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    I calculated the gradient of the tangent at P to be (-(6)^(1/2))/6
    Therefore i thought that the gradient of the normal would be (6)/((6)^(1/2)).
    Can you please explain how this is wrong.
    https://5c59854d0ccd29d489c9e5e689a8...20C4%20OCR.pdf

    Thanks.
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    (Original post by SamuelN98)
    I calculated the gradient of the tangent at P to be (-(6)^(1/2))/6
    Therefore i thought that the gradient of the normal would be (6)/((6)^(1/2)).
    Can you please explain how this is wrong.
    https://5c59854d0ccd29d489c9e5e689a8...20C4%20OCR.pdf

    Thanks.
    What makes you think it's wrong?
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    (Original post by SamuelN98)
    I calculated the gradient of the tangent at P to be (-(6)^(1/2))/6
    Therefore i thought that the gradient of the normal would be (6)/((6)^(1/2)).
    Can you please explain how this is wrong.
    https://5c59854d0ccd29d489c9e5e689a8...20C4%20OCR.pdf

    Thanks.
    It is correct..

    Maybe you just need to think about what 6/root6 is
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    (Original post by SamuelN98)
    I calculated the gradient of the tangent at P to be (-(6)^(1/2))/6
    Therefore i thought that the gradient of the normal would be (6)/((6)^(1/2)).
    Can you please explain how this is wrong.
    https://5c59854d0ccd29d489c9e5e689a8...20C4%20OCR.pdf

    Thanks.
    It might help if you realise that \frac{6}{\sqrt{6}} = \frac{6^1}{6^{1/2}} = 6^{1-1/2} = 6^{1/2} =  \sqrt{6}
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    Yep, nevermind...
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    Thanks for the quick responses!
    I didn't think of it like that, if there equivalent is it suitable in an answer to continue using 6/((6)^(1/2)).
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    (Original post by SamuelN98)
    Thanks for the quick responses!
    I didn't think of it like that, if there equivalent is it suitable in an answer to continue using 6/((6)^(1/2)).
    I don't see why not, as a simplified answer is not necessary for this show that question, that said, why would you bother?
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    (Original post by SamuelN98)
    Thanks for the quick responses!
    I didn't think of it like that, if there equivalent is it suitable in an answer to continue using 6/((6)^(1/2)).
    I suppose so, but it's more difficult to do or see things than if you just simplify it straight away.
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    I`m not sure id see the relationship straight away. Should do now though.
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    (Original post by SamuelN98)
    I`m not sure id see the relationship straight away. Should do now though.
    That's how it was for all of us, it'll become amazingly transparent after some practice!
 
 
 
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