1. lenght of a curve

Find the length of the curve defined by
y=3x32–11
from x=2 to x=10.

ok i think that

L= integral from 2 to 10 sqrt((dx/dt)^2+(dy/dt)^2)dt

dx/dt would be 3x^3/2-11

do i get dy/dt by just solving for x then plug them both in because it doesnt seem to be working for me

2. Try this:

$\displaystyle L = \int_a^b \sqrt{1 + (\frac{dy}{dx})^2}$

3. I'm assuming you want the length of curve of $\displaystyle y=3x^\frac {3}{2}-11$ from 2 to 10?

This is the formula to find the length of the curve.

$\displaystyle L=\int_a^b \sqrt {1+\bigg( \frac{dy}{dx}\bigg )^2}dx$

4. Originally Posted by chengbin
I'm assuming you want the length of curve of $\displaystyle y=3x^\frac {3}{2}-11$ from 2 to 10?

This is the formula to find the length of the curve.

$\displaystyle L=\int_a^b \sqrt {1+\bigg( \frac{dy}{dx}\bigg )^2}dx$

so its intergral from 2 to 10 sqrt(1+(3x^3/2-11)^2)

5. $\displaystyle L = \int\limits_2^{10} {\sqrt {1 + {{\left( {\frac{d}{{dx}}\left( {3{x^{3/2}} - 11} \right)} \right)}^2}}\, dx} = \int\limits_2^{10} {\sqrt {1 + {{\left( {\frac{9}{2}{x^{1/2}}} \right)}^2}}\, dx} = \int\limits_2^{10} {\sqrt {1 + \frac{{81}}{4}x}\, dx} =$

$\displaystyle = \frac{1}{2}\int\limits_2^{10} {\sqrt {4 + 81x}\, dx} = \frac{1}{{162}}\int\limits_2^{10} {\sqrt {4 + 81x}\, d\left( {4 + 81x} \right)} = \left. {\frac{1}{{162}} \cdot \frac{2}{3}\sqrt {{{\left( {4 + 81x} \right)}^3}} } \right|_2^{10} =$

$\displaystyle = \frac{1}{{243}}\left( {\sqrt {{{814}^3}} - \sqrt {{{166}^3}} } \right) = \frac{2}{{243}}\left( {407\sqrt {814} - 83\sqrt {166} } \right) \approx 86.77044{\text{ }}\left( {{\text{linear units}}} \right).$