1. ## arc length

What is the arc length of y=2x^(3/2), from x=1 to x=3?

2. Formula is L=Intagral from a to b (square root of 1+ derivative of f(x) squared dx.

3. Originally Posted by chevy900ss
What is the arc length of y=2x^(3/2), from x=1 to x=3?
${L_y} = \int\limits_a^b {\sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} dx} = \int\limits_1^3 {\sqrt {1 + {{\left( {\frac{d}
{{dx}}2{x^{3/2}}} \right)}^2}} dx} = \int\limits_1^3 {\sqrt {1 + {{\left( {3{x^{1/2}}} \right)}^2}} dx} =$

$= \int\limits_1^3 {\sqrt {1 + 9x} dx} = \frac{1}{9}\int\limits_1^3 {{{\left( {1 + 9x} \right)}^{1/2}}d\left( {1 + 9x} \right)} = \left. {\frac{1}
{9}\left( {\frac{2}{3}{{\left( {1 + 9x} \right)}^{3/2}}} \right)} \right|_1^3 =$

$= \left. {\frac{2}{{27}}{{\left( {1 + 9x} \right)}^{3/2}}} \right|_1^3 = \frac{2}{{27}}\left( {\sqrt {{{28}^3}} - \sqrt {{{10}^3}} } \right) = \frac{2}{{27}}\left( {\sqrt {{{28}^2} \cdot 28} - \sqrt {{{10}^3}} } \right) =$

$= \frac{2}{{27}}\left( {\sqrt {{{28}^2} \cdot {2^2} \cdot 7} - \sqrt {{{10}^2} \cdot 10} } \right) = \frac{2}{{27}}\left( {56\sqrt 7 - 10\sqrt {10} } \right) =$

$= \frac{4}{{27}}\left( {28\sqrt 7 - 5\sqrt {10} } \right) \approx 8.632540504$

4. thats awesome, i was on the right track. I also have 3 more questions. Should i post them on this same forum or make a new post for them. Or is there an email that i can email them to someone and they can help me. They are basically having to do with the same thing.

5. Originally Posted by DeMath
${L_y} = \int\limits_a^b {\sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} dx} = \int\limits_1^3 {\sqrt {1 + {{\left( {\frac{d}
{{dx}}2{x^{3/2}}} \right)}^2}} dx} = \int\limits_1^3 {\sqrt {1 + {{\left( {3{x^{1/2}}} \right)}^2}} dx} =$

$= \int\limits_1^3 {\sqrt {1 + 9x} dx} = \frac{1}{9}\int\limits_1^3 {{{\left( {1 + 9x} \right)}^{1/2}}d\left( {1 + 9x} \right)} = \left. {\frac{1}
{9}\left( {\frac{2}{3}{{\left( {1 + 9x} \right)}^{3/2}}} \right)} \right|_1^3 =$

$= \left. {\frac{2}{{27}}{{\left( {1 + 9x} \right)}^{3/2}}} \right|_1^3 = \frac{2}{{27}}\left( {\sqrt {{{28}^3}} - \sqrt {{{10}^3}} } \right) = \frac{2}{{27}}\left( {\sqrt {{{28}^2} \cdot 28} - \sqrt {{{10}^3}} } \right) =$

$= \frac{2}{{27}}\left( {\sqrt {{{28}^2} \cdot {2^2} \cdot 7} - \sqrt {{{10}^2} \cdot 10} } \right) = \frac{2}{{27}}\left( {56\sqrt 7 - 10\sqrt {10} } \right) =$

$= \frac{4}{{27}}\left( {28\sqrt 7 - 5\sqrt {10} } \right) \approx 8.632540504$
De Math, I think understand the priciples behind arc length.

1. partition

2. sum the distances between each $f(x_i)$ and $f(x_{i-1})$

3. take the limit

4. Define the limit as an integral

So, I have some questions:

1. is what I stated above correct?

2. If it is, could you provide a complete worked example?

6. Originally Posted by chevy900ss
. I also have 3 more questions.
Generally, we start a new thread if the questions doesn't relate the this one specifically.