# Length of Curve

1. Mar 8, 2013

### whatlifeforme

1. The problem statement, all variables and given/known data
Find the length of the curve between x=0 and x=1. Note: can this be done without a calculator?

2. Relevant equations
y = sqrt(4-x^2)

3. The attempt at a solution

x=2sin∅
dx = 2cos∅ d∅

sqrt(4-4(sin∅)^2) ---> 2cos∅

integral (0 to 1) sqrt(1+(-2sin∅)^2)

integral (0 to 1) sqrt(1+4(sin∅)^2) (How do i integrate this without using a calculator?)

Last edited: Mar 8, 2013
2. Mar 8, 2013

### Zondrina

The length of a non-parametrized curve such as yours is given by :

$L = \int_{a}^{b} \sqrt{1 + (f')^2} dx$

Where f' is the derivative of your function with respect to whatever variable it may be.

3. Mar 8, 2013

### LCKurtz

@whatlifeforme: It's too early to think about a substitution before you have the integral set up correctly, as Zondrina points out. Also, where did you get your limits of 0 to 1?

4. Mar 8, 2013

### whatlifeforme

sorry. it should say Find the length of the curve between x=0 and x=1. .

5. Mar 8, 2013

### whatlifeforme

is this correct integral for arc length?

integral (0 to 1) sqrt(1+4(sin∅)^2)

or does this one look correct?

integral (0 to 1) sqrt(1+1/(4*(4-x^2))

either way, they both look hard to integrate.

6. Mar 8, 2013

### tiny-tim

hi whatlifeforme!

i haven't followed what you've done

start again, writing everything out carefully

7. Mar 8, 2013

### Zondrina

Use the formula I gave you.... it really shouldn't be to difficult to compute this.

8. Mar 8, 2013

### whatlifeforme

so starting over:
find the length of the curve from x=0 to x=1.

y=sqrt(4-x^2)

arc length formula.
$\displaystyle\int_0^1 {\sqrt{1+F'(x)^2} dx}$

$\frac{dy}{dx} = \frac{1}{2\sqrt{4-x^2}} * -2x$

$\frac{dy^2}{dx^2} = \frac{x^2}{4-x^2}$

$\displaystyle\int_0^1 {\sqrt{1+[(x^2)/(4-x^2)]} dx}$

does this look correct so far, any tips to go from here?

Last edited: Mar 8, 2013
9. Mar 8, 2013

### LCKurtz

Sure. What are you waiting for? Simplify it.

10. Mar 8, 2013

### whatlifeforme

how? common denominator?

$\displaystyle\int_0^1 {\sqrt{1+[(x^2)/(4-x^2)]} dx}$

$\sqrt{\frac{4-x^2+x^2}{4-x^2}}$

$\sqrt{\frac{4}{4-x^2}}$

$\frac{2}{\sqrt{4-x^2}}$

$2\displaystyle\int_0^1 {\frac{dx}{\sqrt{4-x^2}}}$

2arcsin(x/2)]$^{1}_{0}$

2(∏/6) - 2(0) = ∏/3

Last edited: Mar 8, 2013
11. Mar 8, 2013

### iRaid

Use the trig substitution x=2sinθ when you get $\int \frac{4}{\sqrt{4-x^2}}dx$

12. Mar 8, 2013

### whatlifeforme

that would be the same as recognizing that it is the derivative of arcsin. however, if you were to do the substitution you would still end with the values i have, correct?

13. Mar 8, 2013

### Dick

Yes, but 2(∏/6) - 2(0) isn't equal to ∏/2. Typo?

14. Mar 8, 2013

### Bacle2

How about squaring both sides of the original :

y=${\sqrt{4-x^2}}$

to see what curve y is ?

15. Mar 8, 2013

### whatlifeforme

sorry. fixed. it's ∏/3.

16. Mar 8, 2013

### Dick

Then it's correct. Nice TeX by the way.

17. Mar 9, 2013

### whatlifeforme

thanks. i'm trying to get the hang of it.

18. Mar 9, 2013

### Bacle2

Well, since you already did the full problem, my idea was to use the fact that y is a circle of radius 2, centered at (0,0) , and that (1,√3) is a point corresponding to π/3 radians (easier to see after normalizing by dividing each term by 2 ), and (0,2) corresponds to π/2 radians (there is some ambiguity on wether for x=0 we choose y=2 , or y=-2 ). The length of an arc of (π/2 -π/3) radians in a circle of radius 2 is 2(π/2 -π/3) = π/3 .

19. Mar 9, 2013