Taniyah Estrada
2022-06-25
Answered

What is the distance between the following polar coordinates?: $(7,\frac{5\pi}{4}),(3,\frac{13\pi}{8})$

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robegarj

Answered 2022-06-26
Author has **24** answers

Step 1

using the polar version of the distance formula

${d}^{2}={r}_{1}^{2}+{r}_{2}^{2}-\left[2{r}_{1}{r}_{2}\mathrm{cos}({\theta}_{2}-{\theta}_{1})\right]$

Let $({r}_{1},{\theta}_{1})=(7,\frac{5\pi}{4})\phantom{\rule{1ex}{0ex}}\text{and}\phantom{\rule{1ex}{0ex}}({r}_{2},{\theta}_{2})=(3,\frac{13\pi}{8})$

Step 2

$\Rightarrow {d}^{2}={7}^{2}+{3}^{2}-[2\times 7\times 3\mathrm{cos}(\frac{13\pi}{8}-\frac{5\pi}{4})]$

${d}^{2}=49+9-\left(42\mathrm{cos}\left(\frac{3\pi}{8}\right)\right)$

$\Rightarrow d=\sqrt{58-42\mathrm{cos}\left(\frac{3\pi}{8}\right)}\approx 6.475$

using the polar version of the distance formula

${d}^{2}={r}_{1}^{2}+{r}_{2}^{2}-\left[2{r}_{1}{r}_{2}\mathrm{cos}({\theta}_{2}-{\theta}_{1})\right]$

Let $({r}_{1},{\theta}_{1})=(7,\frac{5\pi}{4})\phantom{\rule{1ex}{0ex}}\text{and}\phantom{\rule{1ex}{0ex}}({r}_{2},{\theta}_{2})=(3,\frac{13\pi}{8})$

Step 2

$\Rightarrow {d}^{2}={7}^{2}+{3}^{2}-[2\times 7\times 3\mathrm{cos}(\frac{13\pi}{8}-\frac{5\pi}{4})]$

${d}^{2}=49+9-\left(42\mathrm{cos}\left(\frac{3\pi}{8}\right)\right)$

$\Rightarrow d=\sqrt{58-42\mathrm{cos}\left(\frac{3\pi}{8}\right)}\approx 6.475$

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