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Find the length of the space curve given by $$r(t) = 2t\,\mathbf{i} + 5\cos(t)\,\mathbf{j} + 5\sin(t)\,\mathbf{k}$$ over the interval $[0,2]$.

I did this and I got the answer as 10.77

Did I get the right answer? please help me i'm not very good at this.

M.H
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jain smit
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  • What method did you use? Here $\vec{r}=(2t,5\cos{t},5\sin{t})$, and calculating $\int_0^2|r'(t)|\mathrm{d}t$ gives you the arc length of the curve (what you are after), and this turns out to be roughly $10.77$, as you got. – Tim Jun 09 '13 at 17:29
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    Your answer is inccorect. The correct answer is $2\sqrt{29}$. Although $2\sqrt{29} \approx 10.77$, $2\sqrt{29} \neq 10.77$. The answer $10.77$ is correct to two decimal places. It is wrong for infinitely many more decimal places, – Fly by Night Jun 09 '13 at 17:36
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    It seems your work is fine, and that the only point of contention here is about the need to be careful about giving answers that are mere approximations. Unless otherwise told , $2 \sqrt{29}$ cannot be further simplified and is the exact solution. Unless otherwise told, use the exact form of the solution and not its approximation $\approx 10.77$ – amWhy Jun 09 '13 at 18:41
  • Dear bob: I would encourage you to take a few moments to review answers you have received, and accept one answer per question you've asked. (An "asker" can only accept one answer to any given question asked.) To accept an answer, just click on the $\large \checkmark$ to the left of the answer you'd like to accept. May I also note that you can now, also, upvote as many answers as you'd like! – amWhy Jun 09 '13 at 18:45

2 Answers2

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Yes and no. The correct answer is $2\sqrt{29} \approx 10.77$.

$$\int_a^b \sqrt{\dot{x}^2+\dot{y}^2+\dot{z}^2} \, \operatorname{d}\! t $$

In your example, $x(t) = 2t$, $y(t) = 5 \cos t$ and $z(t) = 5 \sin t$. Hence $\dot{x}(t) = 2$, $\dot{y}(t) = -5\sin t $ and $\dot{z}(t) = 5\cos t$. In turn \begin{array}{ccc} \dot{x}^2+\dot{y}^2+\dot{z}^2 &=& 2^2 + (-5\sin t)^2 + (5 \cos t)^2 \\ &\equiv& 4 + 25\sin^2 t + 25\cos^2t \\ &\equiv& 4 + 25(\sin^2t + \cos^2 t) \\ &\equiv& 4 + 25 \\ &=& 29 \end{array} The length of the curve is then: $$\int_0^2 \sqrt{\dot{x}^2+\dot{y}^2+\dot{z}^2} \, \operatorname{d}\!t = \int_0^2 \sqrt{29} \, \operatorname{d}\!t = 2\sqrt{29}.$$

Fly by Night
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Hint:$$L=\int_0^{2}|r'(t)|dt$$ then $\int_0^{2}|r'(t)|dt=\int_0^{2}\sqrt{25+4}dt=2\sqrt{29}\sim 10.7703$

M.H
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