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    y=sqrt(x) makes an angle of a with the line y=0
    find a
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    (Original post by gomc)
    y=sqrt(x) makes an angle of a with the line y=0
    find a
    Why don't you try plotting the graph and seeing what arises.

    Hint:
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    Think tangents of \sqrt{x}
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    (Original post by Zacken)
    Why don't you try plotting the graph and seeing what arises.

    Hint:
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    Think tangents of \sqrt{x}
    so differentiate?
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    (Original post by gomc)
    y=sqrt(x) makes an angle of a with the line y=0
    find a
    Have you written the question properly, because, upon reflection we have that y' = \frac{1}{2\sqrt{x}} which is surely undefined at (0,0) - unless, of course the angle is the trivial a = \frac{\pi}{2} = 90^{\circ} because the tangent is essentially the y-axis?
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    (Original post by Zacken)
    Why don't you try plotting the graph and seeing what arises.

    Hint:
    Spoiler:
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    Think tangents of \sqrt{x}

    Whenever there's a thread to do with maths, I could almost guess you'd be the first one to post on it lool
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    (Original post by ravioliyears)
    Whenever there's a thread to do with maths, I could almost guess you'd be the first one to post on it lool
    Haha, I've got some free time, so I've just come to the maths forum and blaze out a few helping hands. :rofl:
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    (Original post by Zacken)
    Have you written the question properly, because, upon reflection we have that y' = \frac{1}{2\sqrt{x}} which is surely undefined at (0,0) - unless, of course the angle is the trivial a = \frac{\pi}{2} = 90^{\circ} because the tangent is essentially the y-axis?
    yes thats all thats given
    and its a non-calc question if that helps
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    (Original post by ravioliyears)
    Whenever there's a thread to do with maths, I could almost guess you'd be the first one to post on it lool
    Or TeeEm!!
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    (Original post by gomc)
    yes thats all thats given
    and its a non-calc question if that helps
    I actually do think the angle is 90^{\circ} using my calculus method.

    The only other method I can think of is using the tangent formula for angles between lines but \sqrt{x} certainly isn't a line... what chapter/topic is this question from?

    Edit: I'm sorry to have given you the full solution, but I only did so because I was uncertain about the question itself, many apologies.
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    (Original post by Zacken)
    I actually do think the angle is 90^{\circ} using my calculus method.

    The only other method I can think of is using the tangent formula for angles between lines but \sqrt{x} certainly isn't a line... what chapter/topic is this question from?

    Edit: I'm sorry to have given you the full solution, but I only did so because I was uncertain about the question itself, many apologies.
    i mean non calculator
    its a core 1 extension question
    no worries haha
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    (Original post by gomc)
    i mean non calculator
    its a core 1 extension question
    no worries haha
    In that case, I think the intended method was y' = \frac{1}{2\sqrt{x}}, so tangent at (0,0) is the y-axis, angle between x-axis and y-axis is 90 degrees. Boom. Done.
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    (Original post by ghostwalker)
    Since you want the angle with the y-axis, turn your graph on its side and express x as a function of y. Answer becomes obvious.

    Note: You want the angle with the y-axis, and it's not 90.
    I'm probably being thick but isn't y=0 the x-axis?
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    (Original post by Zacken)
    I'm probably being thick but isn't y=0 the x-axis?
    Yep - ignore me.
 
 
 
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