Find the length of the parametric curve defined over the given intervalx=3t-6,y=6t+1,0<=t<=1

Tyra 2021-09-14 Answered
Find the length of the parametric curve defined over the given interval \(\displaystyle{x}={3}{t}-{6},{y}={6}{t}+{1},{0}\le{t}\le{1}\)

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Expert Answer

delilnaT
Answered 2021-09-15 Author has 7412 answers
\(\displaystyle{x}={3}{t}-{6},{y}={6}{t}+{1},{0}\le{t}\le{1}\)
length=\(\displaystyle{\int_{{a}}^{{b}}}\sqrt{{{\left(\frac{{\left.{d}{x}\right.}}{{\left.{d}{t}\right.}}\right)}^{{2}}+{\left(\frac{{\left.{d}{y}\right.}}{{\left.{d}{t}\right.}}\right)}^{{2}}}}{\left.{d}{t}\right.}\)
\(\displaystyle\frac{{\left.{d}{x}\right.}}{{\left.{d}{t}\right.}}=\frac{{d}}{{\left.{d}{t}\right.}}{\left[{3}{t}-{6}\right]}={3}\frac{{\left.{d}{t}\right.}}{{\left.{d}{t}\right.}}-{d}\frac{{6}}{{\left.{d}{t}\right.}}={3}{x}{1}-{0}={3}\)
\(\displaystyle\frac{{\left.{d}{y}\right.}}{{\left.{d}{t}\right.}}=\frac{{d}}{{\left.{d}{t}\right.}}{\left[{6}{t}-{1}\right]}={6}\frac{{\left.{d}{t}\right.}}{{\left.{d}{t}\right.}}+{d}\frac{{1}}{{\left.{d}{t}\right.}}={6}{x}{1}+{0}={6}\)
length=\(\displaystyle{\int_{{0}}^{{1}}}\sqrt{{{3}^{{2}}+{6}^{{2}}}}{\left.{d}{t}\right.}\)
\(\displaystyle={\int_{{0}}^{{1}}}\sqrt{{45}}{\left.{d}{t}\right.}=\sqrt{{45}}{\in_{{0}}^{{1}}}{\left.{d}{t}\right.}\)
\(\displaystyle=\sqrt{{45}}{{\left[{t}\right]}_{{0}}^{{1}}}\)
\(\displaystyle=\sqrt{{45}}{\left[{1}-{0}\right]}=\sqrt{{45}}\)
length=\(\displaystyle\sqrt{{45}}={3}\sqrt{{5}}\)
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