Complicated Arc length problem

[missingmyname]

Homework Statement

The length L or the curve given by
[tex]\frac{3y^{4}}{2}+\frac{1}{48y^{2}}-5[/tex]
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:
[tex]f'(y)=6y^{3}-\frac{1}{24y^{3}}[/tex]
Then I plugged all the given data into the formula
[tex]L=\int_{1}^{2}\sqrt{1+(6y^{3}-\frac{1}{24y^{3}})^{2}}~dy[/tex]
I expanded out the binomial and simplified to
[tex]L=\int_{1}^{2}(36y^{6}+\frac{1}{48y^{6}}+\frac{1}{2})^{1/2}~dy[/tex]
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
[tex]u=\frac{1}{x^{6}}~\therefore~du=-\frac{6}{x^{7}}dy[/tex]
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 [itex]\frac{1}{x^{6}}[/itex] for ##u## was chosen.

 

[LCKurtz]

Homework Statement

The length L or the curve given by
[tex]\frac{3y^{4}}{2}+\frac{1}{48y^{2}}-5[/tex]
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:
[tex]f'(y)=6y^{3}-\frac{1}{24y^{3}}[/tex]
Then I plugged all the given data into the formula
[tex]L=\int_{1}^{2}\sqrt{1+(6y^{3}-\frac{1}{24y^{3}})^{2}}~dy[/tex]
I expanded out the binomial and simplified to
[tex]L=\int_{1}^{2}(36y^{6}+\frac{1}{\color{red}{48}y^{6}}+\frac{1}{2})^{1/2}~dy[/tex]
.

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
[tex]\frac{3y^{4}}{2}+\frac{1}{48y^{2}}-5[/tex]
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:
[tex]f'(y)=6y^{3}-\frac{1}{24y^{3}}[/tex]
Then I plugged all the given data into the formula
[tex]L=\int_{1}^{2}\sqrt{1+(6y^{3}-\frac{1}{24y^{3}})^{2}}~dy[/tex]
I expanded out the binomial and simplified to
[tex]L=\int_{1}^{2}(36y^{6}+\frac{1}{48y^{6}}+\frac{1}{2})^{1/2}~dy[/tex]
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
[tex]u=\frac{1}{x^{6}}~\therefore~du=-\frac{6}{x^{7}}dy[/tex]
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 [itex]\frac{1}{x^{6}}[/itex] for ##u## was chosen.

It's not clear how you went from:

[itex]{1+(6y^{3}-\frac{1}{24y^{3}})^{2}}[/itex] to

[itex](36y^{6}+\frac{1}{48y^{6}}+\frac{1}{2})[/itex]

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

 

[Ray Vickson]

Homework Statement

The length L or the curve given by
[tex]\frac{3y^{4}}{2}+\frac{1}{48y^{2}}-5[/tex]
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:
[tex]f'(y)=6y^{3}-\frac{1}{24y^{3}}[/tex]
Then I plugged all the given data into the formula
[tex]L=\int_{1}^{2}\sqrt{1+(6y^{3}-\frac{1}{24y^{3}})^{2}}~dy[/tex]
I expanded out the binomial and simplified to
[tex]L=\int_{1}^{2}(36y^{6}+\frac{1}{48y^{6}}+\frac{1}{2})^{1/2}~dy[/tex]
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
[tex]u=\frac{1}{x^{6}}~\therefore~du=-\frac{6}{x^{7}}dy[/tex]
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 [itex]\frac{1}{x^{6}}[/itex] for ##u## was chosen.

When I do the simplification (using Maple) I get
[tex] L = \int_1^2 \sqrt{ 36y^6+\frac{1}{576 y^6}+\frac{1}{2}} \, dy [/tex]
(because ##24^2 = 576##). It turns out that
[tex] \sqrt{ 36y^6+\frac{1}{576 y^6}+\frac{1}{2}} [/tex]
has the nice form of
[tex] a + b y^3 + \frac{c}{y^3} [/tex]
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
[tex] L = \int_1^2 \sqrt{ 36y^6+\frac{1}{576 y^6}+\frac{1}{2}} \, dy [/tex]
(because ##24^2 = 576##). It turns out that
[tex] \sqrt{ 36y^6+\frac{1}{576 y^6}+\frac{1}{2}} [/tex]
has the nice form of
[tex] a + b y^3 + \frac{c}{y^3} [/tex]
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:
[tex] L = \int_1^2 \sqrt{ 36y^6+\frac{1}{576 y^6}+\frac{1}{2}} \, dy [/tex]
I am still having difficulty despite this advice:
[tex] a + b y^3 + \frac{c}{y^3} [/tex]
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:
[tex] L = \int_1^2 \sqrt{ 36y^6+\frac{1}{576 y^6}+\frac{1}{2}} \, dy [/tex]
I am still having difficulty despite this advice:
[tex] a + b y^3 + \frac{c}{y^3} [/tex]
What exactly is it that I am missing?

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

 

[LCKurtz]

Homework Statement

The length L or the curve given by
[tex]\frac{3y^{4}}{2}+\frac{1}{48y^{2}}-5[/tex]
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:
[tex]f'(y)=6y^{3}-\frac{1}{24y^{3}}[/tex]
Then I plugged all the given data into the formula
[tex]L=\int_{1}^{2}\sqrt{1+(6y^{3}-\frac{1}{24y^{3}})^{2}}~dy[/tex]
I expanded out the binomial and simplified to
[tex]L=\int_{1}^{2}(36y^{6}+\frac{1}{48y^{6}}+\frac{1}{2})^{1/2}~dy[/tex]

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:
[tex] L = \int_1^2 \sqrt{ 36y^6+\frac{1}{576 y^6}+\frac{1}{2}} \, dy [/tex]
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.)