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Math Help - Tangents and Normals

  1. #1
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    Tangents and Normals

    Hi i need to work out the equation of the tangent and norm of the following function.
    -(25-x^2)^1/2 where x=4

    I'm pretty sure i have messed up but here are my solutions that i got.
    y= +/- 3
    F'(x)= (-4*9^1/2)/9
    norm being 9/(4*9^1/2)

    If i am correct up to here my equations should be right. Help is greatly appreciated
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  2. #2
    Moo
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    Hello,

    Quote Originally Posted by Webby View Post
    Hi i need to work out the equation of the tangent and norm of the following function.
    -(25-x^2)^1/2 where x=4

    I'm pretty sure i have messed up but here are my solutions that i got.
    y= +/- 3
    F'(x)= (-4*9^1/2)/9
    norm being 9/(4*9^1/2)

    If i am correct up to here my equations should be right. Help is greatly appreciated
    When differentiating a function, you can't just put the values... especially here, where it's a nonsense to put the values of y in f'(x)


    The equation of a tangent where x=a is : y=f'(a)(x-a)+f(a)

    So calculate f'(x) :
    f(x)=-(25-x^2)^{1/2}

    f'(x)=x(25-x^2)^{-1/2}=\frac{x}{\sqrt{25-x^2}} (I'll let you do the calculation... just tell if you can't).

    --> f'(4)=\frac{4}{\sqrt{25-4^2}}=\frac 43

    f(4)=-(25-4^2)^{1/2}=-\sqrt{25-4^2}=-3, and not \pm 3, because a square root is always positive. So -sqrt(25-4)<0
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  3. #3
    Senior Member nikhil's Avatar
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    Even I am confused

    Quote Originally Posted by Moo View Post
    because a square root is always positive. So -sqrt(25-4)<0
    [x^2]^1/2 = x if x>0 and -x if x<0 e.g [(-3)^2]^1/2=-3 but
    (-3)^2=9 so if we put this talue in previous eq we get
    9^1/2=-3 so we get square root may give a negative value.
    2]I think square root of a number means to find a number say a which when multiplied to itself gives the same number e.g 4^1/2 means to find a number say a which when multiplied to itself gives 4 so a)2x2=4 and b)-2x-2=4 so 4^1/2 may also give -2 which is not positive.
    Though -(x)^1/2 is <0 but it does not mean square root is always positive.if i am wrong please point my mistake out. Thanks
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  4. #4
    Moo
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    I think there is something wrong...

    Yes, x^2=9 \implies x=\pm 3, but it doesn't mean that \sqrt{9}=\pm 3.

    (Of course, I'm not talking about the imaginary numbers and all this stuff)

    This is the curve of \sqrt{x}




    Quote Originally Posted by Wikipedia
    If otherwise unqualified, "the square root" of a number refers to the principal square root
    Quote Originally Posted by Wikipedia
    Every non-negative real number x has a unique non-negative square root, called the principal square root and denoted with a radical symbol as √x.
    ----------------------

    Quote Originally Posted by nikhil View Post
    [x^2]^1/2 = x if x>0 and -x if x<0 e.g [(-3)^2]^1/2=-3 but
    (-3)^2=9 so if we put this talue in previous eq we get
    9^1/2=-3 so we get square root may give a negative value.
    2]I think square root of a number means to find a number say a which when multiplied to itself gives the same number e.g 4^1/2 means to find a number say a which when multiplied to itself gives 4 so a)2x2=4 and b)-2x-2=4 so 4^1/2 may also give -2 which is not positive.
    Though -(x)^1/2 is <0 but it does not mean square root is always positive.if i am wrong please point my mistake out. Thanks
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  5. #5
    Senior Member nikhil's Avatar
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    Still confused

    You have drawn the function
    y=x^1/2 but if you draw the graph of relation y=x^1/2 then for each value x two values will be obtained one positive and other negative.
    its true that square root of every non negative number is positive. Thats why i created the example [(-3)^2]^1/2=-3
    but (-3)^2=9 so if we substitute this value we get
    9^1/2=-3 (which is correct)
    if I just ask 9^1/2=? Without knowing whether this 9 represent 3^2 or (-3)^2 saying the answer is +3 will be wrong.so if we get x^1/2
    in a question as a subpart I don't think it should only be treated as a function,it may also be used as relation.so if You take x^1/2 as relation you will get two values which is against saying that square root is always positive.what do you think??
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  6. #6
    Moo
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    If I draw x^(1/2), it just yield the same graph

    \sqrt{x}=x^{1/2} is not define (over reals) for negative numbers.

    When you write ((-3)^2)^{1/2}, I guess you do first (-3)^2, which gives 9, then, power it to 1/2. And this yields a positive number : 3.
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  7. #7
    Senior Member nikhil's Avatar
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    not enough

    Quote Originally Posted by Moo View Post
    If I draw x^(1/2), it just yield the same graph

    \sqrt{x}=x^{1/2} is not define (over reals) for negative numbers.

    When you write ((-3)^2)^{1/2}, I guess you do first (-3)^2, which gives 9, then, power it to 1/2. And this yields a positive number : 3.
    Actually when I said draw the graph of y=x^1/2 I did not meant to involve negative value (as we are talking about real values in real plane not about imaginary quantities in argand plane) I meant you will get graph similar to y^2=4ax where a>0 (for just a rough visual begaviour) that is for x>0 you will get two values like 4^1/2
    =2 and -2 which is impossible for a function(as a function can not have two values for same element x) but this is not a problem if x^1/2 is treated as relation because getting two values for same element x will be permissible so again 9^1/2 may also mean -3.
    Dude like your guess that (-3)^2 gives 9 and 9^1/2 gives +3
    BUT [(-3)^2]^1/2 is actually =-3 but not=3 so the precedure you gave is not correct.
    Again think if -2x-2=4 then why 4^1/2 can not yield -2 which is a negative quantity against the saying square root is always positive.
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  8. #8
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    If a problem is *given* to you as f(x)=\sqrt{x} then only nonnegative answers are in the range...that's the definition of \sqrt{x} as Moo pointed out.

    Of course

    4 = x^2

    for example, there is no such restriction... x can be positive or negative, that's why when you "take the root" of both sides, you must account for both possibilities:

    2 = |\sqrt{x^2}|

    2 = |x|

    \pm2 = x

    but notice that \sqrt{x^2} is still defined as nonnegative.

    So this should even apply to subproblems...as long as you correctly account for +/- when you take the root, \sqrt{x} will always be positive.

    Eric
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  9. #9
    Senior Member nikhil's Avatar
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    Welcome Eric

    Hi! Eric, its very nice to have you in the forum.
    So do you mean [(-3)^2]^1/2 is =3 that is if we square a negative number and that square root it, the number will become positive?
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  10. #10
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by nikhil View Post
    Hi! Eric, its very nice to have you in the forum.
    So do you mean [(-3)^2]^1/2 is =3 that is if we square a negative number and that square root it, the number will become positive?
    Yes \sqrt{x^2}\equiv|x|

    If x<0 then |x|=-x

    and since x is negative -x>0
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  11. #11
    Senior Member nikhil's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    Yes \sqrt{x^2}\equiv|x|
    Hey! That will definetly help alot. But if -2x-2=4
    Then why can't 4^1/2 can not be taken as -2 .
    NOTE:considering y=x^1/2
    as a function is not necessary,poster may take it as a relation also]
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  12. #12
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by nikhil View Post
    Hey! That will definetly help alot. But if -2x-2=4
    Then why can't 4^1/2 can not be taken as -2 .
    NOTE:considering y=x^1/2
    as a function is not necessary,poster may take it as a relation also]
    Because it is a number, not a variable, it is convention to do as such.

    Someone else might have a more enlightening response.
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  13. #13
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    Quote Originally Posted by nikhil View Post
    Hey! That will definetly help alot. But if -2x-2=4
    Then why can't 4^1/2 can not be taken as -2 .
    NOTE:considering y=x^1/2
    as a function is not necessary,poster may take it as a relation also]
    We usually use the notation x^{m/n} to mean \sqrt[n]{x^m}, i.e., the principal n^{\rm th} root.

    Thus, 4^{1/2} = \sqrt4 = 2. But if you were solving the equation x^2 = 4 you would have to take into account the possibility of x being negative, so the solution is x = \pm\sqrt4 = \pm2.
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