problem 1: arc length\nprovide the general arc length formula. use the general formula to find the exact…

Explanation:

Step1: Recall arc - length formula

The arc - length formula for a function $y = f(x)$ from $x=a$ to $x = b$ is $L=\int_{a}^{b}\sqrt{1+(y')^{2}}dx$.

Step2: Find the derivative of $y$

Given $y=\frac{1}{4}x^{2}-\frac{1}{2}\ln x$. Using the power rule and the derivative of the natural - logarithm function, we have $y'=\frac{d}{dx}(\frac{1}{4}x^{2}-\frac{1}{2}\ln x)=\frac{1}{4}\times2x-\frac{1}{2}\times\frac{1}{x}=\frac{x}{2}-\frac{1}{2x}$.

Step3: Calculate $(y')^{2}$

$(y')^{2}=(\frac{x}{2}-\frac{1}{2x})^{2}=\frac{x^{2}}{4}- \frac{1}{2}+\frac{1}{4x^{2}}$.

Step4: Calculate $1+(y')^{2}$

$1+(y')^{2}=1+\frac{x^{2}}{4}-\frac{1}{2}+\frac{1}{4x^{2}}=\frac{x^{2}}{4}+\frac{1}{2}+\frac{1}{4x^{2}}=(\frac{x}{2}+\frac{1}{2x})^{2}$.

Step5: Calculate the arc - length integral

$L=\int_{1}^{2}\sqrt{1+(y')^{2}}dx=\int_{1}^{2}\sqrt{(\frac{x}{2}+\frac{1}{2x})^{2}}dx=\int_{1}^{2}(\frac{x}{2}+\frac{1}{2x})dx$.

Step6: Evaluate the integral

$\int_{1}^{2}(\frac{x}{2}+\frac{1}{2x})dx=\left[\frac{1}{2}\times\frac{x^{2}}{2}+\frac{1}{2}\ln x\right]{1}^{2}=\left(\frac{x^{2}}{4}+\frac{1}{2}\ln x\right)\big|{1}^{2}$. $=\left(\frac{2^{2}}{4}+\frac{1}{2}\ln 2\right)-\left(\frac{1^{2}}{4}+\frac{1}{2}\ln 1\right)$. Since $\ln 1 = 0$, we have $L = 1+\frac{1}{2}\ln 2-\frac{1}{4}=\frac{3}{4}+\frac{1}{2}\ln 2$.

Answer:

$\frac{3}{4}+\frac{1}{2}\ln 2$