
2022-01-10T17:56:05-0500

"L= \\int \\sqrt{1+(y')^2}dx\\\\\ny=\\sqrt{2x}\\\\\ny'=\\frac{1}{\\sqrt{2x}}\\\\\nL= \\int_0^\\frac{1}{2}\\sqrt{1+\\frac{1}{2x}}dx\\\\"

This can also be written asย "L= \\int_0^\\frac{1}{2}\\sqrt{\\frac{1}{2x}+1}dx\\\\"

we apply linearity

"\\frac{1}{\\sqrt{2}}\\int \\sqrt{\\frac{1}{x}+2}dx\\\\"

let u =ย "\\sqrt{\\frac{1}{x}+2}\\\\"ย , dx = -2"\\sqrt{\\frac{1}{x}+2x^2}du, dx=-2\\int \\frac{u^2}{(u^2-2)}du\\\\"

we integrateย "\\int \\frac{u^2}{(u^2-2)}du = \\int (\\frac{u^2-2}{(u^2-2)^2}+\\frac{2}{(u^2-2)^2})du=\\\\"

"\\int (\\frac{1}{(u^2-2)}+\\frac{2}{(u^2-2)^2})du\\\\"

we split and integrate differently.

"\\int (\\frac{1}{(u^2-2)}du \\implies \\frac{Ln(u-\\sqrt{2})}{2^{\\frac{3}{2}}}- \\frac{Ln(u+\\sqrt{2})}{2^{\\frac{3}{2}}}"

also

"\\int\\frac{2}{(u^2-2)^2}du \\implies \\int (\\frac{1}{(u^2-2)}du \\implies \\frac{Ln(u-\\sqrt{2})}{2^{\\frac{3}{2}}}- \\frac{Ln(u-\\sqrt{2})}{2^{\\frac{3}{2}}} - \\frac{1}{4(u+\\sqrt{2})}- \\frac{1}{4(u-\\sqrt{2})}\\\\"

but substituting the solution of the integrations back, we get

"\\frac{Ln(u-\\sqrt{2})}{2^{\\frac{3}{2}}}- \\frac{Ln(u+\\sqrt{2})}{2^{\\frac{3}{2}}}+ \\frac{Ln(u-\\sqrt{2})}{2^{\\frac{3}{2}}}- \\frac{Ln(u-\\sqrt{2})}{2^{\\frac{3}{2}}} - \\frac{1}{4(u+\\sqrt{2})}- \\frac{1}{4(u-\\sqrt{2})}"

recall, u=ย "\\sqrt{\\frac{1}{x}+2}\\\\"ย then we substitute u in terms of x back and introduce the lower and upper limits

"\\frac{Ln(u-\\sqrt{2})}{2^{\\frac{3}{2}}}- \\frac{Ln(u+\\sqrt{2})}{2^{\\frac{3}{2}}}+ \\frac{Ln(u-\\sqrt{2})}{2^{\\frac{3}{2}}}- \\frac{Ln(u-\\sqrt{2})}{2^{\\frac{3}{2}}} - \\frac{1}{4(u+\\sqrt{2})}- \\frac{1}{4(u-\\sqrt{2})}|_0^\\frac{1}{2}"

We have.

"\\frac{1}{\\sqrt{2}}+\\frac{Ln(\\sqrt{2}+1)-Ln(\\sqrt{2}-1)}{4}"

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