# Arc length

#### Fernando Revilla

##### Well-known member
MHB Math Helper
I quote a question from Yahoo! Answers

Find arc length for the curve c(t)= (t,t,t^2) from 1<=t<=2?
I understand that I find C'(t) and integrate the length of it.

C'(t)= (1, 1, 2t) and so the length is sqrt{ 1 + 1 + 4t^2 } = sqrt{2+4t^2}

Now when integrating this....would I use the sqrt{ x^2 + a^2 } identity where x = 2t and a = sqrt{2}????

Help. Answer is (6-sqrt{3})/sqrt{2} + 1/2 log ( [2sqrt{2}+3]/[sqrt{2}+sqrt{3}] )
I have given a link to the topic there so the OP can see my response.

#### Fernando Revilla

##### Well-known member
MHB Math Helper
Using integration by parts with $u=\sqrt{x^2+a}$ and $dx=dx$, we get $du=\dfrac{x}{\sqrt{x^2+a}}$ and $v=x.$ Then,
\begin{aligned} I&=\int\sqrt{x^2+a}\;dx\\ &=x\sqrt{x^2+a}-\int \frac{x^2}{\sqrt{x^2+a}}dx\\ &=x\sqrt{x^2+a}-\int \frac{x^2+a-a}{\sqrt{x^2+a}}dx\\ &=x\sqrt{x^2+a}-\int \frac{x^2+a}{\sqrt{x^2+a}}dx+a\int \dfrac{dx}{\sqrt{x^2+a}}\\ &=x\sqrt{x^2+a}-I+a\log \left|\;x+\sqrt{x^2+a}\;\right|\\ &\Rightarrow I=\frac{x\sqrt{x^2+a}}{2}+\dfrac{a}{2}\log \left|\;x+\sqrt{x^2+a}\;\right|+C \end{aligned}

Now, using the above formula with $a=1/2$, you can easily find the length $L$ of the curve:
$$L=\int_1^2\sqrt{4x^2+2}\;dx=2\int_1^2\sqrt{x^2+1/2}\;dx=...$$