1. The problem asks to find the arc length of the given functions over the specified intervals.
2. Recall the arc length formula for a function $y=f(x)$ from $x=a$ to $x=b$:
$$L=\int_a^b \sqrt{1+\left(\frac{dy}{dx}\right)^2} \, dx$$
3. For parametric or implicit forms where $x$ is given as function of $y$, use:
$$L=\int_c^d \sqrt{\left(\frac{dx}{dy}\right)^2+1} \, dy$$
### Problem 9
Function: $y=\frac{2}{3}x^{3/2},\ 0\leq x\leq 2$
- Derivative:
$$\frac{dy}{dx} = \frac{2}{3} \cdot \frac{3}{2} x^{1/2} = x^{1/2} = \sqrt{x}$$
- Arc length integral:
$$L_9 = \int_0^2 \sqrt{1 + (\sqrt{x})^2} \, dx = \int_0^2 \sqrt{1+x} \, dx$$
- Compute integral:
$$\int \sqrt{1+x} \, dx = \frac{2}{3}(1+x)^{3/2} + C$$
- Evaluate:
$$L_9 = \left[\frac{2}{3}(1+x)^{3/2}\right]_0^2 = \frac{2}{3}[(3)^{3/2} - 1] = \frac{2}{3}(3\sqrt{3} - 1)$$
### Problem 10
Function: $y=(x+4)^{3/2},\ 0\leq x \leq 4$
- Derivative:
$$\frac{dy}{dx} = \frac{3}{2}(x+4)^{1/2}$$
- Arc length integral:
$$L_{10} = \int_0^4 \sqrt{1 + \left(\frac{3}{2}(x+4)^{1/2}\right)^2} \, dx = \int_0^4 \sqrt{1+ \frac{9}{4}(x+4)} \, dx = \int_0^4 \sqrt{1 + \frac{9}{4}x + 9} \, dx = \int_0^4 \sqrt{\frac{9}{4}x + 10} \, dx$$
- Set $u = \frac{9}{4}x + 10$, so $du = \frac{9}{4} dx \Rightarrow dx = \frac{4}{9} du$
- When $x=0, u=10$; when $x=4, u= \frac{9}{4} \times 4 + 10 = 9 + 10 =19$
- Substitution:
$$L_{10} = \int_{10}^{19} \sqrt{u} \cdot \frac{4}{9} du = \frac{4}{9} \int_{10}^{19} u^{1/2} du = \frac{4}{9} \cdot \frac{2}{3} [u^{3/2}]_{10}^{19} = \frac{8}{27} (19^{3/2} - 10^{3/2})$$
### Problem 11
Function: $y= \frac{2}{3}(1+x^2)^{3/2},\ 0\leq x\leq 1$
- Derivative:
$$\frac{dy}{dx} = \frac{2}{3} \cdot \frac{3}{2} (1+x^2)^{1/2} \cdot 2x = 2x \sqrt{1+x^2}$$
- Arc length integral:
$$L_{11} = \int_0^1 \sqrt{1 + [2x \sqrt{1+x^2}]^2} dx = \int_0^1 \sqrt{1 + 4x^2(1+x^2)} dx = \int_0^1 \sqrt{1 + 4x^2 + 4x^4} dx = \int_0^1 \sqrt{(2x^2+1)^2} dx = \int_0^1 (2x^2 +1) dx$$
- Evaluate:
$$L_{11} = \int_0^1 (2x^2 + 1) dx = \left[ \frac{2}{3} x^3 + x \right]_0^1 = \frac{2}{3} + 1 = \frac{5}{3}$$
### Problem 12
Equation: $36 y^2 = (x^2 -4)^3$, $2\leq x \leq 3$, $y\geq 0$
- Express $y$ explicitly:
$$y = \frac{(x^2 -4)^{3/2}}{6}$$
- Derivative:
$$\frac{dy}{dx} = \frac{3}{2} \cdot \frac{(x^2 -4)^{1/2} \cdot 2x}{6} = \frac{3x \sqrt{x^2 -4}}{6} = \frac{x \sqrt{x^2 -4}}{2}$$
- Arc length integral:
$$L_{12} = \int_2^3 \sqrt{1 + \left(\frac{x \sqrt{x^2 -4}}{2}\right)^2} dx = \int_2^3 \sqrt{1 + \frac{x^2 (x^2 -4)}{4}} dx = \int_2^3 \sqrt{1 + \frac{x^4 -4x^2}{4}} dx = \int_2^3 \sqrt{\frac{4 + x^4 - 4x^2}{4}} dx$$
$$= \int_2^3 \frac{\sqrt{x^4 - 4x^2 + 4}}{2} dx = \frac{1}{2} \int_2^3 \sqrt{(x^2 - 2)^2} dx = \frac{1}{2} \int_2^3 |x^2 - 2| dx$$
- For $x\geq 2$, $x^2 - 2 \geq 2$ positive, so:
$$L_{12} = \frac{1}{2} \int_2^3 (x^2 - 2) dx = \frac{1}{2} \left[ \frac{x^3}{3} - 2x \right]_2^3 = \frac{1}{2} \left(\frac{27}{3} - 6 - \left(\frac{8}{3} - 4\right)\right) = \frac{1}{2} \left(9 - 6 - \frac{8}{3} + 4\right)$$
$$= \frac{1}{2} \left(7 - \frac{8}{3}\right) = \frac{1}{2} \cdot \frac{21 - 8}{3} = \frac{13}{6}$$
### Problem 13
Function: $y = \frac{x^3}{3} + \frac{1}{4x},\ 1 \leq x \leq 2$
- Derivative:
$$\frac{dy}{dx} = x^2 - \frac{1}{4x^2}$$
- Square derivative:
$$\left(\frac{dy}{dx}\right)^2 = \left(x^2 - \frac{1}{4x^2}\right)^2 = x^4 - \frac{x^2}{2x^2} + \frac{1}{16 x^4} = x^4 - \frac{1}{2} + \frac{1}{16 x^4}$$
- Arc length integrand:
$$\sqrt{1 + \left(\frac{dy}{dx}\right)^2} = \sqrt{1 + x^4 - \frac{1}{2} + \frac{1}{16 x^4}} = \sqrt{\frac{1}{2} + x^4 + \frac{1}{16 x^4}}$$
- The integral is:
$$L_{13} = \int_1^2 \sqrt{\frac{1}{2} + x^4 + \frac{1}{16 x^4}} dx$$
- This integral is complicated; it may require numerical methods or advanced techniques.
### Problem 14
Function: $x = \frac{y^4}{8} + \frac{1}{4y^2},\ 1 \leq y \leq 2$
- Derivative:
$$\frac{dx}{dy} = \frac{4y^3}{8} - \frac{2}{4 y^3} = \frac{y^3}{2} - \frac{1}{2 y^3}$$
- Arc length formula in terms of $y$:
$$L_{14} = \int_1^2 \sqrt{1 + \left(\frac{dx}{dy}\right)^2} dy = \int_1^2 \sqrt{1 + \left(\frac{y^3}{2} - \frac{1}{2 y^3}\right)^2} dy$$
- Square:
$$\left(\frac{y^3}{2} - \frac{1}{2 y^3}\right)^2 = \frac{y^6}{4} - \frac{y^3}{2 y^3} + \frac{1}{4 y^6} = \frac{y^6}{4} - \frac{1}{2} + \frac{1}{4 y^6}$$
- Square root:
$$\sqrt{1 + \frac{y^6}{4} - \frac{1}{2} + \frac{1}{4 y^6}} = \sqrt{\frac{1}{2} + \frac{y^6}{4} + \frac{1}{4 y^6}}$$
- Integral:
$$L_{14} = \int_1^2 \sqrt{\frac{1}{2} + \frac{y^6}{4} + \frac{1}{4 y^6}} dy$$
- This integral may also require numerical methods.
### Problem 15
Function: $y = \frac{1}{2} \ln(\sin 2x), \frac{\pi}{8} \leq x \leq \frac{\pi}{6}$
- Derivative:
$$\frac{dy}{dx} = \frac{1}{2} \cdot \frac{\cos 2x \cdot 2}{\sin 2x} = \frac{\cos 2x}{\sin 2x} = \cot 2x$$
- Arc length integral:
$$L_{15} = \int_{\pi/8}^{\pi/6} \sqrt{1 + (\cot 2x)^2} dx = \int_{\pi/8}^{\pi/6} \sqrt{1 + \frac{\cos^2 2x}{\sin^2 2x}} dx = \int_{\pi/8}^{\pi/6} \sqrt{\frac{\sin^2 2x + \cos^2 2x}{\sin^2 2x}} dx = \int_{\pi/8}^{\pi/6} \frac{1}{\sin 2x} dx$$
- Simplify:
$$L_{15} = \int_{\pi/8}^{\pi/6} \csc 2x \, dx$$
- Use substitution $u=2x$, $du=2 dx$, $dx = \frac{du}{2}$
- Limits: when $x=\pi/8, u=\pi/4$; when $x=\pi/6, u=\pi/3$
- Integral:
$$L_{15} = \frac{1}{2} \int_{\pi/4}^{\pi/3} \csc u \, du = \frac{1}{2} [-\ln|\csc u + \cot u|]_{\pi/4}^{\pi/3}$$
- Evaluate:
$$= \frac{1}{2} \left(-\ln(\csc \frac{\pi}{3} + \cot \frac{\pi}{3}) + \ln(\csc \frac{\pi}{4} + \cot \frac{\pi}{4})\right) = \frac{1}{2} \ln \frac{\csc \frac{\pi}{4} + \cot \frac{\pi}{4}}{\csc \frac{\pi}{3} + \cot \frac{\pi}{3}}$$
- Substitute values:
$$\csc \frac{\pi}{4} = \sqrt{2}, \cot \frac{\pi}{4} =1$$
$$\csc \frac{\pi}{3} = \frac{2}{\sqrt{3}}, \cot \frac{\pi}{3} = \frac{1}{\sqrt{3}}$$
$$L_{15} = \frac{1}{2} \ln \frac{\sqrt{2}+1}{\frac{2}{\sqrt{3}} + \frac{1}{\sqrt{3}}} = \frac{1}{2} \ln \frac{\sqrt{2}+1}{\frac{3}{\sqrt{3}}} = \frac{1}{2} \ln \frac{\sqrt{2}+1}{\sqrt{3}}$$
### Problem 16
Function: $y = \ln(\cos x),\ 0 \leq x \leq \frac{\pi}{3}$
- Derivative:
$$\frac{dy}{dx} = \frac{-\sin x}{\cos x} = -\tan x$$
- Square derivative:
$$(\frac{dy}{dx})^2 = \tan^2 x$$
- Arc length integral:
$$L_{16} = \int_0^{\pi/3} \sqrt{1 + \tan^2 x} dx = \int_0^{\pi/3} \sec x \, dx$$
- Integral:
$$\int \sec x \, dx = \ln |\sec x + \tan x| + C$$
- Evaluate:
$$L_{16} = \left[ \ln(\sec x + \tan x) \right]_0^{\pi/3} = \ln(\sec \frac{\pi}{3} + \tan \frac{\pi}{3}) - \ln(\sec 0 + \tan 0)$$
$$= \ln(2 + \sqrt{3}) - \ln(1 + 0) = \ln(2 + \sqrt{3})$$
Arc Length
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