Integrating Without a Calculator: Finding Length of Curve

[whatlifeforme]

Homework Statement

Find the length of the curve between x=0 and x=1. Note: can this be done without a calculator?

Homework Equations

y = sqrt(4-x^2)

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?)

 

[STEMucator]

Homework Statement

Find the length of the curve. Note: can this be done without a calculator?

Homework Equations

y = sqrt(4-x^2)

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?)

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.

 

[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?

 

[whatlifeforme]

@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?

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

 

[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.

 

[tiny-tim]

hi whatlifeforme!

i haven't followed what you've done

start again, writing everything out carefully ​

 

[STEMucator]

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.

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

 

[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?

 

[LCKurtz]

Sure. What are you waiting for? Simplify it.

 

[whatlifeforme]

Sure. What are you waiting for? Simplify it.

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

 

[iRaid]

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

 

[whatlifeforme]

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

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?

 

[Dick]

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

 

[Bacle2]

How about squaring both sides of the original :

y={\sqrt{4-x^2}} to see what curve y is ?

 

[whatlifeforme]

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

sorry. fixed. it's ∏/3.

 

[Dick]

sorry. fixed. it's ∏/3.

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

 

[whatlifeforme]

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

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

 

[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 .

 

[tiny-tim]

How about squaring both sides of the original :

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


to see what curve y is ?

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