f480forever2rz
2021-11-20
Answered

Find the exact length of the polar curve.

$r={\theta}^{2},\text{}0\le \theta \le 2\pi$

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Opeance1951

Answered 2021-11-21
Author has **26** answers

Step 1

It is given that

$r={\theta}^{2}$

Differentiate using the power rule

$\frac{dr}{d\theta}=2\cdot {\theta}^{2-1}$

$\frac{dr}{d\theta}=2\theta$

Formula 6 from this section:

Length of a curve with polar equation$r=f\left(\theta \right),\text{}a\le \theta \le b$ is given by

$L={\int}_{a}^{b}\sqrt{{r}^{2}+{\left[\frac{dr}{d\theta}\right]}^{2}}d\theta$

$={\int}_{0}^{2\pi}\sqrt{{\left({\theta}^{2}\right)}^{2}+{\left(2\theta \right)}^{2}}d\theta$

$={\int}_{0}^{2\pi}\sqrt{{\theta}^{4}+4{\theta}^{2}}d\theta$

Note that:$\sqrt{{\theta}^{4}+4{\theta}^{2}}=\sqrt{{\theta}^{2}({\theta}^{2}+4)}=\sqrt{{\theta}^{2}}\cdot \sqrt{{\theta}^{2}+4}=\theta \cdot \sqrt{{\theta}^{2}+4}$

Since we are integrating over$[0,\text{}2\pi ],\text{}\theta$ is positive and $\sqrt{{\theta}^{2}=\left|\theta \right|=\theta}$

$={\int}_{0}^{2\pi}\theta \cdot \sqrt{{\theta}^{2}+4}d\theta$

$={\int}_{0}^{2\pi}\sqrt{{\theta}^{2}+4}\left[\theta \text{}d\theta \right]$

We will now substitute

${\theta}^{2}+\mathrm{\$}=u}$

Step 2

On differentiation, we get

$2\theta ,d\theta =du\Rightarrow \theta d\theta =\frac{1}{2}du$

The limits of integration will change from$\int}_{0}^{2\pi$ to $\int}_{{0}^{2}+4}^{4{\pi}^{2}+4$

$={\int}_{4}^{4{\pi}^{2}+4}\sqrt{u}\left[\frac{1}{2}du\right]$

$=\frac{1}{2}{\int}_{4}^{4{\pi}^{2}+4}{u}^{\frac{1}{2}}du$

$=\frac{1}{2}{\left[\frac{{u}^{\left(\frac{1}{2}\right)+1}}{\left(\frac{1}{2}\right)+1}\right]}_{4}^{4{\pi}^{2}+4}$

$=\frac{1}{}$###### Not exactly what you’re looking for?

It is given that

Differentiate using the power rule

Formula 6 from this section:

Length of a curve with polar equation

Note that:

Since we are integrating over

We will now substitute

Step 2

On differentiation, we get

The limits of integration will change from

huckelig75

Answered 2021-11-22
Author has **11** answers

Step 1

Given:$r={\theta}^{2}$

$l={\int}_{a}^{b}\sqrt{{r}^{2}+{\left(\frac{dr}{d\theta}\right)}^{2}d\theta}$ - formula for length of a curve.

$\frac{dr}{d\theta}=2\theta$

$L={\int}_{0}^{2\pi}\sqrt{{\left({\theta}^{2}\right)}^{2}+{\left(2\theta \right)}^{2}}d\theta$

$={\int}_{0}^{2\pi}\sqrt{{\theta}^{4}+4{\theta}^{2}}d\theta$

$={\int}_{0}^{2\pi}\sqrt{{\theta}^{2}({\theta}^{2}+4)}d\theta$

$={\int}_{0}^{2\pi}\theta \sqrt{{\theta}^{2}+4}d\theta$

Substitution:

$t={\theta}^{2}+4$

$dt=2\theta d\theta \to \theta d\theta =\frac{dt}{2}$

$0\to 4$

$2\pi \to 4{\pi}^{2}+4$

$L=\frac{1}{2}{\int}_{4}^{4{\pi}^{2}+4}\sqrt{tdt}=\frac{1}{2}\left(\frac{2}{3}{t}^{\frac{3}{2}}\right){\mid}_{4}^{4{\pi}^{2}+4}$

$=\frac{1}{3}({(4{\pi}^{2}+4)}^{\frac{3}{2}}-{4}^{\frac{3}{2}})\approx 92.89$

$L\approx 92.89$

Given:

Substitution:

user_27qwe

Answered 2021-11-29
Author has **208** answers

Step 1

The length of a polar function

The length of the curve is

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