p.730 19,21 or 20,22\nlengths of polar curves\nfind the lengths of the curves in exercises 19 - 27.\n19. the…

Explanation:

Step1: Recall arc - length formula for polar curves

The arc - length formula for a polar curve $r = f(\theta)$ from $\theta=a$ to $\theta = b$ is $L=\int_{a}^{b}\sqrt{r^{2}+\left(\frac{dr}{d\theta}\right)^{2}}d\theta$.

Step2: Take the derivative of $r$ with respect to $\theta$ for problem 19

Given $r = \theta^{2}$, then $\frac{dr}{d\theta}=2\theta$.

Step3: Substitute $r$ and $\frac{dr}{d\theta}$ into the arc - length formula

$L=\int_{0}^{\sqrt{5}}\sqrt{(\theta^{2})^{2}+(2\theta)^{2}}d\theta=\int_{0}^{\sqrt{5}}\sqrt{\theta^{4} + 4\theta^{2}}d\theta=\int_{0}^{\sqrt{5}}\theta\sqrt{\theta^{2}+4}d\theta$.

Step4: Use substitution

Let $u=\theta^{2}+4$, then $du = 2\theta d\theta$. When $\theta = 0$, $u = 4$; when $\theta=\sqrt{5}$, $u=9$. So the integral becomes $\frac{1}{2}\int_{4}^{9}\sqrt{u}du$.

Step5: Integrate $\sqrt{u}$

$\frac{1}{2}\int_{4}^{9}u^{\frac{1}{2}}du=\frac{1}{2}\times\frac{2}{3}u^{\frac{3}{2}}\big|_{4}^{9}=\frac{1}{3}(9^{\frac{3}{2}}-4^{\frac{3}{2}})$.

Step6: Calculate the result

$\frac{1}{3}(27 - 8)=\frac{19}{3}$.

Answer:

For problem 19, the length of the curve is $\frac{19}{3}$.

(If you want solutions for other problems, we can follow the same general steps of finding $\frac{dr}{d\theta}$, substituting into the arc - length formula, and integrating.)