Finding Arc Length

[Jason76]

POST EDITED - 5/5/2013

y = L = length of arc

Formula:

\(\displaystyle \int dy = \int\limits_a^b [1 + ( (\dfrac{dy}{dx}(x)))^{2}]^{1/2} dx\)


Between the x value points 0 (lower bound) and 1 (upper bound):

\(\displaystyle y = x^{3/2}\)

\(\displaystyle \dfrac{dy}{dx} = x (\dfrac{3}{2})^{1/2}\)

\(\displaystyle \int dy = \int\limits_0^1 [1 + ( (\dfrac{3}{2})x^{1/2})^{2}]^{1/2} dx \)

\(\displaystyle \int dy =\int [1 + (\dfrac{9}{4}) x]^{1/2} dx\)

\(\displaystyle y + C_y = (2/3) [1 + (\dfrac{9}{4})x]^{3/2} (\dfrac{4}{9}) + C_x \)

\(\displaystyle y + C_y - C_y = (\dfrac{8}{27}) [1 + (\dfrac{9}{4})x]^{3/2} + C_x - C_y\)

\(\displaystyle y = (\dfrac{8}{27}) [1 + (\dfrac{9}{4})x]^{3/2} + C\)

Now on to evaluating the integral at the lower and upper bound:

\(\displaystyle (\dfrac{8}{27}) [1 + (\dfrac{9}{4})x]^{3/2}\)

Lower bound:

\(\displaystyle (\dfrac{8}{27}) [1 + (\dfrac{9}{4})(0)]^{3/2} = 8/27\)

Upper bound:

\(\displaystyle (\dfrac{8}{27}) [1 + (\dfrac{9}{4})(1)]^{3/2} = (2,197)^{1/2}/27\) - How did the exponents (via addition etc..) come out to be 1/2 ?

Final Answer:

y = \(\displaystyle \dfrac{(2,197)^{1/2}}{27} - \dfrac{8}{27} = 1.440\) units (This is L or the length of the arc)

 

[lookagain]

Now on to evaluating the integral at the lower and upper bound:

\(\displaystyle (\dfrac{8}{27}) [1 + (\dfrac{9}{4})x]^{3/2}\)

Lower bound:

\(\displaystyle (\dfrac{8}{27}) [1 + (\dfrac{9}{4})(0)]^{3/2} = 8/27\)

Upper bound:

\(\displaystyle (\dfrac{8}{27}) [1 + (\dfrac{9}{4})(1)]^{3/2} = 2,197^{1/2}/27\) - How did the exponents (via addition etc..) come out to be 1/2 ?

Final Answer:

L (length of Arc) = \(\displaystyle \dfrac{2,197^{1/2}}{27} - \dfrac{8}{27} = 1,440\) Please redo the arithmetic to the left.units

I calculated the length of arc \(\displaystyle \approx 1.4397 \ \ units.\)

 

[Jason76]

I calculated the length of arc \(\displaystyle \approx 1.4397 \ \ units.\)

The book says "A calculator can approximate to 1.440 units. That's between the constraints we derived: a minimum of \(\displaystyle 2^{1/2}\) approximately 1.414 and a maximum of 2. (but I have no clue what there talking about, but your answer comes close to the approximation).

Anyhow, how about the exponents in the line above it?

 

[DrPhil]

The book says "A calculator can approximate to 1.440 units. That's between the constraints we derived: a minimum of \(\displaystyle 2^{1/2}\) approximately 1.414 and a maximum of 2. (but I have no clue what there talking about, but your answer comes close to the approximation).

Anyhow, how about the exponents in the line above it?

You confused me by using the name "y" for two different things. \(\displaystyle y = x^{3/2} \) is the function, but everywhere else you seem to use y to mean the length. For instance, I the starting formula should be


\(\displaystyle \displaystyle L = \int dL = \int_a^b \left[ 1 + \left( \dfrac{dy}{dx} \right) ^2 \right] ^{1/2} dx \)


which is derived from \(\displaystyle (dL)^2 = (dx)^2 + (dy)^2 \)

So I will mentally change "y" to "L" where needed. For instance, instead of \(\displaystyle y =\), it should be

\(\displaystyle \displaystyle L = \int_0^1 \left[ 1 + \dfrac{9}{4} x \right] ^{1/2} dx \)

Then I can see what you are doing - until you get to the upper bound. That really looks strange. Please fix it!

Now lets look at the statement from the book you found confusing.

If you drew a straight line between the endpoints of the integral, the length would be sqrt(2) .. so that is the "minimum" that L could be. On the other hand, if you take the path first from (0,0) to (1,0) than then from (1,0) to (1,1), then the "maximum" that L could possibly be is 2. So without doing any integration or calculation at all, you already know some constraints:
\(\displaystyle \sqrt{2} < L < 2 \)

Your integral is exact, but when you use a calculator to evaluate you are making an approximation. To three decimal places, the calculator gives 1.440.

\(\displaystyle L = \dfrac{8}{27} \left[ \left( \dfrac{13}{4} \right)^{3/2} - 1 \right] \approx 1.400 \)

You can check that the answer really does lie within the constraints. Does that explain what you quoted from the book?

 

[lookagain]

3/2}[/tex]



L (length of Arc) = \(\displaystyle \dfrac{(2,197)^{1/2}}{27} - \dfrac{8}{27} = > > 1,440 < < \) units

Here is why I stated to check your arithmetic. You have the equivalent of
one thousand four hundred forty instead of one point four four zero,
because you have a comma instead of a decimal point.

 

[Jason76]

You confused me by using the name "y" for two different things. \(\displaystyle y = x^{3/2} \) is the function, but everywhere else you seem to use y to mean the length. For instance, I the starting formula should be


\(\displaystyle \displaystyle L = \int dL = \int_a^b \left[ 1 + \left( \dfrac{dy}{dx} \right) ^2 \right] ^{1/2} dx \)


which is derived from \(\displaystyle (dL)^2 = (dx)^2 + (dy)^2 \)

So I will mentally change "y" to "L" where needed. For instance, instead of \(\displaystyle y =\), it should be

\(\displaystyle \displaystyle L = \int_0^1 \left[ 1 + \dfrac{9}{4} x \right] ^{1/2} dx \)

Then I can see what you are doing - until you get to the upper bound. That really looks strange. Please fix it!

Now lets look at the statement from the book you found confusing.

If you drew a straight line between the endpoints of the integral, the length would be sqrt(2) .. so that is the "minimum" that L could be. On the other hand, if you take the path first from (0,0) to (1,0) than then from (1,0) to (1,1), then the "maximum" that L could possibly be is 2. So without doing any integration or calculation at all, you already know some constraints:
\(\displaystyle \sqrt{2} < L < 2 \)

Your integral is exact, but when you use a calculator to evaluate you are making an approximation. To three decimal places, the calculator gives 1.440.

\(\displaystyle L = \dfrac{8}{27} \left[ \left( \dfrac{13}{4} \right)^{3/2} - 1 \right] \approx 1.400 \)

You can check that the answer really does lie within the constraints. Does that explain what you quoted from the book?

Ok, good point. I will edit the post and change L to y etc.. I did notice, myself, that I what I wrote was strange in that way.

 

[soroban]

Hello, Jason76!

With a little algebra, you can make the problem easier.

\(\displaystyle \text{Find the arc length of }\,y \,=\,x^{\frac{3}{2}}\,\text{ on the interval }[0,\,1]\)

\(\displaystyle \displaystyle\text{Formula: }\: L\;=\;\int^b_a\sqrt{1+\left(\tfrac{dy}{dx}\right)^2}\,dx\)


\(\displaystyle \text{We have: }\:y \:=\: x^{\frac{3}{2}} \quad\Rightarrow\quad \frac{dy}{dx} \:=\:\frac{3}{2}x^{\frac{1}{2}}\)

. . . . . . . \(\displaystyle 1+\left(\frac{dy}{dx}\right)^2 \;=\;1 + \left(\frac{3}{2}x^{\frac{1}{2}}\right)^2 \;=\; 1 + \frac{9x}{4} \;=\;\frac{4+9x}{4}\)

. . . . . . \(\displaystyle \sqrt{1+\left(\frac{dy}{dx}\right)^2} \;=\;\sqrt{\dfrac{9x+4}{4}} \;=\;\dfrac{\sqrt{9x+4}}{2}\)


\(\displaystyle \displaystyle\text{Hence: }\;L \;=\;\int^1_0\frac{\sqrt{9x+4}}{2}\,dx \;=\;\tfrac{1}{2}\int^1_0(9x+4)^{\frac{1}{2}}\,dx \)


\(\displaystyle \text{Let }\,u \,=\, 9x+4 \quad\Rightarrow\quad du \,=\, 9\,dx \quad\Rightarrow\quad dx \,=\,\frac{1}{9}\,du\)

\(\displaystyle \displaystyle\text{Substitute: }\:L \;=\;\tfrac{1}{2}\int u^{\frac{1}{2}}\left(\tfrac{1}{9}\,du\right) \;=\; \tfrac{1}{18}\int u^{\frac{1}{2}}du \;=\;\tfrac{1}{27}u^{\frac{3}{2}} + C \)

\(\displaystyle \text{Back-substitute: }\:L \;=\;\tfrac{1}{27}(9x+4)^{\frac{3}{2}}\,\bigg]^1_0 \)

. . \(\displaystyle L \;=\;\frac{1}{27}(13)^{\frac{3}{2}} - \frac{1}{27}(4)^{\frac{3}{2}} \;=\; \dfrac{13\sqrt{13} - 8}{27}\)

 

[lookagain]

Final Answer:

y = \(\displaystyle \dfrac{(2,197)^{1/2}}{27} - \dfrac{8}{27} = 1,440\) units (This is L or the length of the arc)

Jason76, you still need to edit your post at this point to "1.440" to be correct. As you left it at "1,440" it is still not correct.

 

[Jason76]

Jason76, you still need to edit your post at this point to "1.440" to be correct. As you left it at "1,440" it is still not correct.

Now corrected.