# Thread: Solve for y (with arctan)...

1. ## Solve for y (with arctan)...

Have I gone wrong with this?

$\displaystyle \frac{1}{y}=\frac{1}{3}arctan(\frac{x}{3})+C$

Making y the subject:

$\displaystyle y=\frac{3}{arctan(\frac{x}{3})+C}$

Can you do that with arctan?

2. And if that is the case (I hope it is), would:

$\displaystyle C=\frac{3}{y}-arctan(\frac{x}{3})$

3. Originally Posted by MaverickUK82
Have I gone wrong with this?

$\displaystyle \frac{1}{y}=\frac{1}{3}arctan(\frac{x}{3})+C$

Making y the subject:

$\displaystyle y=\frac{3}{arctan(\frac{x}{3})+C}$

Can you do that with arctan?
Originally Posted by MaverickUK82
And if that is the case (I hope it is), would:

$\displaystyle C=\frac{3}{y}-arctan(\frac{x}{3})$

What are you trying to do? Do you want an equation in y = format?

4. The problem is not the arctan but that "C". Just taking the reciprocal of both sides, $\displaystyle y= \frac{1}{\frac{1}{3}arctan(\frac{x}{3})+ C}$.

Now, multiplying both numerator and denominator by 3, $\displaystyle y= \frac{3}{arctan(\frac{x}{3})+ 3C}$.

Of course, if "C" is some undetermined constant, so is 3C. You could write it as $\displaystyle y= \frac{3}{arctan(\frac{x}{3})+ C'}$
but you should understand that "C" and "C' " are not the same number.

5. Originally Posted by MaverickUK82
And if that is the case (I hope it is), would:

$\displaystyle C=\frac{3}{y}-arctan(\frac{x}{3})$
Replace your "C" on the left by 3C, the divide both sides by 3.

6. Sorry for lateness of reply - been away at work.

That was quite sloppy of me not to see that, so:

$\displaystyle \frac{1}{y}=\frac{1}{3}arctan(\frac{x}{3})+C$

$\displaystyle y=\frac{1}{\frac{1}{3}arctan(\frac{x}{3})}+\frac{1 }{C}$

$\displaystyle y=\frac{3}{arctan(\frac{x}{3})}+\frac{3}{3C}$

$\displaystyle y=\frac{3}{arctan(\frac{x}{3})+3C}$

7. Originally Posted by MaverickUK82
Sorry for lateness of reply - been away at work.

That was quite sloppy of me not to see that, so:

$\displaystyle \frac{1}{y}=\frac{1}{3}arctan(\frac{x}{3})+C$

$\displaystyle y=\frac{1}{\frac{1}{3}arctan(\frac{x}{3})}+\frac{1 }{C}$

Mr F says: The above is wrong. Is that what post #4 says y is equal to? What you have written is equivalent to saying that if $\displaystyle \, \frac{1}{a} = b + c$ then $\displaystyle a = \frac{1}{b} + \frac{1}{c}$. I hope you realise how wrong this is.

$\displaystyle y=\frac{3}{arctan(\frac{x}{3})}+\frac{3}{3C}$

$\displaystyle y=\frac{3}{arctan(\frac{x}{3})+3C}$ Mr F says: Post #4 clearly says this. Why are you asking if it's correct?

8. Mr. Fantastic,

No. I did not realise it was wrong.

9. Originally Posted by MaverickUK82
Mr. Fantastic,

No. I did not realise it was wrong.
I gave you the explanation why it was wrong: $\displaystyle \frac{1}{a} = b + c$ does not imply $\displaystyle a = \frac{1}{b}+ \frac{1}{c}$. eg. 1/2 = 1/4 + 1/4 does not imply 2 = 4 + 4.

You are advised to review and/or take care with the basic algebra (the fact that you're working at the level of inverse trig functions means that the algebra I have explained is assumed knowledge).

10. Originally Posted by MaverickUK82
Have I gone wrong with this?

$\displaystyle \frac{1}{y}=\frac{1}{3}arctan(\frac{x}{3})+C$

Making y the subject:

$\displaystyle y=\frac{3}{arctan(\frac{x}{3})+C}$

Can you do that with arctan?
$\displaystyle \frac{1}{2}=\frac{2}{4}\Rightarrow\ 1(4)=2(2)$

$\displaystyle \frac{1}{3}=\frac{2}{6}\Rightarrow\ 1(6)=2(3)$

and so on.

$\displaystyle \frac{a}{b}=\frac{c}{d}\Rightarrow\frac{b}{a}=\fra c{d}{c}$

You can use either method, but first write both sides as a single fraction.

$\displaystyle \frac{1}{y}=\frac{arctan(\frac{x}{3})}{3}+\frac{3C }{3}=\frac{arctan(\frac{x}{3})+3C}{3}$

Now you can invert the fractions (fastest) or cross-multiply

$\displaystyle 1(3)=y\left(arctan(\frac{x}{3})+3C\right)$

Now divide both sides by the multiplier of y.