Complicated Arc length problem

[missingmyname]

Homework Statement

The length L or the curve given by
$$\frac{3y^{4}}{2}+\frac{1}{48y^{2}}-5$$
from y=1 to y=2

Homework Equations

The Attempt at a Solution

Setting up the formula is easy. First I found the derivative of f(y) which is:
$$f'(y)=6y^{3}-\frac{1}{24y^{3}}$$
Then I plugged all the given data into the formula
$$L=\int_{1}^{2}\sqrt{1+(6y^{3}-\frac{1}{24y^{3}})^{2}}~dy$$
I expanded out the binomial and simplified to
$$L=\int_{1}^{2}(36y^{6}+\frac{1}{48y^{6}}+\frac{1}{2})^{1/2}~dy$$
It is from here I get stuck. It looks to me like I need to use a u-substitution. I checked out the step-by-step solution from wolfram alpha's integral calculator and it uses the following
$$u=\frac{1}{x^{6}}~\therefore~du=-\frac{6}{x^{7}}dy$$
I could go on and list each next step (another substitution is used down the line), but that would do me little good. I do not understand how the value $\frac{1}{x^{6}}$ for $u$ was chosen.

 

[LCKurtz]

Homework Statement

The length L or the curve given by
$$\frac{3y^{4}}{2}+\frac{1}{48y^{2}}-5$$
from y=1 to y=2

Homework Equations

The Attempt at a Solution

Setting up the formula is easy. First I found the derivative of f(y) which is:
$$f'(y)=6y^{3}-\frac{1}{24y^{3}}$$
Then I plugged all the given data into the formula
$$L=\int_{1}^{2}\sqrt{1+(6y^{3}-\frac{1}{24y^{3}})^{2}}~dy$$
I expanded out the binomial and simplified to
$$L=\int_{1}^{2}(36y^{6}+\frac{1}{\color{red}{48}y^{6}}+\frac{1}{2})^{1/2}~dy$$
.

That number is wrong. Fix it and you should get a perfect square under the square root sign.

 

[SteamKing]

Homework Statement

The length L or the curve given by
$$\frac{3y^{4}}{2}+\frac{1}{48y^{2}}-5$$
from y=1 to y=2

Homework Equations

The Attempt at a Solution

Setting up the formula is easy. First I found the derivative of f(y) which is:
$$f'(y)=6y^{3}-\frac{1}{24y^{3}}$$
Then I plugged all the given data into the formula
$$L=\int_{1}^{2}\sqrt{1+(6y^{3}-\frac{1}{24y^{3}})^{2}}~dy$$
I expanded out the binomial and simplified to
$$L=\int_{1}^{2}(36y^{6}+\frac{1}{48y^{6}}+\frac{1}{2})^{1/2}~dy$$
It is from here I get stuck. It looks to me like I need to use a u-substitution. I checked out the step-by-step solution from wolfram alpha's integral calculator and it uses the following
$$u=\frac{1}{x^{6}}~\therefore~du=-\frac{6}{x^{7}}dy$$
I could go on and list each next step (another substitution is used down the line), but that would do me little good. I do not understand how the value $\frac{1}{x^{6}}$ for $u$ was chosen.

It's not clear how you went from:

${1+(6y^{3}-\frac{1}{24y^{3}})^{2}}$ to

$(36y^{6}+\frac{1}{48y^{6}}+\frac{1}{2})$

There seems to be some algebra which doesn't look quite right when expanding the squared expression.
$(a + b)^{2} = a^{2} + 2ab + b^{2}$

 

[Ray Vickson]

Homework Statement

The length L or the curve given by
$$\frac{3y^{4}}{2}+\frac{1}{48y^{2}}-5$$
from y=1 to y=2

Homework Equations

The Attempt at a Solution

Setting up the formula is easy. First I found the derivative of f(y) which is:
$$f'(y)=6y^{3}-\frac{1}{24y^{3}}$$
Then I plugged all the given data into the formula
$$L=\int_{1}^{2}\sqrt{1+(6y^{3}-\frac{1}{24y^{3}})^{2}}~dy$$
I expanded out the binomial and simplified to
$$L=\int_{1}^{2}(36y^{6}+\frac{1}{48y^{6}}+\frac{1}{2})^{1/2}~dy$$
It is from here I get stuck. It looks to me like I need to use a u-substitution. I checked out the step-by-step solution from wolfram alpha's integral calculator and it uses the following
$$u=\frac{1}{x^{6}}~\therefore~du=-\frac{6}{x^{7}}dy$$
I could go on and list each next step (another substitution is used down the line), but that would do me little good. I do not understand how the value $\frac{1}{x^{6}}$ for $u$ was chosen.

When I do the simplification (using Maple) I get
$$L = \int_1^2 \sqrt{ 36y^6+\frac{1}{576 y^6}+\frac{1}{2}} \, dy$$
(because $24^2 = 576$). It turns out that
$$\sqrt{ 36y^6+\frac{1}{576 y^6}+\frac{1}{2}}$$
has the nice form of
$$a + b y^3 + \frac{c}{y^3}$$
for some constants $a,b,c$ (again thanks to Maple). Just find these constants and you are essentially done.

 

[missingmyname]

When I do the simplification (using Maple) I get
$$L = \int_1^2 \sqrt{ 36y^6+\frac{1}{576 y^6}+\frac{1}{2}} \, dy$$
(because $24^2 = 576$). It turns out that
$$\sqrt{ 36y^6+\frac{1}{576 y^6}+\frac{1}{2}}$$
has the nice form of
$$a + b y^3 + \frac{c}{y^3}$$
for some constants $a,b,c$ (again thanks to Maple). Just find these constants and you are essentially done.

Now I feel retarded. That's what I get for going at coursework for 8 hours straight while tired. I clearly have 576, not 48 written the line above on my scratch paper.

 

[missingmyname]

I fear I am going to have to slap myself again, but with this corrected equation:
$$L = \int_1^2 \sqrt{ 36y^6+\frac{1}{576 y^6}+\frac{1}{2}} \, dy$$
I am still having difficulty despite this advice:
$$a + b y^3 + \frac{c}{y^3}$$
What exactly is it that I am missing?

 

[Ray Vickson]

I fear I am going to have to slap myself again, but with this corrected equation:
$$L = \int_1^2 \sqrt{ 36y^6+\frac{1}{576 y^6}+\frac{1}{2}} \, dy$$
I am still having difficulty despite this advice:
$$a + b y^3 + \frac{c}{y^3}$$
What exactly is it that I am missing?

Have you tried solving
$$\left(a + b y^3 + \frac{c}{y^3}\right)^2 \equiv \frac{1}{2}+36 y^6 + \frac{1}{576 y^6}$$
for $a, b,c$?

 

[LCKurtz]

Homework Statement

The length L or the curve given by
$$\frac{3y^{4}}{2}+\frac{1}{48y^{2}}-5$$
from y=1 to y=2

Homework Equations

The Attempt at a Solution

Setting up the formula is easy. First I found the derivative of f(y) which is:
$$f'(y)=6y^{3}-\frac{1}{24y^{3}}$$
Then I plugged all the given data into the formula
$$L=\int_{1}^{2}\sqrt{1+(6y^{3}-\frac{1}{24y^{3}})^{2}}~dy$$
I expanded out the binomial and simplified to
$$L=\int_{1}^{2}(36y^{6}+\frac{1}{48y^{6}}+\frac{1}{2})^{1/2}~dy$$

Look at what you got when you squared that expression before you simplified it by adding the 1.

I fear I am going to have to slap myself again, but with this corrected equation:
$$L = \int_1^2 \sqrt{ 36y^6+\frac{1}{576 y^6}+\frac{1}{2}} \, dy$$
I am still having difficulty...

Now look at what changed after you added the 1. It's very similar... (It might help you recognize it if you put the 1/2 in the middle of your expression.)