Homework Help: 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

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