Find the exact length of the polar curve. r=\theta^{2},\ 0\le\theta\le2

f480forever2rz 2021-11-20 Answered
Find the exact length of the polar curve.
\(\displaystyle{r}=\theta^{{{2}}},\ {0}\le\theta\le{2}\pi\)

Expert Community at Your Service

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

Plainmath recommends

  • Ask your own question for free.
  • Get a detailed answer even on the hardest topics.
  • Ask an expert for a step-by-step guidance to learn to do it yourself.
Ask Question

Expert Answer

Opeance1951
Answered 2021-11-21 Author has 574 answers
Step 1
It is given that
\(\displaystyle{r}=\theta^{{{2}}}\)
Differentiate using the power rule
\(\displaystyle{\frac{{{d}{r}}}{{{d}\theta}}}={2}\cdot\theta^{{{2}-{1}}}\)
\(\displaystyle{\frac{{{d}{r}}}{{{d}\theta}}}={2}\theta\)
Formula 6 from this section:
Length of a curve with polar equation \(\displaystyle{r}={f{{\left(\theta\right)}}},\ {a}\le\theta\le{b}\) is given by
\(\displaystyle{L}={\int_{{{a}}}^{{{b}}}}\sqrt{{{r}^{{{2}}}+{\left[{\frac{{{d}{r}}}{{{d}\theta}}}\right]}^{{{2}}}}}{d}\theta\)
\(\displaystyle={\int_{{{0}}}^{{{2}\pi}}}\sqrt{{{\left(\theta^{{{2}}}\right)}^{{{2}}}+{\left({2}\theta\right)}^{{{2}}}}}{d}\theta\)
\(\displaystyle={\int_{{{0}}}^{{{2}\pi}}}\sqrt{{\theta^{{{4}}}+{4}\theta^{{{2}}}}}{d}\theta\)
Note that: \(\displaystyle\sqrt{{\theta^{{{4}}}+{4}\theta^{{{2}}}}}=\sqrt{{\theta^{{{2}}}{\left(\theta^{{{2}}}+{4}\right)}}}=\sqrt{{\theta^{{{2}}}}}\cdot\sqrt{{\theta^{{{2}}}+{4}}}=\theta\cdot\sqrt{{\theta^{{{2}}}+{4}}}\)
Since we are integrating over \(\displaystyle{\left[{0},\ {2}\pi\right]},\ \theta\) is positive and \(\displaystyle\sqrt{{\theta^{{{2}}}={\left|\theta\right|}=\theta}}\)
\(\displaystyle={\int_{{{0}}}^{{{2}\pi}}}\theta\cdot\sqrt{{\theta^{{{2}}}+{4}}}{d}\theta\)
\(\displaystyle={\int_{{{0}}}^{{{2}\pi}}}\sqrt{{\theta^{{{2}}}+{4}}}{\left[\theta\ {d}\theta\right]}\)
We will now substitute
\(\displaystyle\theta^{{{2}}}+\$={u}\)
Step 2
On differentiation, we get
\(\displaystyle{2}\theta,{d}\theta={d}{u}\Rightarrow\theta{d}\theta={\frac{{{1}}}{{{2}}}}{d}{u}\)
The limits of integration will change from \(\displaystyle{\int_{{{0}}}^{{{2}\pi}}}\) to \(\displaystyle{\int_{{{0}^{{{2}}}+{4}}}^{{{4}\pi^{{{2}}}+{4}}}}\)
\(\displaystyle={\int_{{{4}}}^{{{4}\pi^{{{2}}}+{4}}}}\sqrt{{{u}}}{\left[{\frac{{{1}}}{{{2}}}}{d}{u}\right]}\)
\(\displaystyle={\frac{{{1}}}{{{2}}}}{\int_{{{4}}}^{{{4}\pi^{{{2}}}+{4}}}}{u}^{{\frac{{1}}{{2}}}}{d}{u}\)
\(\displaystyle={\frac{{{1}}}{{{2}}}}{{\left[{\frac{{{u}^{{{\left({\frac{{{1}}}{{{2}}}}\right)}+{1}}}}}{{{\left({\frac{{{1}}}{{{2}}}}\right)}+{1}}}}\right]}_{{{4}}}^{{{4}\pi^{{{2}}}+{4}}}}\)
\(\displaystyle={\frac{{{1}}}{{{2}}}}{\left[{u}^{{{\frac{{{3}}}{{{2}}}}}}\right\rbrace}{\left\lbrace{\frac{{{3}}}{{{2}}}}\right\rbrace}{{]}_{{{4}}}^{{{4}\pi^{{{2}}}+{4}}}}\)
\(\displaystyle={\frac{{{1}}}{{{2}}}}{{\left[{\frac{{{2}}}{{{3}}}}\cdot{u}^{{{\frac{{{3}}}{{{2}}}}}}\right]}_{{{4}}}^{{{4}\pi^{{{2}}}+{4}}}}\)
\(\displaystyle={\frac{{{1}}}{{{2}}}}{\left[{\frac{{{2}}}{{{3}}}}\cdot{\left({4}\pi^{{{2}}}+{4}\right)}^{{{\frac{{{3}}}{{{2}}}}}}-{\frac{{{2}}}{{{3}}}}\cdot{4}^{{{\frac{{{3}}}{{{2}}}}}}\right]}\)
\(\displaystyle={\frac{{{1}}}{{{2}}}}\cdot{\left({2}^{{{2}}}\right)}^{{{\frac{{{3}}}{{{2}}}}}}\cdot{\left(\pi^{{{2}}}+{1}\right)}^{{{\frac{{{3}}}{{{2}}}}}}-{\frac{{{1}}}{{{3}}}}\cdot{\left({2}^{{{2}}}\right)}^{{{\frac{{{3}}}{{{2}}}}}}\)
\(\displaystyle={\frac{{{1}}}{{{3}}}}\cdot{2}^{{{3}}}\cdot{\left(\pi^{{{2}}}+{1}\right)}^{{{\frac{{{3}}}{{{2}}}}}}-{\frac{{{1}}}{{{3}}}}\cdot{2}^{{{3}}}\)
\(\displaystyle={\frac{{{8}}}{{{3}}}}\cdot{\left[{\left(\pi^{{{2}}}+{1}\right)}^{{{\frac{{{3}}}{{{2}}}}}}-{1}\right]}\)
\(\displaystyle{L}={\frac{{{8}}}{{{3}}}}\cdot{\left[{\left(\pi^{{{2}}}+{1}\right)}^{{{\frac{{{3}}}{{{2}}}}}}-{1}\right]}\)
Have a similar question?
Ask An Expert
0
 
huckelig75
Answered 2021-11-22 Author has 292 answers
Step 1
Given: \(\displaystyle{r}=\theta^{{{2}}}\)
\(\displaystyle{l}={\int_{{{a}}}^{{{b}}}}\sqrt{{{r}^{{{2}}}+{\left({\frac{{{d}{r}}}{{{d}\theta}}}\right)}^{{{2}}}{d}\theta}}\) - formula for length of a curve.
\(\displaystyle{\frac{{{d}{r}}}{{{d}\theta}}}={2}\theta\)
\(\displaystyle{L}={\int_{{{0}}}^{{{2}\pi}}}\sqrt{{{\left(\theta^{{{2}}}\right)}^{{{2}}}+{\left({2}\theta\right)}^{{{2}}}}}{d}\theta\)
\(\displaystyle={\int_{{{0}}}^{{{2}\pi}}}\sqrt{{\theta^{{{4}}}+{4}\theta^{{{2}}}}}{d}\theta\)
\(\displaystyle={\int_{{{0}}}^{{{2}\pi}}}\sqrt{{\theta^{{{2}}}{\left(\theta^{{{2}}}+{4}\right)}}}{d}\theta\)
\(\displaystyle={\int_{{{0}}}^{{{2}\pi}}}\theta\sqrt{{\theta^{{{2}}}+{4}}}{d}\theta\)
Substitution:
\(\displaystyle{t}=\theta^{{{2}}}+{4}\)
\(\displaystyle{\left.{d}{t}\right.}={2}\theta{d}\theta\rightarrow\theta{d}\theta={\frac{{{\left.{d}{t}\right.}}}{{{2}}}}\)
\(\displaystyle{0}\rightarrow{4}\)
\(\displaystyle{2}\pi\rightarrow{4}\pi^{{{2}}}+{4}\)
\(\displaystyle{L}={\frac{{{1}}}{{{2}}}}{\int_{{{4}}}^{{{4}\pi^{{{2}}}+{4}}}}\sqrt{{{t}{\left.{d}{t}\right.}}}={\frac{{{1}}}{{{2}}}}{\left({\frac{{{2}}}{{{3}}}}{t}^{{{\frac{{{3}}}{{{2}}}}}}\right)}{{\mid}_{{{4}}}^{{{4}\pi^{{{2}}}+{4}}}}\)
\(\displaystyle={\frac{{{1}}}{{{3}}}}{\left({\left({4}\pi^{{{2}}}+{4}\right)}^{{{\frac{{{3}}}{{{2}}}}}}-{4}^{{{\frac{{{3}}}{{{2}}}}}}\right)}\approx{92.89}\)
\(\displaystyle{L}\approx{92.89}\)
0

Expert Community at Your Service

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

Plainmath recommends

  • Ask your own question for free.
  • Get a detailed answer even on the hardest topics.
  • Ask an expert for a step-by-step guidance to learn to do it yourself.
Ask Question
...