# Thread: Find the length of a graph

1. ## Find the length of a graph

Find the length of the graph of y= from(1, - ) to (4, )

2. Originally Posted by evansf
Find the length of the graph of y= from(1, - ) to (4, )
$y = \frac{1}{3}x^{3/2} - x^{1/2}$

The arc length formula is
$L = \int_a^b \sqrt{1 + \left ( y^{\prime } \right ) ^2 }~dx$

So
$L = \int_1^4 \sqrt{1 + \left ( \frac{1}{2}x^{1/2} - \frac{1}{2}x^{-1/2} \right ) ^2 }~ dx$

$L = \int_1^4 \sqrt{1 + \frac{1}{4} \cdot \left ( x - 2 + \frac{1}{x} \right ) }~ dx$

$L = \int_1^4 \sqrt{\frac{1}{2} + \frac{1}{4} \cdot \left ( x + \frac{1}{x} \right ) }~ dx$

$L = \int_1^4 \sqrt{\frac{1}{2} + \frac{1}{4} \cdot \left ( x + \frac{1}{x} \right ) }~ dx$

Unfortunately, this form shows that the integral is going to turn out to be an elliptic integral. It cannot be calculated by hand, only approximated.

-Dan

3. Originally Posted by topsquark
$L = \int_1^4 \sqrt{\frac{1}{2} + \frac{1}{4} \cdot \left ( x + \frac{1}{x} \right ) }~ dx$

Unfortunately, this form shows that the integral is going to turn out to be an elliptic integral. It cannot be calculated by hand, only approximated.
¿?

It can be computed

4. Hello, evansf!

Find the length of the graph of: . $y \:= \:\frac{1}{3}x^{\frac{3}{2}} - x^{\frac{1}{2}}$ . from $\left(1,\,-\frac{2}{3}\right)$ to $\left(4,\,\frac{2}{3}\right)$
Arc Length Formula: . $\boxed{L \;=\;\int^b_a\sqrt{1 + \left(\frac{dy}{dx}\right)^2}\,dx}$

We have: . $y \;=\;\frac{1}{3}x^{\frac{3}{2}} - x^{\frac{1}{2}}$

Then: . $\frac{dy}{dx}\;=\;\frac{1}{2}x^{\frac{1}{2}} - \frac{1}{2}x^{-\frac{1}{2}}
$

And: . $1 + \left(\frac{dy}{dx}\right)^2\;=\;1 + \left(\frac{1}{2}x^{\frac{1}{2}} - \frac{1}{2}x^{-\frac{1}{2}}\right)^2 \;=\;1 + \frac{1}{4}x^{\frac{1}{4}} - \frac{1}{2} + \frac{1}{4}x^{-\frac{1}{4}}$

. . . . . . $=\;\frac{1}{4}x^{\frac{1}{4}} + \frac{1}{2} + \frac{1}{4}x^{-\frac{1}{4}} \;=\;\left(\frac{1}{2}x^{\frac{1}{2}} + \frac{1}{2}x^{-\frac{1}{2}}\right)^2$

Hence: . $\sqrt{1 + \left(\frac{dy}{dx}\right)^2} \;=\; \sqrt{\left(\frac{1}{2}x^{\frac{1}{2}} + \frac{1}{2}x^{-\frac{1}{2}}\right)^2 } \;=\;\frac{1}{2}x^{\frac{1}{2}} + \frac{1}{2}x^{-\frac{1}{2}} \;=\;\frac{1}{2}\left(x^{\frac{1}{2}} + x^{-\frac{1}{2}}\right)
$

Therefore, we have: . $L \;=\;\frac{1}{2}\int^4_1\left(x^{\frac{1}{2}} + x^{-\frac{1}{2}}\right)\,dx$

. . . Can you finish it?

5. Originally Posted by Soroban

And: . $1 + \left(\frac{dy}{dx}\right)^2\;=\;1 + \left(\frac{1}{2}x^{\frac{1}{2}} - \frac{1}{2}x^{-\frac{1}{2}}\right)^2 \;=\;1 + \frac{1}{4}x^{\frac{1}{4}} - \frac{1}{2} + \frac{1}{4}x^{-\frac{1}{4}}$
You might want to take a look at your FOILing here.

-Dan

6. Originally Posted by Krizalid
¿?

It can be computed
Are you are saying that I goofed and it is not an elliptic integral, or are you merely saying that it can be approximated (fairly) easily?

-Dan

7. Put it at the integrator

--

Ohhh, Soroban made a mistake there