Solved on Feb 20, 2024

Calculate the arc length of y=14x212lnxy=\frac{1}{4} x^{2}-\frac{1}{2} \ln x on [1,3e][1,3 e].

STEP 1

Assumptions
1. The function given is y=14x212lnxy = \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][1, 3e].
3. The formula for the arc length ss of a function y=f(x)y = f(x) from aa to bb is given by: s=ab1+(dydx)2dx 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 yy with respect to xx.
dydx=ddx(14x212lnx) \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.
dydx=142x121x \frac{dy}{dx} = \frac{1}{4} \cdot 2x - \frac{1}{2} \cdot \frac{1}{x}

STEP 4

Simplify the derivative.
dydx=x212x \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.
(dydx)2=(x212x)2 \left(\frac{dy}{dx}\right)^2 = \left(\frac{x}{2} - \frac{1}{2x}\right)^2

STEP 6

Expand the square of the binomial.
(dydx)2=(x2)22x212x+(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

STEP 7

Simplify the expression.
(dydx)2=x2412+14x2 \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][1, 3e] and the expression for (dydx)2\left(\frac{dy}{dx}\right)^2.
s=13e1+x2412+14x2dx 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=13e14x2+x24+12dx 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=13e1+x4+2x24x2dx 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=13ex4+2x2+12xdx 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.
x4+2x2+1=(x2+1)2 x^4 + 2x^2 + 1 = (x^2 + 1)^2

STEP 13

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

STEP 14

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

STEP 15

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

STEP 16

Evaluate the integral from 11 to 3e3e.
s=(123e+12ln3e)(121+12ln1) 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 ln1=0\ln|1| = 0.
s=3e2+12ln(3e)12 s = \frac{3e}{2} + \frac{1}{2}\ln(3e) - \frac{1}{2}

STEP 18

Use the logarithm property 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}

STEP 19

Since ln(e)=1\ln(e) = 1, further simplify the expression.
s=3e2+12ln(3)+1212 s = \frac{3e}{2} + \frac{1}{2}\ln(3) + \frac{1}{2} - \frac{1}{2}

STEP 20

Combine like terms.
s=3e2+12ln(3) s = \frac{3e}{2} + \frac{1}{2}\ln(3)
This is the arc length of the function y=14x212lnxy = \frac{1}{4} x^{2} - \frac{1}{2} \ln x over the interval [1,3e][1, 3e].
s=3e2+12ln(3) s = \frac{3e}{2} + \frac{1}{2}\ln(3)

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