Higher Derivatives > How'd this problem get solved?

[ThisIsMyName]

Original Problem: 9x^2 = 9

 

Steps:

 

1. 18x + 2yy' = -18x

 

2. 2yy' = -18x

 

3. y'= -9x/y

 

4. y''= -9 (y * - x * y'/ y^2)

 

5. y''= -9 (y - x (-9x/y)/y^2)

 

6. y'' = -9 (y^2+9x^2)/(y^3)

 

Answer: y'' = -81/y^3

 

These are the steps from the answer book. My friends and I got up to about step four with no problem, but we can't really understand what they did when they started on the second derivative.

 

Can someone kinda break down steps 4-6? Like what they did/how/why they did it? Thanks!

:)

[sanctus]

Weel I can't see how from the original problem you get equation 1), but since you get to step 4) I explain you the rest (I didn't check until step 4):

For getting from step 4 to step 5 you replace y' from step 3 in the right side of step 4

 

For getting from step 5 to step six you multiply the right side of step 5 by (y/y)

 

Then to get the answer you replace the orginal equation, which gives you x, in step 6.

 

Hope it is clear now.

[LaurieAG]

Can someone kinda break down steps 4-6? Like what they did/how/why they did it? Thanks!

:beer:

 

Hi ThisIsMyName,

 

Could you retype steps 4, 5 and 6 using '*' instead of '×' for the multiplier as there seems to be an error.

 

4. y''= -9 (y × - x × y'/ y^2)

 

4.1 y''= -9 (y × - x × (-9x/y)/ y^2) Substitute for y' from 3.

 

4.2 y''= -9 (y × - x^2 × (-9/y)/ y^2) consolidate x

 

4.3 y''= -9 (y × - x^2 × -9/ y^3) consolidate y

 

4.4 y''= -9 (y × x^2 × 9/ y^3) cancel negatives

 

4.5 y''= -9x9 (y × x^2 / y^3) consolidate constants

 

4.6 y''= -81 (x^2 / y^2) consolidate y

 

Answer: y''= -81 (x / y)^2 consolidate x and y

 

There is definitely something wrong with the given answer, unless x= the square root of 1/y.

 

I just looked at the top and 9x^2=9 so x^2=1 and therefore x = 1 not SQRT(1/y).

 

Answer: y'' = -81/y^3

[LaurieAG]

For getting from step 5 to step six you multiply the right side of step 5 by (y/y).

 

Hi Sanctus,

 

That depends if you can get 5. from 4.

[sanctus]

Laurie, as I said above this is done by

For getting from step 4 to step 5 you replace y' from step 3 in the right side of step 4

as you have done in 4.1...

Your 4.4 seems wrong to me.

 

Anyway you are right checking all the steps there is something fishy. Thisismyname, here is my solution assuming that we have:

[math]

9x^2=9

[/math]

and

[math]

18x+2yy'=-18x

[/math]

as the starting problem (and I assume ' means derivative with respect to x).

[math]

\begin{split}

& 18x+2yy'=-18x\\

\Leftrightarrow & 2yy'=-36x \text{ put al terms in x to the right}\\

\Leftrightarrow & y'=-18\frac{x}{y} \text{ divide by 2y}\\

\Leftrightarrow & y''=-18\frac{1}{y}+18x\frac{y'}{y^2}\text{ derivative with respect to x of x and y}\\

\Leftrightarrow & y''=-18(\frac{1}{y}-\frac{xy'}{y^2}\text{ clean it up a bit}\\

\Leftrightarrow & y''=-18(\frac{1}{y}+\frac{18x^2}{y^3}\text{ replace $y'=-18\frac{x}{y}$}\\

\Leftrightarrow & y''=-18(\frac{1}{y}+2\cdot\frac{9x^2}{y^3}\\

\Leftrightarrow & y''=-18(\frac{y^2+18}{y^3})\text{ replace $9x^2=9$}

\end{split}

[/math]

[LaurieAG]

Hi Sanctus,

 

as you have done in 4.1...

Your 4.4 seems wrong to me.

 

4.3 y''= -9 (y × - x^2 × -9/ y^3) is in the form

y''= A×(-1)[b ×(-1)×C ×(-1)× A]/(D) where A=9, B=y, C=x^2 and D=y^3

 

which equates to A×(-1)[(-1×-1)×B×C×A]/(D)

 

(-1×-1) = 1 so two of the negatives are cancelled.

 

The really interesting thing about this is that when 9x^2=9 so x^2=1 and therefore x = 1 is equivalent to SQRT(1/y) ONLY when y=1.

 

Unfortunately the point 1,1 is not an area under a curve because it is an infinitessimal point whose width is delta, and as it is in the form y=Ax^0, its derivative is zero. integration or differentiation from this point goes nowhere.

 

This is probably a good example of how an inadvertent imaginary unit (i) has somehow slipped in and caused problems. If you look at the Wiki entry on the Imaginary unit - Wikipedia, the free encyclopedia you will see a Proper Usage section. This could possibly come into play and cause an error due to the obfuscated i along with square roots.

[LaurieAG]

Hi Sanctus,

 

The following doesn't apply to your workings but I do suspect that something similar has ocurred in the derivation of at least one of the other given equations.

 

(-1×-1) = 1 so two of the negatives are cancelled.

...

The really interesting thing about this is that when 9x^2=9 so x^2=1 and therefore x = 1 is equivalent to SQRT(1/y) ONLY when y=1.

...

This is probably a good example of how an inadvertent imaginary unit (i) has somehow slipped in and caused problems. If you look at the Wiki entry on the Imaginary unit - Wikipedia, the free encyclopedia you will see a Proper Usage section. This could possibly come into play and cause an error due to the obfuscated i along with square roots.

 

Just to follow this through, from Wiki.

 

-1 = i x i = SQRT(-1) x SQRT(-1) = SQRT(-1 x -1) = SQRT(1) = 1 (INCORRECT)

 

It's probably best to exclude (-1 x -1) from all square root operations, to be safe, or you may just find your solution incorrectly in imaginary land.

[sanctus]

Hi Sanctus,

 

 

 

4.3 y''= -9 (y × - x^2 × -9/ y^3) is in the form

y''= A×(-1)[b ×(-1)×C ×(-1)× A]/(D) where A=9, B=y, C=x^2 and D=y^3

 

which equates to A×(-1)[(-1×-1)×B×C×A]/(D)

 

 

I see your point, but I still think you do something wrong, simply because in one equation you have a sum of x and y to different powers and in the next you have only a big multiplication...can you write it in latex (or at least without the ×)?

[LaurieAG]

I see your point, but I still think you do something wrong, simply because in one equation you have a sum of x and y to different powers and in the next you have only a big multiplication...can you write it in latex (or at least without the ×)?

 

No worries Sanctus,

 

4. y''= -9 (y * - x * y'/ y^2)

 

4.1 y''= -9 (y * - x * (-9x/y)/ y^2) Substitute for y' from 3.

 

4.2 y''= -9 (y * - x^2 * (-9/y)/ y^2) consolidate x

 

4.3 y''= -9 (y * - x^2 * -9/ y^3) consolidate y

 

4.4 y''= -9 (y * x^2 * 9/ y^3) cancel negatives

 

4.5 y''= -9x9 (y * x^2 / y^3) consolidate constants

 

4.6 y''= -81 (x^2 / y^2) consolidate y

 

y''= A*(-1)[b *(-1)*C *(-1)* A]/(D) where A=9, B=y, C=x^2 and D=y^3

 

which equates to A*(-1)[(-1*-1)*B*C*A]/(D)

 

By removing (-1*-1) at step 4.4 I don't go anywhere near y''=-81[(x^2)*(-1*-1)]/y^2 and a possible improper imaginary unit from the square root of y''.

[sanctus]

Why write y* instead of just y? For example ny equation 4. you mean to say:

[math]

-9(y-\frac{xy'}{y^2})

[/math]

?

[LaurieAG]

Why write y* instead of just y? For example ny equation 4. you mean to say:

[math]

-9(y-frac{xy'}{y^2})

[/math]

?

 

Hi Sanctus, no I don't,

 

That's why I asked for clarifications for the multiplier in my first post as 3. and 4. are stated in the first post of this thread as:-

 

3. y'= -9x/y

 

4. y''= -9 (y * - x * y'/ y^2)

 

You do not get what you think I mean unless you miss the bolded multiplier.

 

And the first post has step 5. missing the multiplier.