Length of a curve

Calculus Level 3

What is the length of the curve, y = 1 3 ( x 2 + 2 ) 3 2 y = \frac{1}{3}(x^2+2)^{\frac{3}{2}} , from x = 0 x = 0 to x = 9 x = 9 ?


The answer is 252.

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1 solution

Tunk-Fey Ariawan
Jan 31, 2014

Length of curve is L = 0 9 1 + ( d y d x ) 2 d x = 0 9 1 + ( x x 2 + 1 ) 2 d x = 0 9 x 4 + 2 x 2 + 1 d x = 0 9 ( x 2 + 1 ) 2 d x = 0 9 ( x 2 + 1 ) d x = 1 3 x 3 + x 0 9 = 252 . \begin{aligned} L&=\int_0^9 \sqrt{1+\left(\frac{dy}{dx}\right)^2} \,dx\\ &=\int_0^9 \sqrt{1+(x\sqrt{x^2+1})^2} \,dx\\ &=\int_0^9 \sqrt{x^4+2x^2+1} \,dx\\ &=\int_0^9 \sqrt{(x^2+1)^2} \,dx\\ &=\int_0^9 (x^2+1) \,dx\\ &=\left. \frac{1}{3}x^3+x\right|_0^9\\ &=\boxed{252}. \end{aligned} # Q . E . D . # \text{\# }\mathbb{Q}.\mathbb{E}.\mathbb{D}.\text{\#}

I have done it in this particular process.

Arghyanil Dey - 7 years, 1 month ago

I too did the same

Mehul Chaturvedi - 6 years, 4 months ago

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