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Find the arc length function for the curve $y=x^{2}-\frac{1}{8} \ln (x)$ taking (1,1) as the starting point.

(1) $y=f(x)=x^{2}-\frac{1}{8} \ln (x)$

(2) $y^{\prime}=2 x-\frac{1}{8} \frac{1}{x}=2 x-\frac{1}{8 x}$

(3) $\left(y^{\prime}\right)^{2}=\left(2 x-\frac{1}{8 x}\right)^{2}=(2 x)^{2}-2(2 x)\left(\frac{1}{8 x}\right)+\left(\frac{1}{8 x}\right)^{2}$

$\left(y^{\prime}\right)^{2}=4 x^{2}-\frac{4 x}{8 x}+\frac{1}{64 x^{2}}$

$\left(y^{\prime}\right)^{2}=4 x^{2}-\frac{1}{2}+\frac{1}{64 x^{2}}$

(4) $1+\left(y^{\prime}\right)^{2}=1+ 4 x^{2}-\frac{1}{2} +\frac{1}{64 x^{2}}=4 x^{2}+\frac{1}{2}+\frac{1}{64 x^{2}}$

$\left(2 x+\frac{1}{8 x}\right)\left(2 x+\frac{1}{8 x}\right)=4 x^{2}+\frac{2 x}{8 x}+\frac{2 x}{8 x}+\frac{1}{64 x^{2}}$

$\left(2 x+\frac{1}{8 x}\right)\left(2 x+\frac{1}{8 x}\right)=\left(2 x+\frac{1}{8 x}\right)^{2}$

(5) $\sqrt{1+\left(y^{\prime}\right)^{2}}=\sqrt{\left(2 x+\frac{1}{8 x}\right)^{2}}=2 x+\frac{1}{8 x}$

arc length $=\int_{1}^{x} \sqrt{1+\left(f^{\prime} x\right)^{2}} d x=\int_{1}^{x}\left(2 x+\frac{1}{8 x}\right) d x$

arc length $=L=\int_{1}^{x}\left(2 t+\frac{1}{8 t}\right) d t=\frac{2 t^{2}}{2}+\frac{1}{8} \ln \left.(t)\right|_{1}^{x}$

$L=t^{2}+\frac{1}{8} \ln\left.(t)\right|_{1}^{x}=x^{2}+\frac{\ln (x)}{8}-\left((1)^{2}+\frac{\ln (1)}{8}\right)$

$L=x^{2}+\frac{\ln (x)}{8}-1$

$x=4 \rightarrow L=(4)^{2}+\frac{\ln (4)}{8}-1$

$=15.173$

Find the arc length of th curve $x=t^{2}$ and $y=t^{3}$ from t = 0 to t = 4

(1) $x=t^{2} \quad , \quad y=t^{3}$

(2) $\frac{d x}{d t}=2 t \qquad \frac{d y}{d t}=3 t^{2}$

(3) $\left(\frac{d x}{d t}\right)^{2}+\left(\frac{d y}{d t}\right)^{2}=(2 t)^{2}+(3t^{2})^{2}=4 t^{2}+9 t^{4}=4t^{2}\left(1+\frac{9}{4} t^{2}\right)$

$L=\int_{0}^{4} \sqrt{4 t^{2}\left(1+\frac{9}{4} t^{2}\right)} d t=\int_{0}^{4} 2 t \sqrt{1+\frac{9}{4} t^{2}} d t$

let $u=1+\frac{9}{4}t^{2} \Rightarrow d u=\frac{9}{2}td t$

$L=\int_{0}^{4} \sqrt{u} \cdot 2 t d t \cdot \frac{9}{4} \cdot \frac{4}{9}$

$=\frac{4}{9} \int_{0}^{4} \sqrt{u} \cdot \frac{9}{2}td t$

$L=\frac{4}{9} \int_{0}^{4} u^{\frac{1}{2}} d u=\frac {4}{9} [\frac {2}{3} {u}^{\frac {3}{2}}]_{0}^{4}$

$L=\frac{4}{9}\left[\frac{2}{3}(4)^{3 / 2}-0\right]=\frac{8}{27}(4)^{\frac{3}{2}}$