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Find the length of the curve defined by $y=6 x^{3/2} - 7$ from $x=1$ to $x = 9$.

I need help with this section. I would really appreciate it. Thank you!

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Where did you get stuck? Did you have trouble computing the integral in the arc length formula? –  David Mitra Dec 17 '12 at 14:18
    
I just realized it's arclength :-D I know how to do this. –  Ceelos Dec 17 '12 at 14:18
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Compute $L=\int_1^9 \sqrt{1+ (y')^2 }\,dx$ –  David Mitra Dec 17 '12 at 14:20
    
@DavidMitra, Yes, thank you (: –  Ceelos Dec 17 '12 at 14:20
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3 Answers

up vote 2 down vote accepted

Use the formula

$$ s = \int_{a}^{b}\sqrt{1+y'(x)^2} dx $$

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Thank you, I don't know why I was blanking out. I know how to do this already. Thank you for answering though (-: –  Ceelos Dec 17 '12 at 14:20
    
@Ceelos: You are welcome. –  Mhenni Benghorbal Dec 17 '12 at 14:22
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Hint: Compute $y'$ and then compute the length $\mathcal{l}$:

$$\mathcal{l} =\int_1^9 \sqrt{1 + (y'(x))^2}dx$$

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+1 Effective hint. Yesterday, I was watching an old film, My Life, starring M. Keaton and N. Kidman and was reflecting about some points deeply. Indeed, Giving and its power will change our monotonous passing life, Amy. –  B. S. Aug 15 '13 at 8:07
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We have $$ y=6x^{(3/2)} -7$$ Let $\mathcal{L}$ denote the length of the curve. We can calculate $\mathcal{L}$ by \begin{align} \mathcal{L} &= \int_1^9 \sqrt{1+(y')^2} \, dx \\ &= \int_1^9 \sqrt{1+((6x^{(3/2)} -7)')^2} \, dx \\ &= \int_1^9 \sqrt{1+(9x^{(1/2)})^2} \, dx \\ &= \int_1^9 \sqrt{1+81x} \, dx \\ &= \Bigl[\frac{2}{243} (1+81x)^{3/2} \Bigr]_1^9 \approx 156.22 \end{align} See Wolfram Alpha for the integration.

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@amWhy as I am also quite new to MSE, could you tell me, why you edited "\begin,end {align}" at the first equation to "$$ "? Is there a optical difference between the two? Thanks in advanvce! –  sonystarmap Dec 17 '12 at 18:24
    
Please know that I rarely edit others' answers. I wanted to "undo" my downvote (I simply dislike handing out full answers right away.) But votes get locked in, after 10 minutes, unless the post is edited. Hence, to "undownvote", I edited you post. Your formatting was perfectly fine!!! –  amWhy Dec 17 '12 at 18:28
    
Ah okay thanks for explaining. I also guess you're right concerning not handling out the answer right away. I'm still learning ;) –  sonystarmap Dec 17 '12 at 18:31
    
macydanim: we're ALL still learning! There's no problem with answers, per se. Some of us prefer to provide suggestions, hints, etc., when questions are tagged "homework" and/or when the asker specifies "any suggestions?" or "any hints?" –  amWhy Dec 17 '12 at 23:52
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