# Arc Length

1. Jul 20, 2006

### suspenc3

$$y^2 = 4(x+4)^3 0<=x<=2 ......y>0$$

Ok..so do I just simplify it to y=etc... and then find the derivative and solve..I tried this and I get a kinda long derivative..just seems like it should be easier?

Thanks

2. Jul 20, 2006

### HallsofIvy

Staff Emeritus
Not all that "long" is it? the derivative of y2, with respect to x, is $\2y \frac{dy}{dt}$ and the derivative of 4(x+4)3 is 12(x+ 4)2 so differentiating both sides you get
$$2y \frac{dy}{dx}= 12(x+4)^2$$
So what is $\frac{dy}{dx}$?

3. Jul 20, 2006

### suspenc3

oh, i did it completely wrong..i took the root of both sides and hen tried to get the derivative

4. Jul 20, 2006

### benorin

Whether you solve it implicitly (HallsofIvy's post) or explicitly (taking the rott, etc.) the result is the same: the implicit method gives

$$2y \frac{dy}{dx}= 12(x+4)^2\Rightarrow \frac{dy}{dx}= \frac{6(x+4)^2}{y} = \frac{6(x+4)^2}{2(x+4)^{\frac{3}{2}}}=3(x+4)^{\frac{1}{2}}$$

since solving the given equation for y gives $$y=2(x+4)^{\frac{3}{2}$$ and the positive root was chosen since y>0. This is the starting point for the explicit method, just differentiate the last expression

$$y=\frac{d}{dx}\left[ 2(x+4)^{\frac{3}{2}}\right] = 3(x+4)^{\frac{1}{2}}$$

which is actually easier.

5. Jul 20, 2006

### suspenc3

oohhh..Ok Thanks, I get it

6. Jul 20, 2006

### benorin

7. Jul 20, 2006

### suspenc3

cant i just do..$$\int_0^2 \sqrt{1+(3(x+4)^1^/^2)^2}$$

$$\int_0^2 \sqrt{37+9x}$$

8. Jul 20, 2006

### benorin

Yes, absolutely: I was in error.

9. Jul 20, 2006

### suspenc3

but wha tyou have there..is $$\sqrt{3(x+4)}^2 = 9(x+4)^2$$..

shouldnt it be: $$\sqrt{3(x+4)}^2 = 8(x+4)$$..which can be simplified and then its an easy substitution.

10. Jul 20, 2006

### suspenc3

oh,,ok..Thanks!

11. Jul 20, 2006

### benorin

use rather u=37+9x so that du=9dx to get

$$\int_0^2 \sqrt{37+9x} dx = \frac{1}{9}\int_{37}^{55} \sqrt{u}du$$

Last edited: Jul 20, 2006
12. Jul 20, 2006

### suspenc3

$$y= \frac{x^5}{5} + \frac{1}{10x^3} 1>=x>=2$$

I found the derivative to be $$\frac{5x^4}{6} - \frac{3}{10x^4}$$..im kinda stuck here..should I just keep going and will I eventually see a substituion?

Last edited: Jul 20, 2006
13. Jul 20, 2006

### suspenc3

yes that is what i got..except du=9dx

14. Jul 20, 2006

### benorin

yep, it was

15. Jul 20, 2006

### suspenc3

if I have $$\int_0^\pi^/^3 sqrt{sec^2x}$$..should I substitute u = sec^2x..or can it just be secx..and find the integral?

Last edited: Jul 20, 2006
16. Jul 21, 2006

Do you mean $$\int_0^\frac{\pi}{3}\sqrt{sec^2x} dx$$?

If yes, note that since secx is positive from 0 to $$\pi$$/3, then $$\sqrt{sec^2x}$$ is equivalent to secx.

There is a standard formula for evaulating $$\int_0^\frac{\pi}{3}secxdx$$. Do you know what it is?

Last edited: Jul 21, 2006
17. Jul 21, 2006

### benorin

The derivative of $$y= \frac{x^5}{5} + \frac{1}{10x^3}$$ is $$y^{\prime}= x^4 - \frac{3}{10x^4}$$

18. Jul 21, 2006

### suspenc3

ohh..sorry its suppose to be $$y= \frac{x^5}{6} + \frac{1}{10x^3} 1>=x>=2$$

19. Jul 21, 2006

### suspenc3

and for pizzasky..no I dont know how..I found some stuff on the internet concerning the integral of secx..I never knew this integral up until now..so im guessing theres another way

20. Jul 21, 2006