# Thread: arc length - stuck with intergration

1. ## arc length - stuck with intergration

how do you solve these questions:

1. find the arc length of Y^3= 8X^2 from x=1 and x=8.

my working: ( i know how to form the equation for intergration but i do not know how to carry out the intergration)

differenciating the qns, i got 3y^2 (dy/dx) = 3/4 X ^(1/3)

then i would like to intergrate ( 1 + (9/16 X ^(2/3)))^(1/2) dx from x=8 to x=1.

but im stuck now..how do i perform intergration on this equation?

2.find the arc length of 6xy = x^4 + 3

by implicit differentiation, i got
(dy/dx)^2 = ( 5/6x^2 + 1/ (2x^2))^2

and then i would have to perform intergration on

(1+ (dy/dx)^2 ) ^(1/2) from x= 2 to x= 1...

but i am stuck as to how i should carry out the intergration.

thank you!

2. ok

$\int_{8}^{1} \left( 1+ \frac{9}{16} x^{\frac{2}{3}} \right)^{\frac{1}{2}} \cdot dx$

let $y = x^{\frac{1}{3}}$

$y^3 = x \Rightarrow 3y^2 dy = dx$

$\int \left(1+ \frac{9}{16} y^2 \right) 3y^2 dy$

let $\tan u = \frac{3y}{4}$

$\tan ^2u = \frac{9y^2}{16} \Rightarrow y^2 = \frac{16\tan ^2u}{9}$

and $\sec u = \sqrt{1 + \frac{9 y^2 }{16}}$

$\sec^2 u\cdot du = \frac{3}{4} dy$

$\int \sec u \cdot 3\left(\frac{16\tan ^2u}{9}\right) \left(\frac{4\sec ^2u}{3} \right) du$

factorize the constant

$\int \sec ^3u \tan ^2 u \cdot du$

$\int sec^4 u \cdot du - \int sec u\cdot du$

can you handle this

3. Originally Posted by alexandrabel90
how do you solve these questions:

1. find the arc length of Y^3= 8X^2 from x=1 and x=8.

my working: ( i know how to form the equation for intergration but i do not know how to carry out the intergration)

differenciating the qns, i got 3y^2 (dy/dx) = 3/4 X ^(1/3)

then i would like to intergrate ( 1 + (9/16 X ^(2/3)))^(1/2) dx from x=8 to x=1.

but im stuck now..how do i perform intergration on this equation?

2.find the arc length of 6xy = x^4 + 3

by implicit differentiation, i got
(dy/dx)^2 = ( 5/6x^2 + 1/ (2x^2))^2

and then i would have to perform intergration on

(1+ (dy/dx)^2 ) ^(1/2) from x= 2 to x= 1...

but i am stuck as to how i should carry out the intergration.

thank you!
For the first one, it might be easier if you introduce parametric equation for your curve

$x = t^3, y = 2 t^2; t = 1 \to 2$

and use $\int \limits_c \sqrt{\dot{x}^2 + \dot{y}^2}\,dt$.

4. how did you think of forming this parametric equation?

5. may i also ask how to solve the second question?
thank you!

6. Hello, alexandrabel90!

Here's my approach to the first one . . .

1. Find the arc length of $y^3\:=\: 8x^2$ . from $x=1$ to $x=8.$

We have: ./ $y \:=\:2x^{\frac{2}{3}}$

Then: . $\frac{dy}{dx} \:=\:\tfrac{4}{3}x^{-\frac{1}{3}}\quad\Rightarrow\quad \left(\frac{dy}{dx}\right)^2 \:=\:\frac{16}{9}x^{-\frac{2}{3}} \quad\Rightarrow\quad 1 + \left(\frac{dy}{dx}\right)^2 \:=\:1 + \frac{16}{9x^{\frac{2}{3}}}$

. . $1 + \left(\frac{dy}{dx}\right)^2 \:=\:\frac{9x^{\frac{2}{3}} + 16}{9x^{\frac{2}{3}}} \quad\Rightarrow\quad \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \:=\:\frac{\sqrt{9x^{\frac{2}{3}} + 16}}{3x^{\frac{1}{3}}}$

We have: . $L \;=\;\int^8_0\frac{\sqrt{9x^{\frac{2}{3}} + 16}}{3x^{\frac{1}{3}}}\,dx$

Let: . $3x^{\frac{1}{3}} \:=\:4\tan\theta \quad\Rightarrow\quad x^{\frac{1}{3}} \:=\:\tfrac{4}{3}\tan\theta \quad\Rightarrow\quad x \:=\:\tfrac{64}{27}\tan^3\!\theta$
. . . $dx \:=\:\tfrac{64}{9}\tan^2\!\theta\sec^2\!\theta\,d\ theta$

Substitute: . $L \;=\;\int\frac{4\sec\theta}{3\cdot\frac{4}{3}\tan\ theta}\left(\tfrac{64}{9}\tan^2\!\theta\sec^2\!\th eta\,d\theta\right) \;=\;\tfrac{64}{9}\int \sec^3\!\theta\tan\theta\,d\theta$

$\text{And we have: }\;L \;=\;\tfrac{64}{9}\int \underbrace{\sec^2\!\theta}_{u^2}\underbrace{(\sec \theta\tan\theta\,d\theta)}_{du}$

Got it?

7. Hello, alexandrabel90!

Okay, here's the second one . . .

2.Find the arc length of: . $6xy \:=\: x^4 + 3$ . . What limits?

We have: . $y \:=\:\tfrac{1}{6}x^3 + \tfrac{1}{2}x^{-1}$

Then: . $\frac{dy}{dx} \:=\:\tfrac{1}{2}x^2 - \tfrac{1}{2}x^{-2} \:=\:\frac{x^2-x^{-2}}{2}$

. . $\left(\frac{dy}{dx}\right)^2 \;=\;\frac{(x^2 - x^{-2})^2}{4} \;=\;\frac{x^4 - 2 + x^{-4}}{4}$

. . $1 + \left(\frac{dy}{dx}\right)^2 \;=\;1 + \frac{x^4 - 2 + x^{-4}}{4} \;=\; \frac{x^4 + 2 + x^{-4}}{4} \;=\;\frac{(x^2 + x^{-2})^2}{4}$

. . $\sqrt{1+\left(\frac{dy}{dx}\right)^2} \;=\;\frac{x^2 + x^{-2}}{2}$

And we have: . $L \;=\;\tfrac{1}{2}\int^b_a\left(x^2 + x^{-2}\right)\,dx$ . . . etc.

8. Originally Posted by alexandrabel90
how did you think of forming this parametric equation?
When learning about arc length we learn about it when we have parametic equation. When I saw $y^3 = 8 x^2$ I saw

$\left(\frac{y}{2}\right)^3 = x^2$ and that's when I thought of it.

If you let $\frac{y}{2} = t^p, x = t^q$ and pick p and q to match the powers. Here $p = 2$ and $q = 3$ work! Also note that when $x = 0$ then $t = 1$ and $x = 8$ then $t = 2$. Then it's a matter of assembling the arc length formula

$\dot{x} = 3x^2, \dot{y} = 4x$ so

$\int_0^1 \sqrt{9x^4 + 16x^2}\,dx = 3\int_0^1 x\sqrt{x^2 + \frac{16}{9}}\,dx$

then we can use the sub $u = x^2 + \frac{16}{9}$.

Using the original formula in terms of x works (as Soroban shows) but I think is a little more difficult.

9. thanks!