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I'm really at the end of my wits on this problem. Basically I'm trying to find arc length. The vector-valued function is: $R=\langle t,\sqrt{t}\rangle$ and $t\ge0$.

We're looking for the length of the arc between $0\leq t\leq3$. I know that you basically have to integrate the magnitude of the tangent vector over the interval specified, and to this end

I got as far as the $\int_0^3 \sqrt(1+1/4t)dt.$ On that note, I cannot for the life of me figure out how to integrate this. I've tried factoring, simplifying, substitution (which seemed to make it worse), and anything else I could think of. Am I going about this the wrong way? Is there an easier method? Any help would be greatly appreciated.

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Hint

Start with a change of variable such that $$\sqrt{\frac{1}{4 t}+1}=u$$ which means $$t=\frac{1}{4 \left(u^2-1\right)}$$ $$dt=-\frac{u}{2 \left(u^2-1\right)^2}du$$ Then $$\int \sqrt{\frac{1}{4 t}+1}~dt=-\int \frac{u^2}{2 \left(u^2-1\right)^2}~du$$ Now use partial fraction decomposition. The result is quite simple.

I am sure that you can take from here.

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  • $\begingroup$ awesome... I didn't think of using partial fraction decomposition. Thanks! $\endgroup$
    – impyre
    Jul 7, 2014 at 9:20
  • $\begingroup$ You are welcome. In reality, all comes from the change of variable. After, it is simple. Cheers :) $\endgroup$ Jul 7, 2014 at 9:23
  • $\begingroup$ Sorry, but I'm stuck again. I believe I have the correct partial fraction decomposition and was about to attempt to integrate it, but i can't figure out how to change the limits of integration. I keep getting a 0 in the denominator. I'm trying to find what u is when t=0 and when t=3, or should I use a different approach? $\endgroup$
    – impyre
    Jul 7, 2014 at 10:33
  • $\begingroup$ Just replace from the definition of $u$ : the lower bound is $\infty$ and the upper bound is $\sqrt {13/12}$. $\endgroup$ Jul 7, 2014 at 10:48
  • $\begingroup$ I'm still completely lost on this part. I need t to go from 0 to 3. I've tried using 1+1/t=u^2 to solve for u when t is 0 and when it's 3, but that's not working. Or am I supposed to put the result of the indefinite integral of u into a definite integral of t? $\endgroup$
    – impyre
    Jul 7, 2014 at 10:54

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