Find the exact length of the curve.MSKx=1/3, (y)^{1/2}(y-3),1\leq y\leq9

ruigE 2021-09-25 Answered

Find the exact length of the curve.

\(\displaystyle{x}=\frac{{1}}{{3}},{\left({y}\right)}^{{\frac{{1}}{{2}}}}{\left({y}-{3}\right)},{1}\leq{y}\leq{9}\)

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Expert Answer

irwchh
Answered 2021-09-26 Author has 20885 answers
Length of the curve y=f(x) between x=a and x=b is
\(\displaystyle{L}={\int_{{a}}^{{b}}}{\left(\sqrt{{{\left({\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{y}\right.}}}}+{1}\right)}^{{2}}}}\right)}\)
\(\displaystyle{x}={\frac{{{1}}}{{{3}}}}\sqrt{{{y}}}{\left({y}-{3}\right)}\) fo \(\displaystyle{1}\leq{y}\leq{9}\)
Differentiate to get
\(\displaystyle{\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{y}\right.}}}}={\frac{{{1}}}{{{3}}}}\times{\frac{{{1}}}{{{2}\sqrt{{{y}}}}}}\times{\left({y}-{3}\right)}+{\frac{{{1}}}{{{3}}}}\times\sqrt{{{y}}}\)
\(\displaystyle{\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{y}\right.}}}}={\frac{{{1}}}{{{3}}}}\times{\frac{{{1}}}{{{2}\sqrt{{{y}}}}}}\times{\left({y}-{3}\right)}+{\frac{{{1}}}{{{3}}}}\times\sqrt{{{y}}}\times{\frac{{{2}\sqrt{{{y}}}}}{{{2}\sqrt{{{y}}}}}}\)
\(\displaystyle{\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{y}\right.}}}}={\frac{{{1}}}{{{3}}}}{\left({\frac{{{y}-{3}+{2}{y}}}{{{2}\sqrt{{{y}}}}}}\right)}\)
\(\displaystyle{\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{y}\right.}}}}={\frac{{{1}}}{{{3}}}}{\left({\frac{{{3}{y}-{3}}}{{{2}\sqrt{{{y}}}}}}\right)}\)
\(\displaystyle{\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{y}\right.}}}}={\left({\frac{{{y}-{1}}}{{{2}\sqrt{{{y}}}}}}\right)}\)
Length of the curve is
\(\displaystyle{L}={\int_{{1}}^{{9}}}\sqrt{{{\left({\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{y}\right.}}}}\right)}^{{2}}+{1}}}{\left.{d}{y}\right.}\)
\(\displaystyle={\int_{{1}}^{{9}}}\sqrt{{{\frac{{{\left({y}-{1}\right)}^{{2}}}}{{{4}{y}}}}+{1}}}{\left.{d}{y}\right.}\)
\(\displaystyle={\int_{{1}}^{{9}}}\sqrt{{{\frac{{{\left({y}-{1}\right)}^{{2}}+{4}{y}}}{{{4}{y}}}}}}{\left.{d}{y}\right.}\)
\(\displaystyle={\int_{{1}}^{{9}}}\sqrt{{{\frac{{{\left({y}^{{2}}+{2}{y}+{1}\right)}}}{{{4}{y}}}}}}{\left.{d}{y}\right.}\)
\(\displaystyle={\int_{{1}}^{{9}}}\sqrt{{{\frac{{{\left({y}+{1}\right)}^{{2}}}}{{{4}{y}}}}}}{\left.{d}{y}\right.}\)
\(\displaystyle={\int_{{1}}^{{9}}}{\frac{{{y}+{1}}}{{{2}\sqrt{{{y}}}}}}{\left.{d}{y}\right.}\)
\(\displaystyle={\int_{{1}}^{{9}}}{\frac{{{y}^{{\frac{{1}}{{2}}}}}}{{{2}}}}+{\frac{{{1}}}{{{2}\sqrt{{{y}}}}}}{\left.{d}{y}\right.}\)
\(\displaystyle={{\left[{\frac{{{y}^{{\frac{{3}}{{2}}}}}}{{{3}}}}+\sqrt{{{y}}}\right]}_{{1}}^{{9}}}={\frac{{{32}}}{{{3}}}}\)
Result: \(\displaystyle{\frac{{{32}}}{{{3}}}}\)
Not exactly what you’re looking for?
Ask My Question
13
 

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more
...