**Length Of A Curve...**

** aka Arclength**

#

**The most amazing and awesome part of calculus.**

**Have do we find the length of the curve of a graph of y=f(x)?**

**And how can we do it if the graph is constantly changing?**

#

**APPROXIMATIONS!!!!!!!**

#

**http://www.bbc.co.uk/schools/gcsebitesize/maths/images/graph_7.gif&imgrefurl=http://www.bbc.co.uk/schools/gcsebitesize/maths/algebra/graphsrev4.shtml&usg=__D1OEZjnCrH0wSdeTn5zo6MizKKc=&h=394&w=546&sz=12&hl=en&start=5&um=1&tbnid=uaN2HTQn0EEq2M:&tbnh=96&tbnw=133&prev=/images%3Fq%3Dy%253D3x%252B1%26um%3D1%26hl%3Den%26rlz%3D1T4GGLL_en%26sa%3DN**

**If we wanted to know the length of y=3x+1 from x=0 to x=1, we would use the PYTHAGOREAN THEOREM!!!**

**WOOHOO!!**

**so...**

**The base leg has a length of 1, and the other leg has a length of 3. Using the pythagorean theorem we do:**

# √((1)²+(3)²)= √(10)

**The Length of f(x) from point (0,1) to point (1,4) is √(10)**

**With non-linear graphs,however, finding the arclength gets more difficult. With a curved graph you can't find the exact length using the pythagorean theorem as we used before. That's why we appoximate. Lets say we get the graph:**

** y=x²+5x**

**We have a non-linear graph. I personally want to know the length of the graph from point (0,0) to point (12,204). Because the pythagorean theorem worked so well before, we use** **it** **again, but as** **an approximation. For y= x² + 5x we use the integral. We take the integral from 0 to 12 of the square root of 1 + f'(x)². A generic formula would be**

**for our example we will plug in our numbers**

**= 204.4382639**

**The integral allows us to make a large amount of very very samll right triangles because the step size between two points are extremely small. We then use the hypotenuses of those small triangels to ≈ an arclength pretty accurately.**

**If we just did the normal pythagorean theorem we would get a much less accurate approximation.**

**Here we have just one triangle to approximate the length. One leg has a height of 204, and the other leg has a length of 12 so we do:**

**√((12²)+(204²))**

**to find the hypotenuse,which equals**

**204.353**

**Although these two values ****are** **almost equal they are not the same. It mostly means I chose a bad example equation... hehe**

**So if the equation is not a straight line and you need the arclength use the formula!**

FIRE RAINBOW!

**pretty!!**

**Practice Problems**

1) Find the arc length of the graph of on the interval [0,5].

http://www.cliffsnotes.com/WileyCDA/CliffsReviewTopic/Arc-Length.topicArticleId-39909,articleId-39908.html

2) Find the arc length of the graph of *f(x*) = ln (sin *x*) on the interval [π/4, π/2].

http://www.cliffsnotes.com/WileyCDA/CliffsReviewTopic/Arc-Length.topicArticleId-39909,articleId-39908.html

3) Find the length of the graph [3,10].

http://archives.math.utk.edu/visual.calculus/5/arclength.1/6.html

4) What is the arc length of the curve of y=ln(cosx)

**A)**

**B)**

**C)**

**D)**

**E) **

**F) **

** **

** **** **

** **

** **

**5) Find the lentgth of **

**http://blue.utb.edu/swyche/m2414a/cal_1_final_a/index.htm**

**A) 9/16**

**B) 335/432**

**C) 335/288**

**D) 335/3**

**6)Find the length of **

**http://blue.utb.edu/swyche/m2414a/cal_1_final_a/index.htm**

**A) 9/4**

**B)** **335/108**

**C) 335/288**

**D) 355/3**

SOLUTIONS!.doc

## Comments (2)

## nbarnett@students.maret.org said

at 3:14 pm on Feb 2, 2009

I liked that your page was very conversational. I thought that it was engaging and interesting. Your explanation was clear and I liked how you started with linear line and then progressed to a non-linear line. I also liked the graphs; they helped to visualize the concept. I would suggest highlighting and explaining more about the formula a little bit more, but i thought you did a great job!

## nbarnett@students.maret.org said

at 7:58 pm on Feb 2, 2009

ps. I liked the fire rainbow. it really helped explain arc length :) - Nora

You don't have permission to comment on this page.