Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
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
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
up vote 2 down vote accepted

Use the formula

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

share|cite|improve this answer
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

Hint: Compute $y'$ and then compute the length $\mathcal{l}$:

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

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.