# What is the length of one cycle of the cardioid: r = 5+5cos (theta)?

Length of one cycle $= 40 \text{ }$units

#### Explanation:

I consider this length of arc using a formula from Calculus with given polar curve equation

$r = 5 + 5 \cos \theta$

$L = \int \mathrm{ds}$

$\mathrm{ds} = \sqrt{{r}^{2} + {\left(\frac{\mathrm{dr}}{d \theta}\right)}^{2}} d \theta$

For one cycle

$L = 2 \cdot {\int}_{0}^{\pi} \mathrm{ds} = 2 \cdot {\int}_{0}^{\pi} \sqrt{{r}^{2} + {\left(\frac{\mathrm{dr}}{d \theta}\right)}^{2}} d \theta$

$L = 2 \cdot {\int}_{0}^{\pi} \sqrt{{\left(5 + 5 \cos \theta\right)}^{2} + {\left(- 5 \sin \theta\right)}^{2}} d \theta$

$L = 2 \cdot {\int}_{0}^{\pi} \sqrt{\left(25 + 50 \cos \theta + 25 {\cos}^{2} \theta + 25 {\sin}^{2} \theta\right)} d \theta$

$L = 2 \cdot {\int}_{0}^{\pi} \sqrt{\left(50 + 50 \cos \theta\right)} d \theta$

$L = 2 \sqrt{50} \cdot {\int}_{0}^{\pi} \sqrt{\left(1 + \cos \theta\right)} d \theta$

$L = 40 \text{ }$units

graph{(x^2+y^2-5x)^2-25(x^2+y^2)=0[-20,20,-10,10]}

God bless....I hope the explanation is useful.