Find the length of the curve defined by
#y=18(4x^2−2ln(x)), x in[4,6]#?

Question

Charles Anctil · Accepted Answer

#L~~1440-36ln(3/2)+1/576ln(143/63)# units is a first-order solution accurate to the 8th decimal place.

#y=18(4x^2−2lnx)=36(2x^2-lnx)# #y'=36(4x-1/x)# Arc length is given by: #L=int_4^6sqrt(1+(y')^2)dx# Rearrange: #L=int_4^6y'sqrt(1+(y')^-2)dx# Take the series expansion: #L=int_4^6y'{sum_(n=0)^oo((1/2),(n))(y')^(-2n)}dx# Isolate the #n=0# term and simplify: #L=int_4^6y'dx+sum_(n=1)^oo((1/2),(n))int_4^6(36(4x-1/x))^(1-2n)dx# Solve for the zeroth-order solution and rearrange: #L=[y]_ 4^6+sum_(n=1)^oo((1/2),(n))1/6^(4n-2)int_4^6(x/(4x^2-1))^(2n-1)dx# Isolate the #n=1# term and simplify: #L=1440-36ln(3/2)+1/72int_4^6(x/(4x^2-1))dx+sum_(n=2)^oo((1/2),(n))1/12^(4n-2)int_4^6(1/(2x-1)+1/(2x+1))^(2n-1)dx# Solve for the first-order correction: #L=1440-36ln(3/2)+1/576[ln|4x^2-1|]_ 4^6+sum_(n=2)^oo((1/2),(n))1/12^(4n-2)int_4^6(1/(2x-1)+1/(2x+1))^(2n-1)dx# Hence: #L=1440-36ln(3/2)+1/576ln(143/63)+sum_(n=2)^oo((1/2),(n))1/12^(4n-2)int_4^6(1/(2x-1)+1/(2x+1))^(2n-1)dx# Further work can be done to the desired level of accuracy.

Charles Anctil · Answer

undefined To find the length of the curve defined by $ y = 18(4x^2 - 2 \ln(x)) $ over the interval $[4, 6]$, we'll use the formula for arc length:

$$ L = \int_{a}^{b} \sqrt{1 + (f'(x))^2} \, dx $$

Where $ f'(x) $ is the derivative of $ f(x) $.

First, let's find the derivative of $ y $:

$$ y' = 18(8x - \frac{2}{x}) $$

Now, we'll plug this into the formula for arc length and integrate over the given interval:

$$ L = \int_{4}^{6} \sqrt{1 + \left(18(8x - \frac{2}{x})\right)^2} \, dx $$

After evaluating this integral, you'll find the length of the curve.

Find the length of the curve defined by #y=18(4x^2−2ln(x)), x in[4,6]#?