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Math Help - Solve for y (with arctan)...

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

    Have I gone wrong with this?

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

    Making y the subject:

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

    Can you do that with arctan?
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  2. #2
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    And if that is the case (I hope it is), would:

    C=\frac{3}{y}-arctan(\frac{x}{3})
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    Quote Originally Posted by MaverickUK82 View Post
    Have I gone wrong with this?

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

    Making y the subject:

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

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

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

    What are you trying to do? Do you want an equation in y = format?
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  4. #4
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    The problem is not the arctan but that "C". Just taking the reciprocal of both sides, y= \frac{1}{\frac{1}{3}arctan(\frac{x}{3})+ C}.

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

    Of course, if "C" is some undetermined constant, so is 3C. You could write it as y= \frac{3}{arctan(\frac{x}{3})+ C'}
    but you should understand that "C" and "C' " are not the same number.
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  5. #5
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    Quote Originally Posted by MaverickUK82 View Post
    And if that is the case (I hope it is), would:

    C=\frac{3}{y}-arctan(\frac{x}{3})
    Replace your "C" on the left by 3C, the divide both sides by 3.
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  6. #6
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    Sorry for lateness of reply - been away at work.

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

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


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

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

    y=\frac{3}{arctan(\frac{x}{3})+3C}
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    Quote Originally Posted by MaverickUK82 View Post
    Sorry for lateness of reply - been away at work.

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

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


    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 \, \frac{1}{a} = b + c then a = \frac{1}{b} + \frac{1}{c}. I hope you realise how wrong this is.


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

    y=\frac{3}{arctan(\frac{x}{3})+3C} Mr F says: Post #4 clearly says this. Why are you asking if it's correct?
    Read post #4.
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  8. #8
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    Mr. Fantastic,

    No. I did not realise it was wrong.
    Last edited by mr fantastic; January 24th 2011 at 02:43 AM.
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  9. #9
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    Quote Originally Posted by MaverickUK82 View Post
    Mr. Fantastic,

    No. I did not realise it was wrong.
    I gave you the explanation why it was wrong: \frac{1}{a} = b + c does not imply 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).
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    Quote Originally Posted by MaverickUK82 View Post
    Have I gone wrong with this?

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

    Making y the subject:

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

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

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

    and so on.

    \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.

    \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

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

    Now divide both sides by the multiplier of y.
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