Question

determine the arc length ( l ) of the curve defined by the equation ( y = \frac{2 + 2x^{\frac{3}{2}}}{3} ) over the interval ( 1 leq x leq 35 ). write the exact answer. do not round.

Explanation:

Step1: Simplify the curve equation

$y = \frac{2}{3} + \frac{2}{3}x^{\frac{3}{2}}$

Step2: Find the derivative $y'$

$y' = \frac{2}{3} \cdot \frac{3}{2}x^{\frac{1}{2}} = x^{\frac{1}{2}}$

Step3: Set up arc length formula

The arc length formula for $a \leq x \leq b$ is:
$$L = \int_{a}^{b} \sqrt{1+(y')^2} dx$$
Substitute $y'=x^{\frac{1}{2}}$, $a=1$, $b=35$:
$$L = \int_{1}^{35} \sqrt{1+(x^{\frac{1}{2}})^2} dx = \int_{1}^{35} \sqrt{1+x} dx$$

Step4: Integrate using substitution

Let $u=1+x$, $du=dx$. When $x=1$, $u=2$; when $x=35$, $u=36$.
$$L = \int_{2}^{36} u^{\frac{1}{2}} du = \frac{2}{3}u^{\frac{3}{2}} \bigg|_{2}^{36}$$

Step5: Evaluate the definite integral

$$L = \frac{2}{3}(36^{\frac{3}{2}} - 2^{\frac{3}{2}}) = \frac{2}{3}(216 - 2\sqrt{2})$$
$$L = 144 - \frac{4\sqrt{2}}{3}$$

Answer:

$144 - \frac{4\sqrt{2}}{3}$