# help finding exact arc length of a curve

• November 30th 2008, 09:26 PM
Matt164
help finding exact arc length of a curve
I need help with the following problem:

Find the exact arc length of the curve y= 2/3 * (x+1)^3/2 where the interval for x is [2,7]. Do the algebra and calculus carefully showing your work. (Final answer should be 38/3).

Here is what I tried to do, but was not able to get a final answer of 38/3:

First I found the derivative: y'=2/3(3/2)(x+1)^1/2 *1 = 1(x+1)^1/2

Then I plugged it into the formula for arc length:

s= ∫[from 2 to 7]Sqrt of 1 + ((x+1)^1/2)^2 dx
s= ∫[from 2 to 7]Sqrt of 1 + x^5/2 + 1^5/2 dx
s= ∫[from 2 to 7]Sqrt of (1+ x^5/2 + 1)^1/2
s= ∫[from 2 to 7]Sqrt of (1+ x^3 + 1)
s= (2 + x^3) | from 2 to 7
s=(2 + 7^3)-(2 + 2^3)
s=345-10
s=335

I'm guessing I must have messed up somewhere since my answer isn't anywhere close to 38/3. I am not sure where I messed up or what I did wrong, but if anyone can explain to me how to do this problem or explain what I am doing wrong I would greatly appreciate it. Thanks in advance to anyone who can help.
• November 30th 2008, 11:10 PM
Scopur
Just some computation stuff
$L= \displaystyle\int^7_2 \sqrt{1+ (x+1)^{1/2\cdot2}}\,dx = \displaystyle\int^7_2 \sqrt{x+2}\,dx$
• November 30th 2008, 11:16 PM
CaptainBlack
Quote:

Originally Posted by Matt164
I need help with the following problem:

Find the exact arc length of the curve y= 2/3 * (x+1)^3/2 where the interval for x is [2,7]. Do the algebra and calculus carefully showing your work. (Final answer should be 38/3).

Here is what I tried to do, but was not able to get a final answer of 38/3:

First I found the derivative: y'=2/3(3/2)(x+1)^1/2 *1 = 1(x+1)^1/2

Then I plugged it into the formula for arc length:

s= ∫[from 2 to 7]Sqrt of 1 + ((x+1)^1/2)^2 dx
s= ∫[from 2 to 7]Sqrt of 1 + x^5/2 + 1^5/2 dx
s= ∫[from 2 to 7]Sqrt of (1+ x^5/2 + 1)^1/2
s= ∫[from 2 to 7]Sqrt of (1+ x^3 + 1)

The next line does not follow from the above line. (and note what scopur says as well)

Quote:

s= (2 + x^3) | from 2 to 7
s=(2 + 7^3)-(2 + 2^3)
s=345-10
s=335

I'm guessing I must have messed up somewhere since my answer isn't anywhere close to 38/3. I am not sure where I messed up or what I did wrong, but if anyone can explain to me how to do this problem or explain what I am doing wrong I would greatly appreciate it. Thanks in advance to anyone who can help.

CB
• December 1st 2008, 12:10 PM
Matt164
Thanks. I see what I did wrong now. However, I am still not able to get the answer of 38/3. When I compute the integral: ∫[from 2 to 7]sqrt of (x+2)

I get: sqrt of(7+2) - sqrt of(2+2)= 3 - 2 = 1

Is 1 the correct answer or did I do something else wrong? I'm thinking maybe 38/3 is wrong and the book made a mistake or something. If anyone can help me I would appreciate it very much. Thanks.
• December 2nd 2008, 03:09 PM
Matt164
$L= \displaystyle\int^7_2 \sqrt{1+ (x+1)^{1/2\cdot2}}\,dx = \displaystyle\int^7_2 \sqrt{x+2}\,dx$
$= \left[\frac{2}{3}(x+2)^{\frac{3}{2}}\right]^7_2 = \frac{2}{3}(9)^{\frac{3}{2}}-\frac{2}{3}(4)^{\frac{3}{2}}= 18-\frac{16}{3} = \frac{38}{3}$