Algebra Problem

[Jason76]

Please help me correct any algebra errors that may be noticeable. Ignore the Calculus element, this is about algebra.

\(\displaystyle 2(x)^{1/2}\dfrac{dy}{dx} = (1 - y^{2})^{1/2}\)

\(\displaystyle (dx) 2(x)^{1/2}\dfrac{dy}{dx} = (1 - y^{2})^{1/2}(dx)\)

\(\displaystyle 2(x)^{1/2} dy= (1 - y^{2})^{1/2}dx\)

\(\displaystyle (\dfrac{1}{x^{1/2}})2(x)^{1/2}dy= (1 - y^{2})^{1/2}(\dfrac{1}{x^{1/2}})dx\)

\(\displaystyle 2 dy= (1 - y^{2})^{1/2}(\dfrac{1}{x^{1/2}})dx\)

\(\displaystyle (\dfrac{1}{2})2dy= (1 - y^{2})^{1/2}(\dfrac{1}{x^{1/2}})(\dfrac{1}{2})dx\)

\(\displaystyle dy= (1 - y^{2})^{1/2}(\dfrac{1}{x^{1/2}})(\dfrac{1}{2})dx\)

\(\displaystyle dy= (1^{1/2} - y)(\dfrac{1}{x^{1/2}})(\dfrac{1}{2})dx\)

\(\displaystyle (\dfrac{1}{ (1^{1/2} - y)})dy= (1^{1/2} - y)(\dfrac{1}{x^{1/2}})(\dfrac{1}{2})(\dfrac{1}{ (1^{1/2} - y)})dx\)

\(\displaystyle (\dfrac{1}{ (1^{1/2} - y)})dy =(\dfrac{1}{x^{1/2}})(\dfrac{1}{2})dx\)

 

[MarkFL]

Please help me correct any algebra errors that may be noticeable. Ignore the Calculus element, this is about algebra.

I can't get the multi-quote to work, but in going from the 4th line from the bottom to the 3rd line from the bottom, you have fallen victim to the so-called "Freshmen's Dream" where you assume:

\(\displaystyle (a+b)^n=a^n+b^n\)

This is not true, but it is such a commonly made mistake, it has been named.

 

[soroban]

Hello, Jason76!

You didn't "separate the variables".

\(\displaystyle 2x^{\frac{1}{2}}\frac{dy}{dx} \:=\: (1-y^2)^{\frac{1}{2}}\)

\(\displaystyle \displaystyle\text{Separate: }\:\frac{2\,dy}{(1-y^2)^{\frac{1}{2}}} \:=\: x^{-\frac{1}{2}}dx\)

\(\displaystyle \displaystyle\text{Integrate: }\:2\int\frac{dy}{(1-y^2)^{\frac{1}{2}}} \:=\:\int x^{-\frac{1}{2}}dx\)

. . . . . . . . . . . . \(\displaystyle 2\arcsin y \;=\; 2x^{\frac{1}{2}} + c\)

. . . . . . . . . . . . . \(\displaystyle \arcsin y \;=\;x^{\frac{1}{2}} + C\)

. . . . . . . . . . . . . . . . . .\(\displaystyle y \;=\;\sin(\sqrt{x} + C)\)

 

[Jason76]

I can't get the multi-quote to work, but in going from the 4th line from the bottom to the 3rd line from the bottom, you have fallen victim to the so-called "Freshmen's Dream" where you assume:

\(\displaystyle (a+b)^n=a^n+b^n\)

This is not true, but it is such a commonly made mistake, it has been named.

Ok, so the distributive property doesn't work with exponents.

 

[MarkFL]

Right, look at:

\(\displaystyle (4+1)^{\frac{1}{2}}\)

If the distributive property applied here, then we would have:

\(\displaystyle \sqrt{5}=\sqrt{4}+\sqrt{1}=2+1=3\)

Squaring both sides, we find:

\(\displaystyle 5=9\)

Something is wrong...

 

[Jason76]

Right, look at:

\(\displaystyle (4+1)^{\frac{1}{2}}\)

If the distributive property applied here, then we would have:

\(\displaystyle \sqrt{5}=\sqrt{4}+\sqrt{1}=2+1=3\)

Squaring both sides, we find:

\(\displaystyle 5=9\)

Something is wrong...

Also, I suppose the reason is because the exponents of multiplied bases are added not multiplied.

As far as your square roots thing goes, it makes sense. In the example \(\displaystyle (6 + 7)3 = 39\) We find it's true whether we add the two middle terms and multiply by 3, or by doing the distributive property. Either way the answer is 39. But when applied to this situation, the square root of the added terms, doesn't equal the same answer given by distributing.

In response to Soraban, how did you get rid of the Arcsin?