MHB Length of Curve $y^2=4(x+4)^3$: 13.5429

[ineedhelpnow]

$y^2=4(x+4)^3$
$0 \le x \le 2$

$y=2(x+4)^{3/2}$

$y'=3(x+4)^{1/2}$

$\int_{0}^{2} \ \sqrt{1+9(x+4)},dx = 13.5429$

is that right?

 

[ineedhelpnow]

wait, no one respond yet. i wrote the question down wrong. i have to back and redo it.

actually that is the right question. did i do it right?

 

[I like Serena]

$y^2=4(x+4)^3$
$0 \le x \le 2$

$y=2(x+4)^{3/2}$

$y'=3(x+4)^{1/2}$

$\int_{0}^{2} \ \sqrt{1+9(x+4)}\,dx = 13.5429$

is that right?

What you wrote is all correct (assuming $y\ge 0$).

What has your final integral got to do with what came before it?

 

[ineedhelpnow]

$\int_{a}^{b} \ \sqrt{1+[f'(x)]^2},dx$

the steps i showed was solving for my y' and then i squared it and plugged it in.

 

[Prove It]

$y^2=4(x+4)^3$
$0 \le x \le 2$

$y=2(x+4)^{3/2}$

$y'=3(x+4)^{1/2}$

$\int_{0}^{2} \ \sqrt{1+9(x+4)},dx = 13.5429$

is that right?

Your setup of the integral is correct. You can probably get an exact answer though.

$\displaystyle \begin{align*} \int_0^2{ \sqrt{ 1 + 9 \left( x + 4 \right) } \,\mathrm{d}x } &= \int_0^2{ \sqrt{ 1 + 9x + 36 }\,\mathrm{d}x } \\ &= \int_0^2{ \sqrt{ 9x + 37 } \, \mathrm{d}x } \\ &= \frac{1}{9} \int_0^2{ 9\,\sqrt{ 9x + 37 } \, \mathrm{d}x } \end{align*}$

and now make the substitution $\displaystyle \begin{align*} u = 9x + 37 \implies \mathrm{d}u = 9\,\mathrm{d}x \end{align*}$ and noting that $\displaystyle \begin{align*} u(0) = 37 \end{align*}$ and $\displaystyle \begin{align*} u(2) = 55 \end{align*}$, the integral becomes

$\displaystyle \begin{align*} \frac{1}{9} \int_0^2{ 9\,\sqrt{ 9x + 37} \, \mathrm{d}x } &= \frac{1}{9} \int_{37}^{55}{ \sqrt{u}\,\mathrm{d}u } \\ &= \frac{1}{9} \int_{37}^{55}{ u^{\frac{1}{2}}\,\mathrm{d}u } \\ &= \frac{1}{9} \left[ \frac{u^{\frac{3}{2}}}{\frac{3}{2}} \right] _{37}^{55} \\ &= \frac{1}{9} \left[ \frac{2}{3} u\,\sqrt{u} \right] _{37}^{55} \\ &= \frac{2}{27} \left( 55\,\sqrt{55} - 37\,\sqrt{37} \right) \end{align*}$

 

[ineedhelpnow]

thats what i got but i just posted the decimal form because it was easier. thanks for working it out though :)