# Arc Length of Curve: Find Solution (-2,2) to (2,4)

• sepah50
In summary: A1=&A2=&A3=&A4=&A5=&A6=&A7=&A8=&A9=&output=tableIn summary, the homework statement is to find the arclength of the section y=x2 between points (-2,2) and (2,4). The Attempt at a Solution found that y'=2x and then used the integral table to find the arclength to be 1.4. The other intergral 4x^2=(2x)^2 can be integrated out by parts (and using some elementary substitutions). Thanks for the advice.
sepah50

## Homework Statement

Find the arclength of the section y=x2 between points (-2,2) and (2,4)

## Homework Equations

L = $$\int\sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2}$$

## The Attempt at a Solution

So what I did first is find the derivative of y=x2 which is y'=2x

So I put that into the formula and get $$\int\sqrt{1+4x^2}$$ from limit point 1 to 4

From this point on I attempted to do u-substitution but it didn;t seem to work.. Anyone have any suggestions o how I can get it done?

Last edited:
Trig substitution works well here.

Char. Limit said:
Trig substitution works well here.

Yes, some people have told me to use Trigonometric substitution but the thing is that my teacher hasn't taught us that yet!

Represent it as:

$$\sqrt{1 + 4 x^{2}} = \frac{1 + 4 x^{2}}{\sqrt{1 + 4 x^{2}}} = \frac{1}{\sqrt{1 + 4 x^{2}}} + \frac{4 x^{2}}{\sqrt{1 + 4 x^{2}}}$$

The integral:

$$\int{\frac{dt}{\sqrt{1 + t^{2}}}} = \sinh^{-1}{(t)} = \ln{(t + \sqrt{1 + t^{2}})}$$

is sometimes quoted as a table integral. To my knowledge, there is no other way to prove except by those substitutions mentioned earlier, or hyperbolic ones.

The other intergral:

$$4 \, \int{\frac{x^{2} \, dx}{\sqrt{1 + 4 x^{2}}}}$$

can be integrated out by parts (and using some elementary substitutions). Good luck!

Well, basically here's how it works. Let's say that...

$$2x=tan(\theta)$$

Then this is also true.

$$2 dx = sec^2(\theta) d\theta$$

Just plug those values in for dx and 4x^2=(2x)^2 and then you have a new integral in theta.

But don't forget to change your bounds too.

Char. Limit said:
Well, basically here's how it works. Let's say that...

$$2x=tan(\theta)$$

Then this is also true.

$$2 dx = sec^2(\theta) d\theta$$

Just plug those values in for dx and 4x^2=(2x)^2 and then you have a new integral in theta.

But don't forget to change your bounds too.

Thanks for the advice, it is much appreciated! What I found out is that I can use the integral table and use this formula: http://www.sosmath.com/tables/integral/integ11/integ11.html" It is #8 of this list. Would that work?

Last edited by a moderator:

## 1. What is the formula for finding the arc length of a curve?

The formula for finding the arc length of a curve is L = ∫ √(1+(dy/dx)^2) dx, where dy/dx represents the derivative of the curve.

## 2. How is the starting and ending point of the curve determined?

The starting and ending point of the curve are determined by the given coordinates, (-2,2) and (2,4) in this case. These points represent the beginning and end of the curve.

## 3. Can you explain the steps for finding the arc length of a curve?

First, find the derivative of the curve. Then, plug the derivative into the formula L = ∫ √(1+(dy/dx)^2) dx. Integrate the expression and substitute in the coordinates of the starting and ending points. Finally, solve for L, which represents the arc length of the curve.

## 4. Is there a shortcut for finding the arc length of a straight line?

Yes, the arc length of a straight line can be found using the Pythagorean Theorem, where L = √((x2-x1)^2 + (y2-y1)^2), where (x1,y1) and (x2,y2) represent the starting and ending points of the line.

## 5. Can the arc length of a curve be negative?

No, the arc length of a curve cannot be negative. It represents the distance along the curve and therefore, is always a positive value.

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