# Thread: Another Arc Length Problem

1. ## Another Arc Length Problem

I've attached the problem I'm working on, I keep getting negative answers. I know I'm left with the square root of 5 next to the integral (if I'm not mistaken that is), which would end up being sqrt[5] times x evaluated from -3 to 1 and I keep getting negative answers. Help a brotha out?

2. ## hi

hi

When you have a curve $y =f(x)$

$\int_{\gamma} \, ds = \int_{a}^{b} \sqrt{1+f'(x)^{2}} \, dx$

So in this case, we get:

$L = \int_{-3}^{1} \sqrt{1+2^{2}}\, dx = 4\sqrt{5}$

3. Originally Posted by Twig
hi

When you have a curve $y =f(x)$

$\int_{\gamma} \, ds = \int_{a}^{b} \sqrt{1+f'(x)^{2}} \, dx$

So in this case, we get:

$L = \int_{-3}^{1} \sqrt{1+2^{2}}\, dx = 4\sqrt{5}$
How did you get that though? Because you'd have sqrt[5] minus the sqrt[5] times -3?

4. I've attached another similar problem, here I don't know how the heck to find the integral of 144x^1/4, it ends up being crazy, what would I do in this situation?

5. Originally Posted by fattydq
I've attached another similar problem, here I don't know how the heck to find the integral of 144x^1/4, it ends up being crazy, what would I do in this situation?
$y = 5 + 8x^{\frac{3}
{2}} \hfill \\$

$y' = 12x^{\frac{1}
{2}} \hfill \\$

$1 + \left( {y'} \right)^2 = 1 + 144x \hfill \\$

${\text{Arc Length}} \hfill \\$

$\int\limits_0^1 {\sqrt {1 + 144x} } {\text{ }}dx \hfill \\$

${\text{Substitue, }}1 + 144x = t^2 \hfill \\$

$\Rightarrow 144dx = 2tdt \Rightarrow dx = \frac{{tdt}}
{{72}} \hfill \\$

${\text{As }}x \to 0,{\text{ }}t \to 1{\text{ and as }}x \to 1,{\text{ }}t \to \sqrt{145} \hfill \\$

$= \frac{1}
{{72}}\int\limits_1^{\sqrt{145}} {t^2 dt} = \left. {\frac{1}
{{72}}\left( {\frac{{t^3 }}
{3}} \right)} \right|_1^{\sqrt{145}} = ....... \hfill \\$

6. Originally Posted by fattydq
I've attached another similar problem, here I don't know how the heck to find the integral of 144x^1/4, it ends up being crazy, what would I do in this situation?
$x = y + y^3 \hfill \\$

$\frac{{dx}}
{{dy}} = 1 + 3y^2 \hfill \\$

$1 + \left( {\frac{{dx}}
{{dy}}} \right)^2 = 1 + \left( {1 + 3y^2 } \right)^2 = 2 + 6y^2 + y^4 \hfill \\$

${\text{Arc Length}} \hfill \\$

$\int\limits_2^5 {\left( {2 + 6y^2 + y^4 } \right)} {\text{ }}dy \hfill \\$

7. Originally Posted by fattydq
How did you get that though? Because you'd have sqrt[5] minus the sqrt[5] times -3?

$\int_{-3}^{1} \sqrt{5}\, dx = \sqrt{5}\, \int_{-3}^{1} \, dx = \sqrt{5}(1-(-3)) = \sqrt{5}(4) = 4\sqrt{5}$