What is the arclength of #f(x)=sqrt((x-1)(2x+2))-2x# on #x in [6,7]#?

Question

What is the arclength of #f(x)=sqrt((x-1)(2x+2))-2x# on #x in     [6,7]#?

Scarlett Allison · Accepted Answer

#1.15037# We have #f(x)=sqrt((x-1)*2*(x+1))-2x# #f(x)=sqrt(2)*sqrt(x^2-1)-2x# So we get the integral #int_6^7sqrt(1+(sqrt(2)*x/sqrt(x^2-1)-2)^2)dx# With a numerical method we find #approx 1.1.5037#

Emma Aldridge · Answer

undefined To find the arc length of the function $ f(x) = \sqrt{(x-1)(2x+2)} - 2x $ over the interval $ x $ in $[6, 7]$, we use the arc length formula:

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

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

First, we find the derivative of $ f(x) $:

$$ f'(x) = \frac{d}{dx} \left( \sqrt{(x-1)(2x+2)} - 2x \right) $$

$$ f'(x) = \frac{1}{2\sqrt{(x-1)(2x+2)}} \left( \frac{1}{2}(2x+2) + (x-1)(2) \right) - 2 $$

$$ f'(x) = \frac{(2x+2) + 2(x-1)}{2\sqrt{(x-1)(2x+2)}} - 2 $$

$$ f'(x) = \frac{2x+2 + 2x - 2}{2\sqrt{(x-1)(2x+2)}} - 2 $$

$$ f'(x) = \frac{4x}{2\sqrt{(x-1)(2x+2)}} - 2 $$

$$ f'(x) = \frac{2x}{\sqrt{(x-1)(2x+2)}} - 2 $$

Now, we plug this into the arc length formula:

$$ L = \int_{6}^{7} \sqrt{1 + \left( \frac{2x}{\sqrt{(x-1)(2x+2)}} - 2 \right)^2} \, dx $$

$$ L \approx \int_{6}^{7} \sqrt{1 + \left( \frac{2x}{\sqrt{6(14)}} - 2 \right)^2} \, dx $$

$$ L \approx \int_{6}^{7} \sqrt{1 + \left( \frac{x}{\sqrt{42}} - 2 \right)^2} \, dx $$

$$ L \approx \int_{6}^{7} \sqrt{1 + \left( \frac{x}{\sqrt{42}} - \frac{2\sqrt{42}}{\sqrt{42}} \right)^2} \, dx $$

$$ L \approx \int_{6}^{7} \sqrt{1 + \left( \frac{x - 2\sqrt{42}}{\sqrt{42}} \right)^2} \, dx $$