Question

find the length of the curve y=(9-x^2/3)^3/2 from x=1 to x=27

Answer 1

Integrate buddy.

Answer 2

\[L=\int\limits \sqrt{1+(\frac{dy}{dx})^2}dx\], right?

Answer 3

Pythagorean distance formula..

Answer 4

Hi roj3og :)

How are you dear?

Its arc length. Right?

Answer 5

@CliffSedge integrate from a to b. lower limit is a and upper is b

Answer 6

yeah that is what the book says to do but I can't seem to find the right dy/dx
arc length correct

Answer 7

@Shayaan_Mustafa , yes, I know what limits are..

Answer 8

Why? It is so simple . Just differentiate w.r.t. to x. Is it difficult my friend?

Answer 9

How far along did you get? Were you able to find dy/dx? Do you know what function you need to integrate?

Answer 10

is this correct for the dy/dx
\[-((9-x ^{2/3})/x ^{1/3} ) ^{1/2}\]

Answer 11

not sure where the x^1/3 came from..

Answer 12

or the x^2/3..

Answer 13

Use chain rule. \[\frac{d}{dx} u(x)^{3/2} = \frac{3}{2}u(x)^{1/2}u'(x) : u=9-\frac{x^2}{3}\]

Answer 14

n.b. u(x)^3/2 means the function, u-of-x raised to 3/2, not u times x to the 3/2.

Answer 15

Am I reading your function incorrectly? Is it\[y=(9-\frac{x^2}{3})^{3/2}\] or \[y=(9-x^{2/3})^{3/2}?\]

Answer 16

If it's the latter, then your derivative was almost correct.
\[y=(9-x^{2/3})^{3/2} \rightarrow \frac{dy}{dx}=-\frac{(9-x^{2/3})^{1/2}}{x^{1/3}}\]

Answer 17

it is the second one.

Answer 18

Were you able to do the integration after finding that derivative?

Answer 19

\[
1+y'(x)^2=\frac{9-x^{2/3}}{x^{2/3}}+1=\frac{9}{x^{2/3}}
\]

Answer 20

\[
\sqrt{1+y'(x)^2}=\sqrt{\frac{9}{x^{2/3}}}
=\frac{3}{\sqrt[3]{x}}
\]

Answer 21

eliassaab, thank you but that is what i finally came up with as well and the homework doesn't give that as an answer. the options they give me are.
28/3, 14/3, 21, or 14

Answer 22

\[
\int_1^{27} \frac{3}{\sqrt[3]{x}} \,
dx=36
\]

Answer 23

Either you typed the problem wrong or the choices are not correct.

Answer 24

I was looking at the answers for another problem, HAHA. eliassaab you saved me on that one. I've been working on this for too long, everything is starting to blur together.

Answer 25

thanks everyone!