Solved on Feb 20, 2024

Calculate the arc length of $y=\frac{1}{4} x^{2}-\frac{1}{2} \ln x$ on $[1,3 e]$ .

STEP 1

Assumptions
1. The function given is $y = \frac{1}{4} x^{2} - \frac{1}{2} \ln x$ .
2. The interval over which we need to calculate the arc length is $[1, 3e]$ .
3. The formula for the arc length $s$ of a function $y = f(x)$ from $a$ to $b$ is given by: $s = \int_{a}^{b} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx$

STEP 2

First, we need to find the derivative of the function $y$ with respect to $x$ .
$\frac{dy}{dx} = \frac{d}{dx} \left(\frac{1}{4} x^{2} - \frac{1}{2} \ln x\right)$

STEP 3

Calculate the derivative using the power rule and the derivative of the natural logarithm.
$\frac{dy}{dx} = \frac{1}{4} \cdot 2x - \frac{1}{2} \cdot \frac{1}{x}$

STEP 4

Simplify the derivative.
$\frac{dy}{dx} = \frac{x}{2} - \frac{1}{2x}$

STEP 5

Now, we need to square the derivative to use in the arc length formula.
$\left(\frac{dy}{dx}\right)^2 = \left(\frac{x}{2} - \frac{1}{2x}\right)^2$

STEP 6

Expand the square of the binomial.
$\left(\frac{dy}{dx}\right)^2 = \left(\frac{x}{2}\right)^2 - 2 \cdot \frac{x}{2} \cdot \frac{1}{2x} + \left(\frac{1}{2x}\right)^2$

STEP 7

Simplify the expression.
$\left(\frac{dy}{dx}\right)^2 = \frac{x^2}{4} - \frac{1}{2} + \frac{1}{4x^2}$

STEP 8

Now, we can set up the integral for the arc length using the interval $[1, 3e]$ and the expression for $\left(\frac{dy}{dx}\right)^2$ .
$s = \int_{1}^{3e} \sqrt{1 + \frac{x^2}{4} - \frac{1}{2} + \frac{1}{4x^2}} \, dx$

STEP 9

Combine the terms under the square root.
$s = \int_{1}^{3e} \sqrt{\frac{1}{4x^2} + \frac{x^2}{4} + \frac{1}{2}} \, dx$

STEP 10

Simplify the expression under the square root by finding a common denominator.
$s = \int_{1}^{3e} \sqrt{\frac{1 + x^4 + 2x^2}{4x^2}} \, dx$

STEP 11

Extract the constant from under the square root.
$s = \int_{1}^{3e} \frac{\sqrt{x^4 + 2x^2 + 1}}{2x} \, dx$

STEP 12

Recognize the expression under the square root as a perfect square.
$x^4 + 2x^2 + 1 = (x^2 + 1)^2$

STEP 13

Simplify the integrand using the perfect square.
$s = \int_{1}^{3e} \frac{x^2 + 1}{2x} \, dx$

STEP 14

Split the integrand into two separate fractions.
$s = \int_{1}^{3e} \frac{1}{2} + \frac{1}{2x} \, dx$

STEP 15

Integrate each term separately.
$s = \left[\frac{1}{2}x + \frac{1}{2}\ln|x|\right]_{1}^{3e}$

STEP 16

Evaluate the integral from $1$ to $3e$ .
$s = \left(\frac{1}{2} \cdot 3e + \frac{1}{2}\ln|3e|\right) - \left(\frac{1}{2} \cdot 1 + \frac{1}{2}\ln|1|\right)$

STEP 17

Simplify the expression, noting that $\ln|1| = 0$ .
$s = \frac{3e}{2} + \frac{1}{2}\ln(3e) - \frac{1}{2}$

STEP 18

Use the logarithm property $\ln(ab) = \ln(a) + \ln(b)$ .
$s = \frac{3e}{2} + \frac{1}{2}(\ln(3) + \ln(e)) - \frac{1}{2}$

STEP 19

Since $\ln(e) = 1$ , further simplify the expression.
$s = \frac{3e}{2} + \frac{1}{2}\ln(3) + \frac{1}{2} - \frac{1}{2}$

STEP 20

Combine like terms.
$s = \frac{3e}{2} + \frac{1}{2}\ln(3)$
This is the arc length of the function $y = \frac{1}{4} x^{2} - \frac{1}{2} \ln x$ over the interval $[1, 3e]$ .
$s = \frac{3e}{2} + \frac{1}{2}\ln(3)$

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Calculate the arc length of y=14x2−12ln⁡xy=\frac{1}{4} x^{2}-\frac{1}{2} \ln xy=41​x2−21​lnx on [1,3e][1,3 e][1,3e].

STEP 1

STEP 2

First, we need to find the derivative of the function yyy with respect to xxx.dydx=ddx(14x2−12ln⁡x) \frac{dy}{dx} = \frac{d}{dx} \left(\frac{1}{4} x^{2} - \frac{1}{2} \ln x\right) dxdy​=dxd​(41​x2−21​lnx)

STEP 3

Calculate the derivative using the power rule and the derivative of the natural logarithm.dydx=14⋅2x−12⋅1x \frac{dy}{dx} = \frac{1}{4} \cdot 2x - \frac{1}{2} \cdot \frac{1}{x} dxdy​=41​⋅2x−21​⋅x1​

STEP 4

Simplify the derivative.dydx=x2−12x \frac{dy}{dx} = \frac{x}{2} - \frac{1}{2x} dxdy​=2x​−2x1​

STEP 5

Now, we need to square the derivative to use in the arc length formula.(dydx)2=(x2−12x)2 \left(\frac{dy}{dx}\right)^2 = \left(\frac{x}{2} - \frac{1}{2x}\right)^2 (dxdy​)2=(2x​−2x1​)2

STEP 6

Expand the square of the binomial.(dydx)2=(x2)2−2⋅x2⋅12x+(12x)2 \left(\frac{dy}{dx}\right)^2 = \left(\frac{x}{2}\right)^2 - 2 \cdot \frac{x}{2} \cdot \frac{1}{2x} + \left(\frac{1}{2x}\right)^2 (dxdy​)2=(2x​)2−2⋅2x​⋅2x1​+(2x1​)2

STEP 7

Simplify the expression.(dydx)2=x24−12+14x2 \left(\frac{dy}{dx}\right)^2 = \frac{x^2}{4} - \frac{1}{2} + \frac{1}{4x^2} (dxdy​)2=4x2​−21​+4x21​

STEP 8

STEP 9

Combine the terms under the square root.s=∫13e14x2+x24+12 dx s = \int_{1}^{3e} \sqrt{\frac{1}{4x^2} + \frac{x^2}{4} + \frac{1}{2}} \, dx s=∫13e​4x21​+4x2​+21​​dx

STEP 10

Simplify the expression under the square root by finding a common denominator.s=∫13e1+x4+2x24x2 dx s = \int_{1}^{3e} \sqrt{\frac{1 + x^4 + 2x^2}{4x^2}} \, dx s=∫13e​4x21+x4+2x2​​dx

STEP 11

Extract the constant from under the square root.s=∫13ex4+2x2+12x dx s = \int_{1}^{3e} \frac{\sqrt{x^4 + 2x^2 + 1}}{2x} \, dx s=∫13e​2xx4+2x2+1​​dx

STEP 12

Recognize the expression under the square root as a perfect square.x4+2x2+1=(x2+1)2 x^4 + 2x^2 + 1 = (x^2 + 1)^2 x4+2x2+1=(x2+1)2

STEP 13

Simplify the integrand using the perfect square.s=∫13ex2+12x dx s = \int_{1}^{3e} \frac{x^2 + 1}{2x} \, dx s=∫13e​2xx2+1​dx

STEP 14

Split the integrand into two separate fractions.s=∫13e12+12x dx s = \int_{1}^{3e} \frac{1}{2} + \frac{1}{2x} \, dx s=∫13e​21​+2x1​dx

STEP 15

Integrate each term separately.s=[12x+12ln⁡∣x∣]13e s = \left[\frac{1}{2}x + \frac{1}{2}\ln|x|\right]_{1}^{3e} s=[21​x+21​ln∣x∣]13e​

STEP 16

Evaluate the integral from 111 to 3e3e3e.s=(12⋅3e+12ln⁡∣3e∣)−(12⋅1+12ln⁡∣1∣) s = \left(\frac{1}{2} \cdot 3e + \frac{1}{2}\ln|3e|\right) - \left(\frac{1}{2} \cdot 1 + \frac{1}{2}\ln|1|\right) s=(21​⋅3e+21​ln∣3e∣)−(21​⋅1+21​ln∣1∣)

STEP 17

Simplify the expression, noting that ln⁡∣1∣=0\ln|1| = 0ln∣1∣=0.s=3e2+12ln⁡(3e)−12 s = \frac{3e}{2} + \frac{1}{2}\ln(3e) - \frac{1}{2} s=23e​+21​ln(3e)−21​

STEP 18

Use the logarithm property ln⁡(ab)=ln⁡(a)+ln⁡(b)\ln(ab) = \ln(a) + \ln(b)ln(ab)=ln(a)+ln(b).s=3e2+12(ln⁡(3)+ln⁡(e))−12 s = \frac{3e}{2} + \frac{1}{2}(\ln(3) + \ln(e)) - \frac{1}{2} s=23e​+21​(ln(3)+ln(e))−21​

STEP 19

Since ln⁡(e)=1\ln(e) = 1ln(e)=1, further simplify the expression.s=3e2+12ln⁡(3)+12−12 s = \frac{3e}{2} + \frac{1}{2}\ln(3) + \frac{1}{2} - \frac{1}{2} s=23e​+21​ln(3)+21​−21​

STEP 20

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Calculate the arc length of $y=\frac{1}{4} x^{2}-\frac{1}{2} \ln x$ on $[1,3 e]$ .

First, we need to find the derivative of the function $y$ with respect to $x$ .
$\frac{dy}{dx} = \frac{d}{dx} \left(\frac{1}{4} x^{2} - \frac{1}{2} \ln x\right)$

Calculate the derivative using the power rule and the derivative of the natural logarithm.
$\frac{dy}{dx} = \frac{1}{4} \cdot 2x - \frac{1}{2} \cdot \frac{1}{x}$

Simplify the derivative.
$\frac{dy}{dx} = \frac{x}{2} - \frac{1}{2x}$

Now, we need to square the derivative to use in the arc length formula.
$\left(\frac{dy}{dx}\right)^2 = \left(\frac{x}{2} - \frac{1}{2x}\right)^2$

Expand the square of the binomial.
$\left(\frac{dy}{dx}\right)^2 = \left(\frac{x}{2}\right)^2 - 2 \cdot \frac{x}{2} \cdot \frac{1}{2x} + \left(\frac{1}{2x}\right)^2$

Simplify the expression.
$\left(\frac{dy}{dx}\right)^2 = \frac{x^2}{4} - \frac{1}{2} + \frac{1}{4x^2}$

Combine the terms under the square root.
$s = \int_{1}^{3e} \sqrt{\frac{1}{4x^2} + \frac{x^2}{4} + \frac{1}{2}} \, dx$

Simplify the expression under the square root by finding a common denominator.
$s = \int_{1}^{3e} \sqrt{\frac{1 + x^4 + 2x^2}{4x^2}} \, dx$

Extract the constant from under the square root.
$s = \int_{1}^{3e} \frac{\sqrt{x^4 + 2x^2 + 1}}{2x} \, dx$

Recognize the expression under the square root as a perfect square.
$x^4 + 2x^2 + 1 = (x^2 + 1)^2$

Simplify the integrand using the perfect square.
$s = \int_{1}^{3e} \frac{x^2 + 1}{2x} \, dx$

Split the integrand into two separate fractions.
$s = \int_{1}^{3e} \frac{1}{2} + \frac{1}{2x} \, dx$

Integrate each term separately.
$s = \left[\frac{1}{2}x + \frac{1}{2}\ln|x|\right]_{1}^{3e}$

Evaluate the integral from $1$ to $3e$ .
$s = \left(\frac{1}{2} \cdot 3e + \frac{1}{2}\ln|3e|\right) - \left(\frac{1}{2} \cdot 1 + \frac{1}{2}\ln|1|\right)$

Simplify the expression, noting that $\ln|1| = 0$ .
$s = \frac{3e}{2} + \frac{1}{2}\ln(3e) - \frac{1}{2}$

Use the logarithm property $\ln(ab) = \ln(a) + \ln(b)$ .
$s = \frac{3e}{2} + \frac{1}{2}(\ln(3) + \ln(e)) - \frac{1}{2}$

Since $\ln(e) = 1$ , further simplify the expression.
$s = \frac{3e}{2} + \frac{1}{2}\ln(3) + \frac{1}{2} - \frac{1}{2}$