Question

Find the arc length (L) of
\[y^2=4(x+3)^3\]
for the closed interval [0,1] (that'll mean a=0 & b=1). Formula for arc length.
\[ L = \int\limits_{a}^{b} \sqrt{1+(\frac{dy}{dx})^2} dx \]
The problem is I get either a messy cube root fraction in terms of Y, or a \(\pm\)root mess (kind of like what you get for circles, ellipses, or hyperbolas). Can somebody show me the steps for the differentiation into integration here? I assume it's probably some tip or trick I've over looked. (+1 medal will be given)

Answer 1

hi agentx5
i dont know if this is right or not
can we write
\[y=\pm 2(x+3)^{3/2}\\ \frac{dy}{dx}=\pm 3(x+3)^{1/2} \\ (\frac{dy}{dx})^2= 9(x+3)\\ then\\ L=2 \int\limits_{0}^{1} \sqrt{1+9(x+3)} dx\]
?

Answer 2

@agentx5

Answer 3

\[2\cdot [\frac{9x^2}{2}+28x+C]_0^1\]
\[2\cdot ((\frac{9(1)^2}{2}+28(1))-(\frac{9(0)^2}{2}+28(0))\]

Hmm the correct answer should be \(\large\frac{104}{3}\)

Answer 4

Nobody? :-/

Answer 5

i will totally work on it after my exam :)

Answer 6

Gracias/Thank You/Schönen Dank :-)

Answer 7

\[L=\int_{0}^{1}\sqrt{1+\left[f'(x)\right]^2}dx\]
\[y^2=4(x+3)^3=f(x)^2 \Rightarrow f(x)=\pm \sqrt{4(x+3)^3}=\pm 2\sqrt{(x+3)^3}\]
\[f'(x)=\pm 3\sqrt{x+3}\]
\[L=\int_{0}^{1}\sqrt{1+9(x+3)}dx=\int_0^1\sqrt{9x+4}dx=\frac{2}{27}(13\sqrt{13}-8)\]
I'm not getting \(\frac{104}{3}. . .\)

Answer 8

I see nothing wrong with @mukushla 's or @Limitless 's answers, so there is a typo somehwere

Answer 9

oh wait, Limitless messed up

Answer 10

\[L=\int_0^1\sqrt{1+9(x+3)}dx=\int_0^1\sqrt{9x+28}dx\]

Answer 11

still don't get 104/3

Answer 12

@mukushla did, as well.

Answer 13

I don't see any reason for the \(2\) outside of the integration.

Answer 14

oh yeah, neither do I

Answer 15

but I think we all agree that the answer to the question as it is given is not 104/3

Answer 16

Indeed, @TuringTest.

Answer 17

@TuringTest
y=2(x+3)^(3/2)
and
y=-2(x+3)^(3/2)
thats why i put 2 outside
is that right or ...

Answer 18

@mukushla, the two disappears after simplification of the derivative.