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    Find the domain and range of z=\sqrt{9-x^2-y^2}

    So I know the domain is {(x,y) |    x^2+y^2\leq9}

    But I don't know about the range!

    I know that z has to be a positive square root;z\geq0
    but the book says the range is {z |    0\leq z\leq3}



    How do they know z is also going to be \leq3 ?


    Thanks
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    (Original post by purplerainn)
    So I know the domain is {(x,y) |    x^2+y^2\leq9}

    But I don't know about the range!

    I know that z has to be a positive square root;z\geq0
    but the book says the range is {z |    0\leq z\leq3}



    How do they know z is also going to be \leq3 ?


    Thanks
    Hint:

    Sketch the circle

    x^2 + y^2 = 9

    Think of what I've put in bold.
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    (Original post by Indeterminate)
    Hint:

    Sketch the circle

    x^2 + y^2 = 9

    Think of what I've put in bold.

    I still don't understand :ashamed:
    So the domain is the set of points in that disk, but I don't know how to relate that info to the range


    The book mentions
    Since z is a positive square root, z\geq0. Also, because 9-x^2-y^2\leq9

    But from finding the domain, I thought that x^2+y^2\leq9...
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    Try squaring z. Looking at the RHS, what do you know about a number that has been squared? Also take the non-negativity of z into account.


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    (Original post by Indeterminate)
    Try squaring z. Looking at the RHS, what do you know about a number that has been squared? Also take the non-negativity of z into account.


    Posted from TSR Mobile

    Thanks, I got it now!

    I have another question:

    Find the Maclaurin series of \frac{x}{\sqrt{4+x^2}}
    So \frac{x}{2} \sum [1+ [\frac{x^2}{4}]^{1/2}]

    Wouldn't all of the derivatives for that MacLaurin series be 0 because of the \frac{x}{2} in each of them? :confused:
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    (Original post by purplerainn)
    Thanks, I got it now!

    I have another question:



    So \frac{x}{2} \sum [1+ [\frac{x^2}{4}]^{1/2}]

    Wouldn't all of the derivatives for that MacLaurin series be 0 because of the \frac{x}{2} in each of them? :confused:
    What you've written in your sum is completely different from the function you've been asked to evaluate!

    the Maclaurin series is basically the appropriate binomial series multiplied by the factor you've taken outside - can you proceed from this?
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    (Original post by davros)
    the Maclaurin series is basically the appropriate binomial series multiplied by the factor you've taken outside - can you proceed from this?
    But isn't that what I did?

    \frac{x}{2} times the Mac Laurin series for [1+ [\frac{x^2}{4}]^{1/2}]

    would give the MacLaurin series for the original function?
    I think the sigma notation should have had no part in my previous post
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    (Original post by purplerainn)
    But isn't that what I did?

    \frac{x}{2} times the Mac Laurin series for [1+ [\frac{x^2}{4}]^{1/2}]

    would give the MacLaurin series for the original function?
    I think the sigma notation should have had no part in my previous post
    No - your index 1/2 should be -1/2 and it should be outside the overall bracket!
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    (Original post by davros)
    No - your index 1/2 should be -1/2 and it should be outside the overall bracket!

    Oh, that's a mistake in my typing and latex.

    [1+ [\frac{x^2}{4}]]^-1/2

    So now, wouldn't each of it's derivatives and hence each of the terms be 0 because of the \frac{x}{2} in each derivative?
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    (Original post by purplerainn)
    Oh, that's a mistake in my typing and latex.

    [1+ [\frac{x^2}{4}]]^-1/2

    So now, wouldn't each of it's derivatives and hence each of the terms be 0 because of the \frac{x}{2} in each derivative?
    Have you actually tried differentiating? You'll find that as you differentiate successively the product rule gives you more and more terms - some will be zero and some will not.

    Note that your original function is an odd function (i.e. f(-x) = -f(x)) so it won't have any even powers of x in its expansion.

    Unless you're specifically asked to use Maclaurin, I would definitely recommend starting with the binomial series and multiplying by the factor x/2.
 
 
 
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